fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES
fiziks Forum for CSIR-UGC JRF/NET, GATE, IIT-JAM/IISc, JEST, TIFR and GRE in PHYSICS & PHYSICAL SCIENCES
Kinetic theory, Thermodynamics
(IIT-JAM/JEST/TIFR/M.Sc Entrance)
Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
[email protected]
fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Thermal & Statistical Mechanics 1. Kinetic theory of gases…………………………………………………………………..(1-29) 1.1 Basic assumption of kinetic theory 1.1.1 Pressure exerted by a gas 1.2 Gas Law for Ideal gases: 1.2.1 Boyle’s Law 1.2.2 Charle’s Law 1.2.3 Avogadro’s Law 1.2.4 Graham’s Law of Diffusion 1.2.5 Ideal Gas Equation: 1.3 Kinetic Interpretation of Temperature 1.4 Maxwell-Boltzmann Distribution Law 1.4.1 The Distribution in term of Magnitude 1.4.2 To Determine Value of β in term of Temperature T 1.4.2 Average Velocity 1.4.3 Root Mean Square Velocity 1.4.4 Most Probable Velocity Questions and Solutions 2. Real Gases……………………………………………………………………….(30-43) 2.1 Andrew’s Experiment on Carbon Dioxide 2.2 van der Waals Equation of State. 2.3 Correction in Ideal Gas Equation to Achieve van der Waals Gas Equation of State. 2.3.1 Correction for Finite Size 2.3.2 Correction for Intermolecular Attraction 2.3.3 Maxwell Equal Area 2.3.4 Critical Point 2.3.5 van der Waals Equation of State and Virial Coefficient Questions and Solutions
Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES
3. Basics of Thermodynamics and Laws of Thermodynamics…………………(44-79) 3.1 Mathematical Formulations of thermodynamics. 3.1.1 Some important Formulas 3.2 Fundamental Concept 3.2.1 System 3.2.2 Isolated System 3.2.3 Thermodynamical State 3.2.4 State Function 3.2.5 Intensive and Extensive Properties 3.3 The Ideal Gas: 3.4 Laws of Thermodynamics 3.4.1 Zeroth law of Thermodynamics: 3.4.2 First law of Thermodynamics: 3.4.3 Work Done during Different Process. 3.4.4 Specific Heat 3.4.5 Heat Capacity of Ideal Gas: 3.4.6 Molar Heat Capacity 3.4.7 Coefficient of Volume Expansion or Expansivity 3.4.8 Isothermal Elasticity and Isothermal Compressibility 3.5 Different Types of Thermo Dynamical Process and use of First Law of Thermodynamics 3.5.1 Isochoric Process: 3.5.2 Isobaric Process 3.5.3 Isothermal Process 3.5.4 Adiabatic Process Questions and Solutions
Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 4. Second Law of Thermodynamics and Entropy……………………………...(80-110) 4.1 Second Law of Thermodynamics 4.2 Heat Engines 4.2.1 Heat Reservoir 4.2.2 Efficiency of Heat Engine (η) 4.2.3 Carnot Cycle 4.3 Entropy 4.3.1 Inequality of Clausius Questions and Solutions 5. Maxwell relation and Thermodynamic Potential………………………..(111-142) 5.1 Maxwell relations 5.2 Different types of thermodynamic potential and Maxwell relation 5.2.1 Internal Energy 5.2.2 Enthalpy 5.2.3 Helmholtz Free Energy 5.2.4 Gibbs Energy 5.3 Application of Maxwell Relation 5.3.1 First T − dS Equation 5.3.2 Second T − dS Equation 5.3.3 Third T-dS Equation: 5.3.4 First Energy Equation 5.3.5 Second Energy Equation Questions and Solutions
Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 6. Phase Transition and Low Temperature Physics………………………...(143-167) 6.1
Third Law of Thermodynamics and Attainable of Low Temperature
6.2
Production of Low Temperature and Joule – Kelvin Expansion:
6.3
Phase Transition 6.3.1
First Order Phase Transition
6.3.2
Equilibrium Between Two Phases
6.3.3
Clapeyron-Clausius Equation
6.3.4
Liquid-Vapour Phase Transition
6.3.5
Properties of First Order Phase Transition
6.3.6
Second Order Phase Transition:
Questions and Solutions
Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES
Chapter - 1 Kinetic Theory of Gases 1.1 Basic Assumption of Kinetic Theory: 1. Any infinitely small volume of a gas contains a large number of molecule. 2. A gas is made up identical molecule which behaves as rigid, perfectly elastic, hard sphere. 3. The molecules continuously move about in random directions. All directions of motion are equally probable. 4. The size of the molecules is much less than the average distance between them. 5. The molecule of a gas exert no force on each other except when they collide. 6. The collision between molecules and with walls are perfectly elastic. 7. The direction of molecular velocities are assumed to be distribute uniformly. 8. The molecules move with all speeds ranging from 0 to ∞. 9. The time of collision is much less than the time between collisions. 1.1.1 Pressure Exerted by a Gas Suppose there are n molecules per cubic meter each of mass m, and its is assumed that ni no. of molecule have velocity vi . Mathematically
∑ ni = n and vi2 = vix2 + viy2 + viz2 where vix viy and viz are x, y, z component of velocity of gases. From assume of kinetic theory of gases v ix2 = viy2 = viz2
=
vi2 3
suppose molecules are kept in the cubic container of parameter L . A molecule moving in the x direction will have momentum mvix normal to face of the cube before collision ΔPix = mv ix − (− mvix ) = 2mvix
ni 2mvix ni 2mvix2 ni mvix2 Force acting on the wall by molecule is f ix = = == 2L L Δt
Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES mn v 2 Pressure exert on the wall of container by molecule Pix = i3 ix L so that pressure in the x direction expected by all group
m ∑ ni vix2 L3
Px = ∑ Pix = Average value of v2 is given by
∑n v = ∑n
2 i ix
v x2
i
=
∑n v
2 i ix
i =1
n
i
For three dimensional system vx2 + v y2 + vz2 = v 2 and for isotropic system
v
2 x
= v
2 y
= v
2 z
=
v2 3
So Px can be written as
1m m n v2 Px = 3 n v x2 , P = Px = 3 L3 L
2 1 mn v P= 3 V
1 PV = mN v 2 3 where V is volume of the container and v 2 is average value of square of velocity. 1.2
Gas Law for Ideal Gases:
1.2.1 Boyle’s Law
At constant temperature (T ) , the pressure ( P ) of a given mass a gas is inversely proportional to its volume (V)
P∝
1 V
1.2.2 Charle’s Law
At constant pressure ( P ) the volume of a given mass of a gas is proportional to its temperature (T)
V ∝T Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 1.2.3 Avogadro’s Law
At the same temperature and pressure, equal volume of all gases contain equal number of molecules (N).
N1 = N2 1.2.4 Graham’s Law of Diffusion
When two gases at the same pressure and temperature are allowed to diffuse into each other, the rate of diffusion (r) at each gas is inversely proportional to square root at density of gas (ρ) r1 = r2
ρ2 ρ1
Dalton’s Law of Partial Pressure: The sum of pressure exerted (P) by each gas occupying the same volume as that of the mixture (P1, P2, P3,….) P = P1 + P2 + P3 +…. 1.2.5 Ideal Gas Equation:
Consider a sample of an Ideal gas at pressure P, volume V and temperature T the gas follows the equation PV = nRT
Where n is number of molecules and R is proportionality constant known as gas constant R = 8.314 J/mol/K
Boltzmann constant K is ratio between R to Avogadro number NA k B =
R 8.314 = N A 6.03 × 1023
k B = 1.3 × 10 −23 J / K
Example: Find the maximum attainable temperature of ideal gas in each process given
by p = p0 − αV 2 ; where p 0 , α and β are positive constants, and V is the volume of one
mole of gas. Solution:
P = P0 − αV 2
(i)
Number of mole of gas = 1
Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES RT We know put in (i) PV = nRT ⇒ P = V
RT = P0 − αV 2 V
P0V αV 3 ⇒T = − R R
(ii)
dT P 3αV 2 =0 ⇒ 0 − =0 R R dV
For T maximum,
V =
P0 3α
put in (ii) one will get Tmax =
P0 2 P0 3 3α
Example: Two thermally insulated vessel 1 and 2 are filled with air. They are connected by a
short tube with a value. The volume of vessels and the pressure and temperate of air in them are (V1 , P1 , T1 )
and ( V2 , P2 , T2 ) respectively. Calculate the air temperate and
pressure established after opening of value if air follow Ideal gas equation. Solution: For vessel (1) P1V1 = n1 RT1
For vessel (2) P2V2 = n2 RT2
n1 =
P1V1 RT1
n2 =
P2V2 RT2
After opening the value let pressure volume and temperature is P, V, T PV = nRT
V = V1 + V2
n = n1 + n2 =
P1V1 P2V2 + RT1 RT2
Hence system is isolated then Energy of (1) + energy of (2) = energy of composite 3 3 3 n1 KT1 + n2 KT2 = (n1 + n2 )KT 2 2 2 n1T1 + n2T2 = (n1 + n2 ) / T .
T=
n1T1 + n2T2 n1 + n2
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES (P V ) P1V1 T1 + 2 2 T2 ( PV RT1 RT2 1 1 + P2V2 ) ⇒ T = T1T2 = P1V1 P2V2 PV 1 1T2 + P2V2T1 + RT1 RT2
P=
PV = nRT
nRT V
P=
P1V1 + P2V2 V1 + V2
Example: A horizontal cylinder closed from one end is rotated with a constant angular velocity
ω about a vertical axis passing through the open end of the cylinder. The outside air pressure is equal to p 0 , the temperature to T , and the molar mass of air to M . Find the air pressure as a function of the distance r from the rotation axis. The molar mass is assumed to be independent of r . Solution: Force equation of dr element. dF = (dm )rω 2
dP =
if S is cross section area then
dF ⎛ dm ⎞ 2 =⎜ ⎟ rω S ⎝ S ⎠
⎛ S ⎞ dm = ⎜ 2 ⎟dP ⎝ rω ⎠
Also we know
⎛ dm ⎞ P(Sdr ) = ⎜ ⎟ RT ⎝M ⎠ PS (dr ) =
ω M, T
RT ⎛ S ⎞ ⎜ ⎟dP M ⎝ rω 2 ⎠
r
P
0
P0
Mω 2 ∫ rdr = RT ∫
dP P
S
P0
r This end is open in air
Mω 2 r 2 P = RT ln P0 2 P = P0 e
Mω 2 r 2 2 RT
Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 2 1 Example: Prove that PA = mN v 2 and E = k BT = kBT in two dimension. 2 2 Solution: A molecule moving in the x direction will have momentum mvix normal to face of the
cube before collision
ΔPix = mvix − (− mvix ) = 2mvix Force acting on the wall by molecule is f ix =
ni 2mvix ni 2mvix2 ni mvix2 = == Δt 2L L
Pressure exert on the wall of container by molecule Pix =
mni vix2 L3
So that pressure in the x direction expected by all group
m ∑ ni vix2 3 L
Px = ∑ Pix = Average value of v 2 is given by
∑n v = ∑n
2 i ix
v x2
=
i
∑nv i =1
i
For two dimensional system v
2 x
+ v
2 y
2 i ix
= v
n 2
and v
2 x
= v
2 y
=
v2 2
So Px can be written as
1 m m Px = 2 n v x2 , P = Px = n v2 2 L2 L PA =
2 1 mn v P= 2 A
1 mN v 2 2
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fiziks 1.3
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Kinetic Interpretation of Temperature
According to assumption of Kinetic theory of gases, there is only translation motion of the molecule and there is not any potential acting between them, so Average energy E of gases are equivalent to Average translation energy of a molecule
E =
1 m v2 2
1 Pressure at P as P = mn v 2 3 2 PV = Vn E 3
=
PV =
2 ⎛1 ⎞ 2 n⎜ m v 2 ⎟ = n E 3 ⎝2 ⎠ 3
2 N E 3
where n =
N number density V
E =
3⎛ R ⎞ 3 RT and E = ⎜ ⎟T 2⎝ NA ⎠ 2 NA
E =
3 K BT where k B is Boltzman constant 2
So average kinetic energy is given by E =
3 k BT where T is absolute temperature. 2
Example: It is possible to treat electromagnetic radiation in container whose wall is mirrors, as a
gas of particle (photons) with a constant speed c and whose energy is related to their momentum p which is directed parallel to their velocity by E = pc .Show that if 1 container is full of radiation the equation of state is PV = E 3 1 1 1 G G Solution: Pressure P = nm v 2 = n mv ⋅ v = n p ⋅ v 3 3 3 For Photon v = c and velocity is parallel to momentum, so 1 P = n Pc 3 PV =
1 Npc 3
1N pc 3V
⇒
P=
⇒
1 PV = E 3
Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES z 1.4 Maxwell-Boltzmann Distribution Law: dv Distribution of Molecular velocity in perfect gas.
Maxwell-Boltzmann distribution law is applicable for Ideal gas where molecules have no vibrational or rotational
v y
energies. In the equilibrium state of the molecules have complete x random motion and probability that a molecule has a given velocity component is independent of other two components. In given figure dv is volume element in velocity space for a molecule at velocity G v ≡ (v x , v y , v z ) . v 2 = v x2 + v y2 + v z2
We need to calculate number of molecules simultaneously having component in the range v x to v x + dv x , v y to v y + dv y and v z to v z + dv z
It is assumptions in Maxwell-Boltzmann distribution law is that probability that molecule selected at random has velocities in a given range is a function purely at the magnitude of velocity and the width of the interval. So fraction of molecule having velocity component in the range v x to v x + dv x , v y to v y + dv y and v z to v z + dv z is f (v x )dv x , f (v y )dv y and f (vz )dvz respectively. dN = f (v x ) f (v y ) f (v z )dv x dv y dv z N
where dN is number of molecule having between velocity v to v + dv and N is total number of molecules. dN = N f (v x ) f (v y ) f (v z )dv x dv y dv z
Number of molecule having velocity vx to vx + dvx, vy to vy + dvy and vz to vz + dvz is same as number of molecule having velocity v to v + dv. So
N f (v x ) f (v y ) f (v z )dv x dv y dv z = N F (v 2 ) dv x dv y dv z
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES G 2 F is some function of v (magnitude of velocity) and for fixed value of v , F ( v 2 ) is
constant. So
dF (v 2 ) = 0 is equivalent to d [ f (v x ) f (v y ) f (v z )] = 0 f ′ ( vx ) dvx f ( v y ) f ( vz ) + f ′ ( v y ) dv y f ( vx ) f ( vz ) + f ′ ( vz ) dvz f ( vx ) f ( v y ) = 0
Dividing both side with f(vx) f(vy) f(vz) f ′ (v x ) f ′ (v y ) f ′ (v z ) dv x + dv y + dv z = 0 f (v x ) f (v y ) f (v z )
(i)
v 2 = constant v x2 + v y2 + v z2 = v 2 vxdvx + vydvy + vzdvz = 0
(ii)
by method of Lagrange’s method of undetermined multiplies multiply by 2β in equation (ii) and add in equation (i) ⎛ f ′ (v y ) ⎞ ⎞ ⎛ f ′ (v z ) ⎞ ⎛ f ′ (v x ) ⎜⎜ + 2 β v y ⎟⎟dv y + ⎜⎜ + 2 β v z ⎟⎟dv z = 0 + 2 β v x ⎟⎟dv x + ⎜⎜ ⎠ ⎝ f (v z ) ⎠ ⎝ f (v x ) ⎝ f (v y ) ⎠
hence v x , v y and v z are independent f ′ ( vy ) f ′ ( vx ) f ′ ( vz ) + 2 β vx = 0 ⇒ + 2 β vz = 0 + 2β vy = 0 ⇒ f ( vx ) f ( vz ) f ( vy )
f (v x ) = Ax e − βvx
f (v y ) = Ay e − βv y
f (v z ) = Az e − βvz
2
2
2
f (v x ), f (v y ), f (v z ) are probability density, so ∞
∫
∞
f (v x )dv x = 1,
∫
−∞
f (v y )dv y = 1,
−∞
∞
∫ f (v )dv z
z
= 1,
−∞
Use the integration ∞
−βv n ∫ e v dv =
0
∞
Ax
1
2
∫
−∞
2β
n +1 ( ) 2
n +1 2 ∞
e − β vx dvx = 1 = Ax 2 ⋅ ∫ e− β vx dvx = 1 2
2
0
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 1/ 2
1/ 2
⎛β ⎞ ⎛β ⎞ Ax = ⎜ ⎟ Similarly, Ay = ⎜ ⎟ ⎝π ⎠ ⎝π ⎠ ⎛β ⎞ f (v x ) = ⎜ ⎟ ⎝π ⎠
dN ⎛ β ⎞ =⎜ ⎟ N ⎝π ⎠
1/ 2
3/ 2
e
− β v x2
⎛β ⎞ , f (v y ) = ⎜ ⎟ ⎝π ⎠
1/ 2
⎛β ⎞ Az = ⎜ ⎟ ⎝π ⎠ 1/ 2
e
− β v 2y
⎛β ⎞ , f (v z ) = ⎜ ⎟ ⎝π ⎠
1/ 2
e − βvz
2
e − β (v x + v y + v z )dv x dv y dv z 2
2
2
where − ∞ < v x < ∞ ,
− ∞ < vz < ∞
− ∞ < vy < ∞ ,
1.4.1 The Distribution in Term of Magnitude v 2 = v x2 + v y2 + v z2 which is equation of sphere and dv x dv y dv z can be replace by 4πv 2 dv
dN ⎛ β ⎞ f (v )dv = =⎜ ⎟ N ⎝π ⎠
3/ 2 2
4πe − βv v 2 dv
0 v p
(b) v rms > v p > v
(c) v > v p > v rms
(d) v rms > v > v p
A hypothetical speed distribution for a sample of N gas particles is shown below. Here P (v ) = 0 for v > 2v0 . How many particles have speeds between 1.2v0 and 1.9v0 ?
P(v )
a
0 (a) Q10.
N 5
(b)
V0 Speed V
7N 15
(c)
2V0 2N 21
(d) None of these
A parallel beam of nitrogen molecules moving with velocity v m/s impinges on a wall at an angle θ to its normal. The concentration of molecules in the beam n cm3. The pressure exerted by the beam on the wall assuming the molecules to scatter in accordance with the perfectly elastic collision law is given by (a) 2nmv 2 cos θ
(b) nmv 2 cosθ
(c) 2nmv 2 sin θ
(d) nmv 2 sin θ
Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
[email protected] 16
fiziks Q11.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES If The mass of each molecule is equal to m then The temperature of a gas will the number
of molecules, whose velocities fall within the given interval from v to v + dv be the greatest (a) T = Q12.
mv 2 kB
(b) T =
mv 2 2k B
(c) T =
mv 2 3k B
(d) T =
mv 2 4k B
Using the Maxwell distribution function, the mean value of the modulus of the modulus of this projection in x direction .e. v x
if the mass of each molecule is equal to m and
the gas temperature T is given by (a) 0 Q13.
(b)
k BT πm
(c)
k BT m
Making use of the Maxwell distribution function, if
(d)
2 k BT πm
1 the mean value of the reciprocal v
of the velocity of molecules in an ideal gas and v is the average velocity at a temperature T ,if the mass of each molecule is equal to m .then which one of the following is correct. (a)
1 1 = v v
(b)
1 4 = v π v
(c)
1 2 = v π v
(d)
1 4π = v v
Q14. If the root mean square velocity of hydrogen molecules exceeds their most probable velocity by Δv m/s then temperature is given by (a) T = kB
(
(c) T = kB
mΔv 2 3− 2
(
)
(b) T =
2
mΔv 3− 2
kB
)
2
(d) T = kB
(
(
mΔv 3− 2 mΔv 3− 2
)
2
)
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fiziks Q15.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES In the case of ideal gaseous in three dimensional the temperature at which the velocities
of the molecules v1 m / s and v 2 m / s are associated with equal values of the Maxwell distribution function F (v ) (a) T = (c) T = Q16.
m ( v22 − v12 ) 4k B ln v2 / v1
(
)
m v22 − v12 4k B ln v1 / v2
(
)
(
)
(b) T =
m v22 + v12 4k B ln v2 / v1
(d) T =
m v22 + v12 4k B ln v1 / v2
A gas consists of molecules of mass m and is at a temperature T in three dimension. Making use of the Maxwell velocity distribution function, the corresponding distribution of the molecules over the kinetic energies E is given by . ⎛ 1 ⎞ ⎟⎟ (a) f ( E )dE = π ⎜⎜ ⎝ πk BT ⎠
3/ 2
⎛ 1 ⎞ ⎟⎟ (c) f ( E )dE = π ⎜⎜ π k T ⎝ B ⎠
Q17.
−
e 3/ 2
E k BT
−
e
E k BT
E .dE
E.dE
⎛ 1 ⎞ ⎟⎟ (b) f ( E )dE = 2π ⎜⎜ ⎝ πk BT ⎠
3/ 2
⎛ 1 ⎞ ⎟⎟ (d) f ( E )dE = 2π ⎜⎜ π k T ⎝ B ⎠
3/ 2
−
E k BT
−
E k BT
e
e
E .dE
E.dE
A gas consists of molecules of mass m and is at a temperature T in two dimensions. Making use of the Maxwell velocity distribution function, the corresponding distribution of the molecules over the momentum p is given by p2
⎛ ⎞ − 2 mk BT 1 ⎟⎟e (a) f ( p ) = ⎜⎜ pdp 2 π mk T B ⎠ ⎝ p2
⎛ ⎞ − 2 mk BT 1 ⎟⎟e (c) f ( p ) = ⎜⎜ dp 2 π mk T B ⎝ ⎠
p2
⎛ 1 ⎞ − 2 mk BT ⎟⎟e (b) f ( p ) = ⎜⎜ pdp mk T ⎝ B ⎠ p2
⎛ 1 ⎞ − 2 mk BT ⎟⎟e (d) f ( p ) = ⎜⎜ dp mk T B ⎝ ⎠
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES MSQ (Multiple Select Questions)
Q18.
Consider the following statements related to kinetic theory of gases are correct: (a) The molecules of a gas are all alike in size and shape and are hard, smooth, spherical particles. (b) The size of the molecules is very small compared to the volume occupied by the gas. (c) The molecules exert no appreciable force on one another except during a collision. (d) The collisions of the molecules with the walls of the containing vessel are inelastic.
Q19.
Consider the following statements. Which of the following is correct? (a) The root mean square velocity of molecules of a gas having Maxwellian distribution of velocities, is higher than their most probable velocity, at any temperature. (b) A very small number of molecules of a gas which posses very large velocities increase the root mean square velocity without affecting the most probable velocity (c) Most probable velocity is lowest among the most probable velocity, average velocity and root mean square velocity. (d) Mean square velocity is equal to square of mean velocity
Q20.
Consider a collision between an oxygen molecule and a hydrogen molecule in a mixture of oxygen and hydrogen kept at room temperature. Which of the following are possible? (a) The kinetic energies of both the molecules increase. (b) The kinetic energies of both the molecules decrease. (c) The kinetic energy of the oxygen molecule increases and that of the hydrogen molecule decreases. (d) The kinetic energy of the hydrogen molecule increases and that of the oxygen molecule decreases.
Q21.
Consider a mixture of oxygen and hydrogen kept at room temperature. As compared to a hydrogen molecule an oxygen molecule an oxygen molecule hits the wall (a) with greater average speed
(b) with smaller average speed
(c) with greater average kinetic energy
(d) with smaller average kinetic energy
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fiziks Q22.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Which of the following quantities is zero on an average for the molecules of an ideal gas
in equilibrium?
Q23.
Q24.
Q25.
(a) any component of momentum
(b) magnitude of momentum
(c) x component of velocity
(d) speed
The average momentum of a molecule in a sample of an ideal gas depends on (a) temperature
(b) number of moles
(c) volume
(d) mass of molecule
Which of the following quantities is the same for all ideal gases at the same temperature? (a) The kinetic energy of 1 mole
(b) The kinetic energy of 1 g
(c) The number of molecules in 1 mole
(d) The number of molecules in 1 g
Which of the following is correct for ideal gas in two dimensional system ⎛ E ⎞ n (a) The energy distribution is f ( E ) ∝ exp− ⎜ ⎟ E the value of n = 1 for two ⎝ k BT ⎠ dimensional function. (b) The average kinetic energy is equal to k BT
(c) The rms velocity of the gas is
3k BT m
(d) The most probable velocity of the gas is Q26.
k BT m
Keeping the number of moles, volume and temperature the same, which of the following are not the same for all ideal gases? (a) rms speed of a molecule
(b) density
(c) pressure
(d) average magnitude of momentum
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES NAT (Numerical Answer Type)
Q27.
At temperature ……………… 0C the rms speed of gaseous hydrogen molecules equal to that of oxygen molecules at 47 0 C .
Q28.
Temperature of an ideal gas is increased such that the most probable velocity of molecules increase by a factor of 4. The rms velocity increase by the factor ……………
Q29.
If density of hydrogen gas is 0.1 kg/m 3 and atmospheric pressure is 1.0 × 10 5 N/m 2 , then root mean square speed of hydrogen molecule is…………m/sec
Q30.
The temperature in Kelvin, at which the average speed of H 2 molecules will be same as that of N 2 molecules at 35 o C , will be……………
Q31.
Consider a Maxwellian distribution of the velocity of the molecules of an ideal gas three dimension. Let vmp and vrms denote the most probable velocity and the root mean square velocity, respectively. The magnitude of the ratio
vrms is………….(Answer must be up to vmp
one decimal point) Q32.
Consider a Maxwellian distribution of the energy of the molecules of an ideal gas in three dimensions. Let Eav and Erms denote the average energy and the root mean square energy, respectively. The magnitude of the ratio
Erms is………….(Answer must be up to Eav
one decimal point). Q33.
Consider a Maxwellian distribution of the velocity of the molecules of an ideal gas in two dimension. Let vav and vrms denote the average velocity and the root mean square velocity, respectively. The magnitude of the ratio
vrms is………….(Answer must be up to vav
two decimal point)
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Solutions MCQ (Multiple Choice Questions)
Ans. 1: (a) Solution: For Maxwellian distribution vmp =
v 2 k BT 3 3k BT , vrms = ⇒ rms = m m vmp 2
Ans. 2: (a) Solution: Area under the f (v ) curve is conserved and the mean velocity shift towards right for higher temperature. Ans. 3: (b) Solution: Ideal gas obeys Maxwell velocity distribution law at equilibrium temperature. Then average value of (α vx − β v y )
(αv
Now
− βy )
2
x
= α 2 v x2 + β 2 v y2 − 2 αβv x v y
v x = 0, v y = 0 and vx2 =
(α v
Then
(α v
x
x
− β vy )
− β vy ) 2
2
2
=α2
k BT = v y2 = vz2 m
= α 2 v x2 + β 2 v y2 − 2αβ v x v y
(
)
k BT k T k T + β2 B = α2 + β2 B m m m
Ans. 4: (b) Solution: Ideal gas obeys Maxwell velocity distribution law at equilibrium temperature. Then average value of (α vx v y ) Now a v
2 x
(α v v ) x y
2
2
= α 2vx2 vy2
k T = B = v y2 = v z2 m
,Then
(α v v ) x y
2
⎛k T ⎞ =α ⎜ B ⎟ ⎝ m ⎠
2
2
Ans. 5: (a) Ans. 6: (b) Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES ⎛ dN ⎞ Solution: By Maxwell’s distribution law the area, of graph between ⎜ ⎟ versus velocity v is ⎝ dv ⎠
same at all temperature. Hence, A I = AII = AIII Ans. 7: (c) Solution: vrms =
3RT m
(i)
Most probable velocity v p =
2 RT m
From Equation (i) and (ii)
⇒ vrms and v p both proportional T Ans. 8: (d) Ans. 9: (b) Solution: Since, total probability is one, hence area of the figure should be one ⇒
1 av 0 + a(2v 0 − v 0 ) = 1 2
⇒
3 2 av 0 = 1 ⇒ a = 2 3v 0
now area between v = 1.2v0 to v = 1.9v0 =
P(v )
a
2 7N 0 × (1.9v 0 − 1.2v 0 ) = 3v 0 15
V0 Speed V
2V0
Ans. 10: (a) Solution: Momentum transfer in one collision = 2mv cos θ Number of molecules collision per second = n (vA ) ⎛ dp ⎞ ⎜ ⎟ = nv(2mv cos θ )A ⎝ dt ⎠ mt
F = 2nmv 2 cos θ
F / A = P = 2nmv 2 cos θ
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Ans. 11: (c) ⎛ m ⎞ ⎟⎟ f (v ) = ⎜⎜ ⎝ 2πk BT ⎠
Solution:
3/ 2
−
e
mv 2 2 k BT
4πv 2
v = constant but T is variable.
Here
Then for f (v) maximum.
df (v ) =0 dT
⎛ m ⎞ ⎟⎟ f (v ) = ⎜⎜ ⎝ 2πk B ⎠
3/ 2
⎛ mv 2 ⎞ ⎛ ⎜ ⎟⎞ ⎜ −3 / 2 −⎜⎝ 2 kBT ⎟⎠ ⎟ 2 ⎜⎜ T e ⎟⎟4πv ⎝ ⎠
mv 2 3 −5 / 2 mv 2 df (v ) = T ⇒T = = 0 , ⇒ T −7 / 2 2k B 2 3k B dT
Ans. 12: (d) 1/ 2
⎛ m ⎞ ∫−∞ −vx N ⎜⎝ 2π kBT ⎟⎠ = 0
Solution: vx
−
e
m 2 vx 2 k BT
dvx + ∫
∞
0
1/ 2
⎛ m ⎞ vx N ⎜ ⎟ ⎝ 2π k BT ⎠
−
e
m 2 vx 2 k BT
dvx = vx =
N
2 k BT πm
Ans. 13: (b) 1 ∫ v (dN ) ∞
1 = v
Solution:
0
N
=
∫
∞
0
1 ⎛ dN ⎞ ⎜ ⎟ v⎝ N ⎠
=∫
∞
0
1⎛ m ⎞ ⎜ ⎟ v ⎜⎝ 2πk BT ⎟⎠
3/ 2
−
e
m 2 v 2 k BT
4πv 2 dv
1 2m 4 = = v πk BT π v
Ans. 14: (a) Solution: v p = ⇒
(
2k BT m
Δv 3− 2
)
and vrms = m = T kB
3k BT 3k BT 2k BT − vrms − v p = Δv = m m m
⇒T = kB
(
mΔv 2 3− 2
)
2
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Ans. 15: (a)
Solution:
⎛ m ⎞ F (v) = ⎜ ⎟ ⎝ 2π k BT ⎠
3/ 2
e − mv / 2 kBT 4π v 2
Let temperature T at which for v1 and v2 , F ( v ) are same. ⎛ m ⎞ ⎜ ⎟ ⎝ 2π k BT ⎠
3/ 2
e
− mv12 / 2 k BT
⎛ m ⎞ 4π v = ⎜ ⎟ ⎝ 2π k BT ⎠ 2 1
3/ 2 2
e − mv2 / 2 kBT 4π v22
Here range will be taken same for both
T=
m ( v22 − v12 )
4k B ln ( v2 / v1 )
Ans. 16: (b) ⎛ m ⎞ ⎟⎟ Solution: f (v) = ⎜⎜ ⎝ 2πk BT ⎠ K .E . = E =
3/ 2
−
e
mv 2 2 k BT
2E Differentiate: dE = mvdv m
1 mv 2 ⇒ v = 2
⎛ m ⎞ ⎟⎟ f ( E )dE = ⎜⎜ ⎝ 2πk BT ⎠
3/ 2
−
e
⎛ 1 ⎞ f ( E ) dE = 2π ⎜ ⎟ ⎝ π k BT ⎠
4πv 2 dv
E k BT
3/ 2
−
e
4π E k BT
⇒ vdv =
dE m
2 E dE . m m
E .dE
Ans. 17: (b) mv 2
⎛ m ⎞ − 2 k BT ⎟⎟e Solution: In two dimensional f (v) = ⎜⎜ 2πvdv ⎝ 2πk BT ⎠ mv 2
⎛ m ⎞ − 2 k BT 1 p2 2 ⎟⎟e dp = mdv putting the value in f (v) = ⎜⎜ 2πvdv K .E. = mv = 2 π k T 2 2m B ⎠ ⎝ p2
⎛ m ⎞ − 2 mkBT p dp ⎟⎟e f ( p ) = ⎜⎜ 2π m m ⎝ 2πk BT ⎠
p2
⎛ 1 ⎞ − 2 mkBT ⎟⎟e ⇒ f ( p ) = ⎜⎜ pdp ⎝ mk BT ⎠
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES MSQ (Multiple Select Questions)
Ans. 18: (b) and (d) Solution: Kinetic theory of gases are based on the following assumptions: (i) All gases are made up of tiny elastic particles known as molecules. (ii) the volume of the molecule is negligible. (iii) The collision between the molecules is elastic. (iv) They exert no force on each other. Ans. 19: (a), (b) and (c) dN dC
Solution:
C av C rms C mp
Ans. 20: (c) and (d) Solution: Momentum will transferred from one molecule to other as higher momentum will change to lower momentum and vice versa . Ans. 21: (b) Ans. 22: (a) and (c) Ans. 23: (a) and (d) Ans. 24: (a) and (c) Ans. 25: (b) and (d) ⎛ E ⎞ Solution: For two dimension f ( E ) ∝ exp− ⎜ ⎟ ⎝ k BT ⎠
n=0
∞
average energy is
∫ Ef ( E )dE 0 ∞
∫ f ( E )dE
= k BT
0
for two dimensional system vrms =
2 k BT m
vmp =
k BT m
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Ans. 26: (a), (b) and (d)
Solution: Only pressure is not function of mass.
NAT (Numerical Answer Type)
Ans. 27:
−2530 C
Solution:
Vrms =
3RT m 3RT0 m 3RTH = = TH = H (T0 ) m0 m0 mH
⇒
m H = 2amu, m0 = 32 amu and T0 = 47 + 273 = 320 0 C so, TH =
2 (320 ) 32
TH = 20 K
Ans. 28:
= TH = (20 − 273) C = −2530 C 0
4
Solution: v rms =
3RT and v p = m
2 RT m
⇒ vrms and v p are both proportional to T ⇒ v rms increases by 4 times Ans. 29:
1710 m/s
Solution: By kinetic theory of gases the pressure exerted by the gas on the wall of container is given as 1 Pressure P = d v 2 3
Here, P = 1 × 10 5 N/m 2 , d = 0.1 kg/m 3 .
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 3P So, by Eq. (i) v = d
3 × 1× 105 = 3 × 10 6 m/s = 1710 m/s 0.1
vrms = Ans. 30:
TH = 22 K
The average speed of a molecule of gas at constant temperature T is given as v av =
T M v 8 RT ⇒ H = H N vN M H TN πM
vH = vN
given,
⇒
TN M H = TH M N
⇒
TH M H 2 = = TN M N 28
⇒
TH =
1 1 × TN = × (273 + 35 ) 14 14
TH = 22 K
Ans. 31:
0.8
Solution: For Maxwellian distribution v mp = Ans. 32:
1.8
Solution:
Eav =
Erms Eav
Ans. 33:
3k BT 2
Erms =
v 2k BT 3k BT 2 ⇒ mp = , vrms = vrms 3 m m
15 .k BT 2
15 2.7 = 2 = = 1.8 3 1.5 2
2.25
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES ∞
mv 2
∞
− m π k BT v.e 2 kBT 2π vdv = Solution: v = vav = ∫ vf (v)dv = ∫ 2π kBT 0 8m 0 ∞
v
2
∞
mv 2
− m 2k T v 2 .e 2 kBT 2π vdv = B = ∫ v f (v)dv = ∫ 2π k BT 0 m 0
vrms =
2
2kBT m
In two dimension system vrms =
vrms = vav
2kBT m
2 1 = 1.4 = 2.25 π .62 8
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES
Chapter - 2 Real Gases 2.1
Andrew’s Experiment on Carbon Dioxide Andrew’s experiment investigated the behaviour of CO2 and analyse the Pressure
( P ) versus volume (V ) at different temperature T . The observations are following: 1. Above a temperature of about (T = 480 C ) the CO2 resemble that of Ideal gas.
2. As temperature lowered, the isotherms exhibit distortion which gradually increases, which is indication from the ideal gas character. 3. At 31.40 C a Kink is observed which suggests the gas can be liquified under
compression. 4. As temperature is lower further the kink spread into a horizontal line, i.e. compression produces liquification. From A to B , CO2 behave as a gas. At the point P
D K
J
condenses at constant pressure from B to C so that liquid and vapour coexist. At C , the gas is completely in the liquid phase.
liquid
B the liquification of the gas just starts. The gas
Q
From C to D the slop is very steep since a liquid
P
gas
condensation C B F
480 C 0 A 31.40 C 21.50 C 13.1 C
V
is almost incompressible. Conclusion: The temperature at which it becomes possible to liquefy a gas under
compression is known as critical temperature
(TC ) = 480 C ],
(TC )
[In Andrew experiment
corresponding pressure and volume is known as critical pressure ( PC )
and critical volume (VC ) . A gas can be liquified only if it cooled upto or below its characteristic critical temperature. Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES There exist a continuity of liquid and gaseous states, i.e. they are two distinct stages at a
continuous physical phenomenon. 2.2
van der Waals equation of state.
The van der Waals equation for real gases are given by a ⎞ ⎛ ⎜ P + 2 ⎟(V − b ) = RT V ⎠ ⎝
for 1 mole of gas and
⎛ n2 a ⎞ P + ⎜ ⎟ (V − nb ) = nRT V2 ⎠ ⎝
for n mole of gas.
Assumption for real gas: 1. Gas molecules have finite size 2. There are weak interaction force, which depends only upon distance between them. 3. The molecular density is small and the number of collisions with the walls of the container are exactly the same point and finite size molecular. 2.3
Correction in Ideal Gas equation to achieve van der Waals gas equation of state.
2.3.1 Correction for finite size: if V is volume available for one mole of gas (volume of
container). If size of molecule take into account then (V − b) is volume available for real gas which is less than V . b is popularly known as covolume which is dependent on the nature of gas. Example: If Vm is molecular volume of real gas then prove that b = 4NVm if N is total number
of molecule in container. The volume available to first molecule = V The volume available to second molecule = V − Vs
d = 2r
Vm
r
r
Where Vs volume of exclusion i.e. around any ⎛ 4π d 3 ⎞ molecule, a spherical volume is Vs = ⎜ ⎟ will ⎝ 3 ⎠
Volume of exclusion
V s = 4π
(2 r )3
Vs = 8Vm
3
be denied to every other molecule. Volume of exclusion Vs =
4π ( 2r )
3
3
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES th Similarly volume available to N molecule = V - (N - 1)Vs
Average volume available for each molecule V =
1 N
N
∑V − ( i − 1)V
s
i =1
(V - b) = V - 4 N Vm
=V −
N ( N − 1) Vs 2N
=V −
N N Vs = V − 8Vm = V − 4 NVm 2 2
so b = 4NVm
2.3.2 Correction for intermolecular attraction:
A molecule in the equally in all direction so that there is no resultant force on it. But for outermost layer close to surface there will be net inward force. So whenever a molecule strikes the walls of container, the momentum exchange will be less than for Ideal gas. There forces are cohesive in nature and proportional to number of molecule. So for real gas change in pressure is
a . So for real gas pressure will be V2
a ⎞ ⎛ ⎜P+ 2 ⎟ V ⎠ ⎝
a ⎞ ⎛ So gas equation reduce to ⎜ P + 2 ⎟ (V − b ) = RT V ⎠ ⎝ Then
P=
RT a − 2 V −b V
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 2.3.3 Maxwell Equal Area 2 James Clerk Maxwell replaced the isotherm
between a and c with a horizontal line positioned so that P
1
the areas of the two hatched regions are equal (means area of adb and bec are equal). The flat line portion of the
isotherm
now
corresponds
to
PV
0
liquid-vapor
e a d
c
b
VL 1
2 VG 3
4
equilibrium. As shown in figure. The portions a − d and c − e are interpreted as metastable states of super-heated liquid and super-cooled vapor respectively. The equal area rule can be expressed as:
PV (VG − VL ) =
VG
∫ PdV
VL
where
PV is the vapor pressure (flat portion of the curve), VL is the volume of the pure
liquid phase at point a on the diagram, and VG is the volume of the pure gas phase at point c on the diagram. The sum of these two volumes will equal the total volume V . Example One mole of a certain gas is contained in a vessel of volume V . At a temperature
T1 the gas pressure is p1 atm and at a temperature T2 the pressure is p2 atm. Find the Van der Waals parameters for this gas. Solution: it is given no of mole n = 1
a ⎞ ⎛ ⎜ P1 + 2 ⎟(V − b ) = RT1 V ⎠ ⎝
(i)
a ⎞ ⎛ ⎜ P2 + 2 ⎟(V − b ) = RT2 V ⎠ ⎝
(ii)
from (i) and (ii)
V 2 (T1 P2 − T2 P1 ) a= (T2 − T1 ) b =V −
R(T2 − T1 ) (P2 − P1 )
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Example: Under what pressure will carbon dioxide of molar mass M have the density ρ at the
temperature T . If given gas is obeying for a Van der Waals gas. Solution: Assume M is molar mass of the carbon dioxide and V is the volume so ρ =
M V
Van der wall equation (for one mole gas):
a ⎞ ⎛ ⎜ P + 2 ⎟(V − b ) = RT V ⎠ ⎝ ⎛ ⎞ ρ 2 a ⎞⎛ M + − b ⎟ = RT P ⎜ 2 ⎟⎜ M ⎠⎝ ρ ⎠ ⎝
⇒P=
ρ 2a ρRT ρ 2a RT − 2 − 2 ⇒P= M − ρb M ⎛M ⎞ M ⎜⎜ − b ⎟⎟ ⎝ρ ⎠
2.3.4 Critical Point
The van der Waals equation of state for a gas is given by a ⎞ ⎛ ⎜ P + 2 ⎟(V − b ) = RT V ⎠ ⎝
where P, V and T represent the pressure, volume and temperature respectively, and a and b are constant parameters. At the critical point, where all the roots of the above cubic equation are degenerate means all roots are equal. In another way mathematically For the critical isotherm is the point of inflection point On basis of above definition one can find the critical volume Vc , critical pressure Pc and critical temperature Tc for van der waal gas. For Van der Waals equation
a ⎞ ⎛ ⎜ P + 2 ⎟ (V − b ) = RT V ⎠ ⎝ P=
RT a − 2 V −b V
⎛ ∂P ⎞ ⎜ ⎟ =0 ⎝ ∂V ⎠T
(i) for extremum point
RT 2a ⎛ ∂P ⎞ + 2 = 0 at V = Vc , T = Tc ⎜ ⎟ =− 2 ⎝ ∂V ⎠T (V − b ) V
(ii)
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 2 ⎛∂ P⎞ ⎜ 2 ⎟ = 0 for inflection point ⎝ ∂V ⎠T 2 RT
(V − b )
3
−
6a =0 V3
at V = Vc , T = Tc
(iii)
solving (ii) and (iii)
Vc = 3b
Tc =
8a 27 Rb
Put value of Vc and Tc one can get Pc =
a 27b 2
RTc 8 = = c c which is popular known as critical coefficient for van der Waals gas. Pc vc 3
2.3.5 van der Waals Equation of State and Virial Coefficient
According to virial theorem the equation of state is given by pV = α +
β V
+
γ V2
+ .....
(i)
Where α , β and γ are first second and third virial coefficient . For the Ideal gas α = RT and other coefficient are zero. Virial coefficient for Van der Waals gas
To put van der Waals equation in virial form we first rewrite it as −1
a ⎛ b⎞ pV = RT ⎜ 1 − ⎟ − V ⎝ V⎠
Using binomial theorem, we have −1
b b2 ⎛ b⎞ 1 1 − = + + + .... ⎜ ⎟ V V2 ⎝ V⎠
Hence pV = RT +
RTb − a RTb 2 + + ..... …. V V2
(ii)
As will be noted, van der Waals equation has only three virial coefficients and a comparison with equation (i) yields Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES α = RT , β = RTb − a and γ = RTb 2
At the Boyle temperature, the second virial coefficient is zero. Hence, RTB b − a = 0
or
TB =
a Rb
From the preceding, section we recall that the critical temperature of a gas obeying van der Waals equation of state is TC =
8a 27 Rb
on comparing these expressions, we get TB =
27 TC = 3.375 TC 8
that is, the Boyle temperature, on the basis of Van der Waals equation, is 3.375 times the critical temperature.
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES MCQ (Multiple Choice Questions)
Q1.
A gas behaves as an ideal gas at: (a) very low pressure and high temperature (b) high pressure and low temperature (c) high temperature and high pressure (d) low pressure and low temperature
Q2.
⎛ a ⎞ In the van der Waals equation, the terms ⎜ 2 ⎟ and ( b ) are introduced to account for the: ⎝V ⎠ (a) inter-molecular attraction and the total volume occupied by the gas (b) molecular size and the size of the containing vessel (c) inter-molecular attraction and the volume of the molecules (d) inter-molecular attraction and the force exerted by the molecules on the walls of the container
Q3.
‘Critical temperature’ is defined as the: (a) lowest temperature at which the gas can be liquefied at constant pressure (b) lowest temperature at which the gas can be liquified by increase of pressure alone (c) highest temperature at which the gas can beliquified by increase of pressure alone (d) highest temperature at which the gas can be liquified at constant pressure
Q4.
The work performed by one mole of a Van Der Waals gas during its isothermal expansion from the volume V1 to V2 at a temperature T is given by . (a) RT ln
⎛ 1 V2 − b 1⎞ + a⎜⎜ − ⎟⎟ V1 − b ⎝ V2 V1 ⎠
(b) RT ln
⎛1 V2 − b 1 ⎞ + a⎜⎜ − ⎟⎟ V1 − b ⎝ V1 V2 ⎠
(c) RT ln
⎛1 V1 − b 1 + a⎜⎜ − V2 − b ⎝ V1 V2
(d) RT ln
⎛ 1 V1 − b 1⎞ + a⎜⎜ − ⎟⎟ V2 − b ⎝ V2 V1 ⎠
⎞ ⎟⎟ ⎠
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES
Q5. D P
A
E
C B
13.1o C
V Consider the above graph in respect of van der Waals equation of state. Which portion of
the graph cannot be explained? (a) AB
(b) BC
(c) DE
(d) BCD
MSQ (Multiple Select Questions)
Q6.
Q7.
Which of the following are important in case of a van der Waals gas? (a) Short range attraction
(b) Long range repulsion
(c) Short range repulsion
(d) Long range attraction
Which of the important results of Andrews’ experiment are correct (a) There exists a temperature called critical temperature, above which a gas cannot be liquefied, however great the applied pressure is. (b) There exists a temperature called critical temperature, below which a gas cannot be liquefied, however great the applied pressure is. (c) For van der Waal gases critical temperature is
a . 27 Rb
(d) Oxygen, nitrogen and hydrogen are permanent gases and they cannot be liquefied. Q8.
a ⎞ ⎛ ⎜ P + 2 ⎟(V − b ) = RT V ⎠ ⎝
In the equation of state for real gases
(a) The critical points are point of inflection. (b) Critical volume is given by Vc = 3b (c) Critical pressure is given by Pc =
a 27b 2
(d) critical temperature is given by Tc =
8a 27 Rb
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES NAT (Numerical Answer Type Questions)
Q9.
If the value of van der Waals constant b for a real gas is 32 cm 3 / mol, then the approximate volume of one molecule of the gas i ……… ×10 −23 cm3 (Avogadro constant = 6.02 × 10 23 ) (Answer must be upto one decimal)
Q10.
⎛ RT van der Waals equation predicts that the critical coefficient of a gas ⎜⎜ c ⎝ PcVc
⎞ ⎟⎟ has the ⎠
value………..(Answer must upto one decimal point ) Q11. If equation of state is given by P =
RT a ⎞ ⎛ exp ⎜ − ⎟ then critical volume V −b ⎝ RTV ⎠
Vc = ……….. b .
Q12.
If equation of state is given by P = Tc = ………..
Q13.
RT a ⎞ ⎛ exp ⎜ − ⎟ then critical volume V −b ⎝ RTV ⎠
a . Rb
If equation of state is given by P =
RT a ⎞ ⎛ exp ⎜ − ⎟ then critical coefficient V −b ⎝ RTV ⎠
RTc = ........ PV c c
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Solutions MCQ (Multiple Choice Questions) Ans. 1: (a) Solution: The relation between PV and P is given as PV = A + BP + CP 2 + DP 2 +
(i)
A, B, C … are virial constant A > B > C > D … . If P is very high then PV ≠ constant.
So, for ideal gas P should be small and T should be large. Ans. 2: (a) Solution: The ideal gas equation is PV = RT whereas gas equation, for real gases is given by a ⎞ ⎛ van der Waals which is ⎜ P + 2 ⎟(V − b ) = RT where, the factor b is due to volume V ⎠ ⎝ ⎛ a ⎞ occupied by the molecules itself and ⎜ 2 ⎟ is due to molecular attractive force. ⎝V ⎠ Ans. 3: (c) Ans. 4: (a) Solution: We know van der Waals gas equation:
a ⎞ ⎛ ⎜ P + 2 ⎟(V − b ) = RT V ⎠ ⎝
(i)
ΔW = ∫ dW = ∫ pdV
from (i):
P=
RT a − 2 V −b V V2
then
a ⎞ ⎛ RT W = ∫⎜ − 2 ⎟dV V −b V ⎠ V1 ⎝ W = RT ln
⎛1 1⎞ V2 − b + a⎜ − ⎟ V1 − b ⎝ V2 V1 ⎠
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Ans. 5: (d) D
Solution: The Van der Waals equation is given as
a ⎞ ⎛ ⎜ P + 2 ⎟(V − b ) = RT V ⎠ ⎝
P
A
C
E
B
where a, b are constant.
13.1o C V
The graph drawn is shown as the curve ABCDE . This does not agree with the experimental isothermals for CO2 as obtained by Andrews However the portion DE has been explained as due to super cooling of vapours and the portion AB due to super heating of the liquid. But the portion BCD cannot be explained because it shows decrease in volume with decrease in pressure.
MSQ (Multiple Select Questions) Ans. 6: (c) and (d)
a ⎞ ⎛ Solution: van der Waals equation ⎜ P + 2 ⎟(V − b ) = RT , where a, b are constant. In this case V ⎠ ⎝ short range of repulsion and long range of attraction is important. Ans. 7: (a) and (d)
Solution: From the result of Andrew’s experiment we can define critical temperature as a temperature above which a gas cannot be liquefied however great the applied pressure is All gases are real gas so they can be liquefy For van der Waals gases, critical temperature is given by Tc =
8a 27 Rb
Ans. 8: (a), (b), (c) and (d) Solution: The van der Waals equation is given as
a ⎞ ⎛ ⎜ P + 2 ⎟(V − b ) = RT V ⎠ ⎝ ⎛ d2P ⎞ ⎛ ∂P ⎞ ⎟ = 0 and ⎜⎜ 2 ⎟⎟ = 0 (point of inflection ) thus, we get Vc = 3b ⎜ ⎝ ∂V ⎠T ⎝ dV ⎠T and Tc =
8a ⇒ 27 Rb
Pc =
a 27b 2
Thus, Vc = 3b
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES NAT (Numerical Answer Type Questions) Ans. 9:
1.3
Solution: The van der Waals constant b is given as b = four times the actual volume of the
molecules. volume of one molecule =
b 4N
volume of one molecule =
32 = 1.3 × 10 − 23 cm 3 23 4 × 6 × 10
Ans. 10:
2.6
Solution: For van der Waal Gases Tc =
8a a , Vc = 3b , Pc = 27b 2 27 Rb
RTc 8 = PcVc 3
Ans. 11:
2
Solution: P =
RT a ⎞ ⎛ exp ⎜ − ⎟ V −b ⎝ RTV ⎠
critical point are point of inflection so at critical points
⎛ d2P ⎞ ⎛ ∂P ⎞ ⎜⎜ 2 ⎟⎟ = 0 (point of inflection) = 0 and ⎟ ⎜ ⎝ ∂V ⎠T ⎝ dV ⎠T
So
Vc = 2b
Ans. 12:
0.25
Solution: P =
RT a ⎞ ⎛ exp ⎜ − ⎟ critical point are point of inflection so at critical points V −b ⎝ RTV ⎠
⎛ d2P ⎞ ⎛ ∂P ⎞ ⎜⎜ 2 ⎟⎟ = 0 (point of inflection ) = 0 and ⎟ ⎜ ⎝ ∂V ⎠T ⎝ dV ⎠T
Vc = 2b, Tc =
a 4bR
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fiziks Ans. 13:
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 3.7
Solution: P =
RT a ⎞ ⎛ exp ⎜ − ⎟ V −b ⎝ RTV ⎠
critical point are point of inflection so at critical points
⎛ d2P ⎞ ⎛ ∂P ⎞ ⎟ = 0 and ⎜⎜ 2 ⎟⎟ = 0 (point of inflection) ⎜ ⎝ ∂V ⎠T ⎝ dV ⎠T
Vc = 2b, Tc =
a a and Pc = 2 2 4bR 4e b
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES
Chapter – 3 Basics of Thermodynamics and Laws of Thermodynamics 3.1
Mathematical Formulations of Thermodynamics.
y = y ( x1 , x2 ,......xn ) then differential dy is said to be exact and one can write dy = ∑ ci dxi i
dy = c1dx1 + cdx2 + ....... = ⎛ ∂y ⎞ ⎛ ∂y ⎞ ⎛ ∂y ⎞ dy = ⎜ ⎟dxi ⎟ dx1 + ⎜ ⎟ dx2 + ....... = ∑ ⎜ i ⎝ ∂xi ⎠ ⎝ ∂x1 ⎠ ⎝ ∂x2 ⎠
and its corresponding ci xi are said to be conjugate to each other. 3.1.1 Some Important Formulas 1.
∂2 y ∂2 y = ∂xk ∂xl ∂xl ∂xk
⎛ ∂x ⎞ ⎛ ∂y ⎞ 2. ⎜⎜ ⎟⎟ = ⎜ ⎟ ⎝ ∂y ⎠ z ⎝ ∂x ⎠ z ⎛ ∂x ⎞ ⎛ ∂y ⎞ ⎛ ∂x ⎞ 3. ⎜ ⎟ = ⎜⎜ ⎟⎟ ⎜ ⎟ ⎝ ∂z ⎠ y ⎝ ∂y ⎠ z ⎝ ∂z ⎠ x ⎛ ∂x ⎞ ⎛ ∂x ⎞ ⎛ ∂z ⎞ 4. ⎜⎜ ⎟⎟ = −⎜ ⎟ ⎜⎜ ⎟⎟ ⎝ ∂z ⎠ y ⎝ ∂y ⎠ x ⎝ ∂y ⎠ z ⎛ ∂x ⎞ ⎛ ∂x ⎞ ⎛ ∂w ⎞ 5. ⎜ ⎟ = ⎜ ⎟ ⎟ ⎜ ⎝ ∂y ⎠ z ⎝ ∂w ⎠ Z ⎝ ∂y ⎠ z ⎛ ∂x ⎞ ⎛ ∂x ⎞ 6. dx = ⎜ ⎟ dy + ⎜ ⎟ dw ⎝ ∂w ⎠ y ⎝ ∂y ⎠ w ⎛ ∂x ⎞ ⎛ ∂x ⎞ ⎛ ∂x ⎞ ⎛ ∂w ⎞ ⎜ ⎟ =⎜ ⎟ +⎜ ⎟ ⎟ ⎜ ⎝ ∂y ⎠ z ⎝ ∂y ⎠ w ⎝ ∂w ⎠ y ⎝ ∂y ⎠ z
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fiziks 3.2
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Fundamental Concept
3.2.1 System: A system can be any object, region of space etc, selected for study and set apart (mentally) from everything else. Which then become surrounding. The system of interest in thermodynamics are finite and macroscopic rather than microscopic. The imaginary envelope which encloses a system and separated it from its surrounding is called the boundary of the system. 3.2.2 Isolated system It can not exchange either matter or energy with the surroundings. If exchange of matter is allowed the system is said to be open; if only energy and not matter is closed (but not isolated). 3.2.3 Thermodynamical state A thermodynamic state is a set of values of properties of a thermodynamic system that must be specified to reproduce the system. Thermodynamic state is the macroscopic condition of a thermodynamic system as described by its particular thermodynamic parameter. Such as temperature (T) pressure (P) volume (V) density (ρ ) . 3.2.4 State function state function also called “State variable” thermodynamic variables describe the momentary condition of thermodynamic system For a continuous process, such variable are exact different also fully determined by their initial and final thermodynamic states. Example includes entropy, pressure, temperature, volume, etc. 3.2.5 Intensive and extensive properties Intensive properties:-it is a physical property of a system that does not depend on the system size or the amount of material in system. It is scale invariant.
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Example: Chemical potential, density, viscosity, resistivity, specific heat capacity, pressure, elasticity, magnetization, velocity, acceleration, temperatures, etc. Solution: Extensive properties:- It is one that is additive for independent non-interacting subsystem. It is directly proportional to the amount of material in the system. Example energy, Entropy, Gibbs energy, mass momentum, volume, change, weight, Note: If f and g are arbitrary intensive variable then fg ,
f df , and f + g will also g dg
intensive variable. If F and G are also two arbitrary extensive variable then F + G will an extensive variable but
F dF and will be intensive. G dG
If F is extensive variable and f is intensive variable the f F and
F dF and is extensive f df
variable. 3.3
The Ideal Gas: The Ideal gas law is the equation of hypothetical Ideal gas. It is derived from kinetic theory and satisfied Boyle’s and Charles’s law. The state of an amount of ideal gas is determined by its pressure (P) volume (V) and temperature (T). The Ideal gas equation of state for n mole is given by PV = nRT Where R is gas constant and given by 8.314 J. K-1 mole-1
3.4
Laws of Thermodynamics
3.4.1 Zeroth Law of Thermodynamics: If two systems 1 and 2 are separately in thermal equilibrium with third 3 they must be in thermal equilibrium with one another. 3.4.2 First Law of Thermodynamics: Energy is conserved when heat is
δW
taken into account. Mathematically If δ Q amount of heated to the system and if system
dU
will do δ W amount of work then change in internal energy is dU given by
δW
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES dU = δ Q − δ W obviously heat exchange and work is dependent on path and internal energy is state function. U is a mathematical abstraction that keeps account of the exchange of energy that be fall
the system. The term δ Q means the amount of that amount of energy added to or remove by conduction of heat or by thermal radiation.
P
B
δ W is amount of energy gained or lost as result of work. 3.4.3 Work Done during Different Process.
A
Work done when process is occur to between A to B
V
B
W = ∫ PdV A
work done is area under the PV diagram. If work is done by the system it has positive sign, and if work is done on the system it has negative sign. 3.4.4 Specific Heat: Heat capacity of a body is numerically equal to quantity of heat required to raise its temperature by 1 unit. C=
Δθ ΔT
The specific heat of a material is numerically equal to quantity of heat required to raise the temperature of unit mass of that materials through 1unit C=
dθ mdT
3.4.5 Heat Capacity of Ideal Gas: if f is degree of freedom of Ideal gas then from equipartition of energy total sum of energy is equivalent to sum of kinetic energy associated with each degree of freedom
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 3.4.6 Molar Heat Capacity: heat capacity defined as the energy required to raise the temperature of one mole of Ideal gas by one Kelvin at constant volume. f N A KT where NA is Avogadro number 2
U =
f fR dU = N AK = dT 2 2 CV =
Rf , 2
for Ideal gas
⎛f ⎞ C P = C V + R = R⎜⎜ + 1⎟⎟ ⎝2 ⎠
γ is defined as ratio of heat capacity at constant pressure to constant volume. γ=
⎛ 2⎞ = ⎜1 + ⎟ Cv ⎝ f ⎠
Cp
3.4.7 Coefficient of Volume Expansion (α ) or expansivity α = ⎛ ∂P ⎞ 3.4.8 Isothermal Elasticity ET = −V ⎜ ⎟ ⎝ ∂V ⎠T
1 ⎛ ∂V ⎞ ⎜ ⎟ V ⎝ ∂T ⎠ P
and Isothermal compressibility β T =
1 ET
Example: Find the isothermal compressibility α of a Van der Waals gas as a function of volume V at temperature T .
Note: By definition, α = −
1 ⎛ ∂V ⎞ ⎜ ⎟ . V ⎜⎝ ∂ p ⎟⎠T
Solution: We know bulk modulus of a gas is given by a ⎞ ⎛ Van der Wall equation: ⎜ P + 2 ⎟(V − b ) = RT V ⎠ ⎝
(i)
If process is isothermal: T = constant −a a ⎞ ⎛ ⎞ ⎛ ⎜ dP + 3 dV ⎟(V − b ) + ⎜ P + 2 ⎟(dV ) = 0 V V ⎠ ⎝ ⎠ ⎝
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES a ⎞ ⎛ dP a ⎞ ⎛ − 3 ⎟(V − b ) + ⎜ P + 2 ⎟ = 0 ⎜ V ⎠ ⎝ dV V ⎠ ⎝
(
)
(
)
dP P + a /V 2 a − PV 3 + aV + a (V − b ) =− + 3 = (V − b ) V dV V 3 (V − b )
α =−
⎤ V 3 (V − b ) V 2 (V − b ) 1⎡ = V ⎢⎣ − PV 3 + aV + a(V − b ) ⎥⎦ PV 3 − aV + a(b − V )
(
[(
)
]
)
(ii)
Put values of P from (i) in (ii) V 2 (V − b ) α= 2 RTV 3 − 2a(V − b )
[
]
3.5 Different Type of Thermodynamical Process and Use of First Law of Thermodynamics. 3.5.1 Isochoric Process: When volume remain constant during the process the process is said to be isochoric process. dV = 0
dW = PdV
dW = 0
P
In an isochoric process work done during the process is zero.
B
From first law of thermodynamics.
dU = δ Q − δ W
A
δW = 0
V
dU = δ Q = nCV dT In isochoric process change in interval is equal to heat exchange. 3.5.2 Isobaric Process: When pressure of the system remain constant during the process. Work done during the process is given by B
∫ PdV = P(V
B
− VA )
A
P
A
B
= nR (TB − T A ) = nR ΔT
where ΔT is change in temperature during the process.
V
δ Q = dU + δ W = nCV ΔT + nRΔT = n ( CV + R ) ΔT Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES δ Q = nCP ΔT CV + R = CP (for the ideal gas) Heat exchange during the process for n mole of gas is equal to nC p ΔT where C p is specific heat capacity at constant process. 3.5.3 Isothermal Process: when temperature remains constant during the process then system is said isothermal process. In isothermal process change in internal energy is zero because ΔT = 0
A P
VB
dV V VA
δ W = PdV = nRT ∫
δ W = nRT ln
⇒
VB VA
B
during the isothermal process heat exchange
δ Q = δW
= nRT ln
V
VB VA
3.5.4 Adiabatic Process: When there is not any heat exchange during the process then process is said to be adiabatic process. Adiabatic process is defined by
PV γ = constant
P
For Ideal gas PV = RT for one mole of gas. So
TV γ −1 = constant V
P1−γ T γ = constant where γ =
CP CV
work done during adiabatic process δ W = PdV k PV = k dW = γ dV V γ
VB
= k ∫ V −γ dV VA
VB
⎛ V 1−γ − VA1−γ ⎞ V −γ +1 k =k⎜ B ⎟ −γ + 1 V ⎝ 1− γ ⎠ A
PBVBγ VB1−γ − PAVAγ VA1−γ PBVB − PAVA = 1− γ 1− γ Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES For n mole Ideal gas
W=
nR (TB − TA ) 1− γ
For Adiabatic process δ Q = dU + δ W Hence δ Q = 0
dU = −δ W
So in adiabatic process change in internal energy is equal to minus of work done i.e.
dU =
nR (TA − TB )
γ −1
Sign convention: Work done by the system is positive, work done on the system is negative, heat absorbed by system dQ is positive. Heat reject by the system is negative. Example: n mole of certain ideal gas at temperature T0 were cooled isochorically so that the gas pressure reduced n times. Then as a result at the isobaric process the gas expanded till its temperature get back to initial value. Find the total amount of heat absorbed by the gas in the process. Solution: Let at state A, the pressure, volume, temperature is P0 , V0 , T0 , n is number of mole. According to question:
P0V0 T0
isochoric
P0 V0 n TB
(A )
isochoric
( B) P0V0 P0V0 = T0 nTB
P0 Vc n T0
(C ) TB =
T0 n
P0 V0 P0 Vc = = Vc = nV n TB n T0
In process A to B ⎛T ⎞ dQ = nCv ΔT = nCv ⎜ 0 − T0 ⎟ ⎝n ⎠
In process B to C Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES T ⎞ T ⎞ ⎛ ⎛ dQ = nCv ΔT + PdV = nCv ⎜ T0 − 0 ⎟ + nR ⎜ T0 − 0 ⎟ n⎠ n⎠ ⎝ ⎝ Total heat absorbed by gas
dQ ( A → B ) + dQ ( B → C ) T ⎞ T ⎞ ⎛T ⎞ ⎛ ⎛ ⎛ 1⎞ = nCv ⎜ 0 − T0 ⎟ + nCv ⎜ T0 − 0 ⎟ + nR ⎜ T0 − 0 ⎟ = nRT0 ⎜ 1 − ⎟ n⎠ n⎠ ⎝ n⎠ ⎝n ⎠ ⎝ ⎝
Example: A sample of an ideal gas is taken though the cyclic process abca as shown in figure. It absorbed 50 J of heat during part ab, no heat during bc and reject 70 J of heat during ca, 40 Jule of work is done on the gas during part bc. (a) Find the internal energy of the gas at b and c if it is 1500 J at a. (b) Calculate the work done by the gas during the part ca. Solution: (a) during a → b V = constant
ΔQ = dU = 50 J U a = 1500J so at U b = 1550J
P
b
Work b → c = - 40 J U b = 1550 J
U c = 1590 (no heat at bc) c
(b) during path c → a
ΔU = 1500 − 1590 = −90 J
a V
ΔQ = −70 J ΔW = ΔQ − ΔU = −70 + 90 = 20 J Example: A sample of an Ideal gas has pressure P0, volume V0 and temperature T0. It is isothermally expand to twice its original volume. It is then compressed at constant pressure to have the original volume V0. Finally, the gas is heated at constant volume to get the original temperature. (a) Show the process in VT diagram. (b) Calculate the heat absorbed in the process. Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Solution: (a) 2V0 V V0
b
c
a
T
(b) Hence the process is cyclic then change in internal energy in cycle is zero. So heat supplied in the cycle is equal to work done. Work done during A → B is isothermal, so WAB = nRT0 ln
2V0 V
= nRT0 ln 2
b → c isobaric, so work done is PdV.
PaVa = PbVb P0V0a = Pb 2V0 Pb =
P0 2
Wb →c =
P0 PV (V0 − 2V0 ) = − 0 0 2 2
Wc → a is isochoric Wc → a = 0
Total work done is given by ⎛ P0V0 ⎞ nRT0 ln 2 + ⎜ − ⎟ 2 ⎠ ⎝
1⎞ ⎛ = nRT0 ⎜ ln 2 − ⎟ 2⎠ ⎝
Ans.
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES MCQ (Multiple Choice Questions) Q1.
Q2.
The first law of thermodynamics is a statement of (a) conservation of heat
(b) conservation of work
(c) conservation of momentum
(d) conservation of energy
If heat is supplied to an ideal gas in an isothermal process, (a) the internal energy of the gas will increase (b) the gas will do positive work (c) the gas will do negative work (d) the said process is not possible
Q3.
Figure given below shows two processes A and B on a system. Let ΔQ1 and ΔQ2 be the heat given to the system in processes A and B respectively. Then P
A
B V
(a) ΔQ1 > ΔQ2 Q4.
(b) ΔQ1 = ΔQ2
(c) ΔQ1 < ΔQ2
(d) ΔQ1 ≤ ΔQ2
Refer to figure in Let ΔU1 and ΔU 2 be the changes in internal energy of the system in the processes A and B . Then P
A
B
(a) ΔU1 > ΔU 2
(b) ΔU1 = ΔU 2
V (c) ΔU1 < ΔU 2
(d) ΔU1 ≠ ΔU 2
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fiziks Q5.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Consider the process on a system shown in figure given below. During the process, the work done by the system
P
V
Q6.
(a) continuously increases
(b) continuously decreases
(c) first increases then decreases
(d) first decreases then increases
Consider the following two statements. (A) If heat is added to a system, its temperature must increase. (B) If positive work is done by a system in a thermodynamic process, its volume must increase.
Q7.
(a) Both A and B are correct
(b) A is correct but B is wrong
(c) B is correct but A is wrong
(d) Both A and B are wrong
An ideal gas goes from the state i to the state f as shown in figure below. The work done by the gas during the process
P
f i T
(a) is positive
(b) is negative
(c) is zero
(d) cannot be obtained from this information
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fiziks Q8.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Consider two processes on a system as shown in figure below. P B
A
T
The volumes in the initial states are the same in the two processes and the volumes in the final states are also the same. Let ΔW1 and ΔW2 be the work done by the system in the processes A and B respectively. (a) ΔW1 > ΔW2
(b) ΔW1 = ΔW2
(c) ΔW1 < ΔW2
(d) Nothing can be said about the relation between ΔW1 and ΔW2 . Q9.
A gas is contained in a metallic cylinder fitted with a piston. The piston is suddenly moved in to compress the gas and is maintained at this position. As time passes the pressure of the gas in the cylinder (a) increases
(b) decreases
(c) remains constant
(d) increases or decreases depending on the nature of the gas. Q10.
7⎞ ⎛ The pressure and density of a gas ⎜ γ = ⎟ change adiabatically from (P1 , d1 ) to (P2 , d 2 ) . 5⎠ ⎝
If
d2 = 32 , then the value of d1
(a) 32
(b) 128
⎛ P2 ⎜⎜ ⎝ P1
⎞ ⎟⎟ ⎠ (c)
1 32
(d)
1 128
Q11. The pressure P volume V and temperature T for a certain material are related by P =
(AT − BT ) where 2
V
A, B are constants. The work done by the materials if the
temperature changes from T to 2T while the pressure remains constant? (a) AT − BT 2
(b) AT − 2BT 2
(c) AT − 3BT 2
(d) 2 AT − 3BT 2
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Q12. The specific heat (c ) of a substance is found to vary with temperature (T ) as c = α + β T 2 . Where T is Celsius temperature. At what temperature does the specific heat of the substance becomes equal to the mean specific heat of the substance in a temperature range between 0 and T0 ? (a) Q13.
T0 2
(b)
T0 2
(c)
T0 3
(d)
T0 3
The equation of state of a gas is given as P(V − b ) = nRT where b is a constant, n is the
number of moles and R is the universal gas constant. When 2 moles of this gas undergo reversible isothermal expansion from volume V to volume 2V , what is the work done by the gas?
Q14.
(a) 2 RT In [(V − b ) / (2V − b )]
(b) 2 RT In [(2V − b ) / (V − b )]
(c) 2 RT In [(V − b ) / (2V )]
(d) 2 RT In [(2V ) / (V − b )]
If (1) represents isothermal and (2) represents adiabatic, which of the graphs given above in respect of an ideal gas are correct?
(1) T
(2)
T
Q15.
P V
P
V I (a) I and II
(1) (2)
(2) (1)
III
II
(b) II and III
(c) I and III
(d) I, II and III
What is the minimum attainable pressure of an ideal gas in the process given by T = a + bV 2 , where a, b are constants and V is the volume of one mole of ideal gas? (a)
ab
(b) R ab
(c) 2 R ab
(d)
a/b
(R is the universal gas constant)
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fiziks Q16.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES One mole of an ideal gas is takes from an initial state (P, V , T ) to a final state
(2P, 2V , 4T ) by two different paths as shown in
the Fig. 1 and 2 given above. If the
changes in internal energy between the final and the initial states of the gas along the paths I and II are denoted by ΔU I and ΔU II respectively, then: V ⎛ ⎞ C ⎜ 2 P, , 4T ⎟ 2 ⎝ ⎠
C (2 P, 2V , 4T ) A(P, V , T )
B
C
C (2 P, 2V , 4T )
P
P ⎛P ⎞ B⎜ , 2V , T ⎟ ⎝2 ⎠
A
V Fig. 2 (Path II)
V Fig. 1 (Path I)
(a) ΔU I = ΔU II
Q17.
(b) ΔU I > ΔU II
B (P , V , T )
(c) ΔU I < ΔU II
(d) ΔU I = 0.66 ΔU II
A thermodynamic system is taken from an initial state A
50
to another state B and back to state A via state C as
40
shown by the path A → B → C → A in the P − V
P
A
30
diagram given then The work done by the system in going ⎛ N ⎞ ⎜ 2 ⎟ 20 ⎝m ⎠ from the state A to the state B is:
Q18.
(a) 35 × 3.0 J
(b) 35 × 3.5 J
(c) 50 × 3.5 J
(d) 50 × 5.0 J
B
C
10
1
3 2 V m3
( )
4
5
6
A system absorbs 1.5 × 10 3 J of energy as heat and produces 500J of work. The change in the internal energy of the system will be: (a) 1500 J
(b) 100 J
(c) - 1500 J
(d) 1000 J
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fiziks Q19.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES A given amount of heat cannot be completely converted into work. However it is possible
to convert a given amount of work completely into heat. This apparently contradictory statement results from the:
Q20.
(a) zeroth law of thermodynamics
(b) first law of thermodynamics
(c) second law of thermodynamics
(d) third law of thermodynamics
If the number of degrees of freedom of a molecule in a gas is n then the ratio of specific heats is given by:
Q21.
(a) 1 +
1 n
(b) 1 +
(c) 1 +
2 n
(d)
The ratio
2n 2n − 1
Slope of isothermal curve is equal to: Slope of adiabatic curve (b) γ
(a) 1 Q22.
1 2n
(c)
1
(d)2
γ
One mole of a perfect gas expands adiabatically. As a result of this, its pressure, temperature and volume change from P1 , T1 , V1 , to P2 , T2 and V2 respectively. if molar specific heat at constant volume is CV , then the work done by the as is:
⎛V (a) 2.303P1V1 log⎜⎜ 2 ⎝ V1 (c) Q23.
⎞ ⎟⎟ ⎠
⎛V (b) RT1 log⎜⎜ 2 ⎝ V1
P12V12 − P22V22 R(T2 − T1 )
⎞ ⎟⎟ ⎠
(d) CV (T1 − T2 )
For a diatomic gas having 3 translational and 2 rotational degrees of freedom, the energy is given by: 5 (a) k BT 2
3 (b) k BT 2
(c)
1 kBT 2
(d) k B T
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fiziks Q24.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES The maximum attainable temperature of ideal gas in each of the following processes
p = p0 e − βV is (a) Tmax = Q25.
P0 eβ R
(b) Tmax =
P0 2e β R
2P0 eβ R
(c) Tmax =
(d) Tmax =
2 P0 3eβ R
A horizontal cylinder closed from one end is rotated with a constant angular velocity ω about a vertical axis passing through the open end of the cylinder. The outside air pressure is equal to p 0 , the temperature to T , and the molar mass of air to M . Find the air pressure as a function of the distance r from the rotation axis. The molar mass is assumed to be independent of r . (a) P = P0 e
Q26.
−
Mω 2 r 2 2 RT
(b) P = P0 e
Mω 2 r 2 2 RT
(c) P = P0 e
−
Mω 2 r 2 RT
(d) P = P0 e
Mω 2 r 2 RT
A thermally insulated vessel containing a gas whose molar mass is equal to M and the ratio of specific heats C p / CV = γ moves with a velocity v . If
vessel is suddenly
stopped then increment of temperature is given by (a) ΔT = (c) ΔT = Q27.
(γ − 1) R
Mv 2
2 ( γ − 1)
R
Mv 2
(b) ΔT =
(γ − 1) Mv 2
(d) ΔT =
2R
( 2γ − 1) R
Mv 2
Gaseous hydrogen contained initially under standard conditions in a sealed vessel of volume V0 was cooled by ΔT . The amount of heat will be lost by the gas if initial pressure is P0 and temperature T0 (a) ΔQ = (c) ΔQ =
P0V0 ΔT T0 (γ − 1) P0V ( γ − 1) ΔT T0
(b) ΔQ = (d) ΔQ =
γP0V0 ΔT T0 (γ − 1) P0V0 (γ − 1)ΔT γT0
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fiziks Q28.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES A given quantity of gas is taken from the state A → C reversibly, by two paths, A → C
directly and A → B → C as shown in the figure. During the process A → C the work done by the gas is 100 J and the heat absorbed is
P
A
150 J . If during the process A → B → C , the work done by the gas is 30 J , then which one of following is correct? (a) heat absorbed in A → B → C 20 J
C
B V
(b) heat absorbed in A → B → C 80 J (c) heat reject in A → B → C 20 J (d) heat reject in A → B → C 80 J Q29.
Let ΔQ be the heat exchange in a quasistatic reversible thermodynamic process. Then which of the following is correct? (a) ΔQ is a perfect differential if the process is isothermal (b) ΔQ is a perfect differential if the process is isochoric (c) ΔQ is always a perfect differential (d) ΔQ cannot be a perfect differential
Q30.
Let ΔW be the work done in a quasistatic reversible thermodynamic process. Which of the following statements about ΔW is correct? (a) ΔW is a perfect differential if the process is isothermal (b) ΔW is a perfect differential if the process is adiabatic (c) ΔW is always a perfect differential (d) ΔW cannot be a perfect differential
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES MSQ (Multiple Select Questions)
Q31.
The pressure p and volume V of an ideal gas both increase in a process. (a) Such a process is not possible (b) The work done by the system is positive (c) The temperature of the system must increase. (d) Heat supplied to the gas is equal to the change in internal energy.
Q32.
In a process on a system, the initial pressure and volume are equal to the final pressure and volume. (a) The initial temperature must be equal to the final temperature. (b) The initial internal energy must be equal to the final internal energy. (c) The net heat given to the system in the process must be zero. (d) The net work done by the system in the process must be zero.
Q33.
Refer to figure below. Let ΔU1 and ΔU 2 be the change in internal energy in processes A and B respectively, ΔQ be the net heat given to the system in process A + B and
ΔW be the net work done by the system in the process A + B . P A B V
Q34.
(a) ΔU1 + ΔU 2 = 0
(b) ΔU1 − ΔU 2 = 0
(c) ΔQ − ΔW = 0
(d) ΔQ + ΔW = 0
The internal energy of an ideal gas decreases by the same amount as the work done by the system. (a) The process must be adiabatic.
(b) The process must be isothermal.
(c) The process must be isobaric.
(d) The temperature must decrease.
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fiziks Q35.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Which of the following is correct statement statements?
(a) The value of
(C
Cp p
+ CV )
for helium is
8 5⎞ ⎛ for ⎜ γ = ⎟ 3⎠ 3 ⎝
(b) value of specific heat capacity at constant volume for helium is (c) The value of
(C (C
p p
− CV )
+ CV )
for H 2 is
1 7⎞ ⎛ for ⎜ γ = ⎟ 5⎠ 6 ⎝
(d) value of heat capacity at constant pressure is Q36.
3R 2
7R 2
The first law of thermodynamics, ΔU = ΔQ − ΔW , indicates that when a system
goes
from its initial state to a final state which of following is correct statement (a) ΔU is the same for every path (b) ΔQ the same for every path in isochoric process (c) ΔW is the same for every path in adiabatic process (d) ΔW and ΔQ are the same for every path Q37.
Two thermally insulated vessels 1 and 2 are filled with air and connected by a short tube equipped with a valve. The volumes of the vessels, the pressures and temperatures of air in them are known (V1 , p1 , T1 and V2 , p 2 , T2 ) . The pressure established after the opening of the valve is given by (a) P =
P1V1 + P2V2 V1 + V2
(b) P =
(c) P =
T1V1 + T2V2 V1 + V2
(d) T =
P1 / V1 + P2 / V2 1 1 + V1 V1 T1T2 (P2V2 + P1V1 ) P1V1T2 + P2V2T1
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Q38. Two moles of certain ideal gas at a temperature T0 cooled isochorically so that the gas
pressure reduced n times. Then, as a result of the isobaric process, the gas expanded till the temperature got back to the initial value. (a) the work is given by RT0 (b) the work is given by 2 RT0 (c) the heat exchange is given by RT0 (d) the heat exchange is given by 2 RT0
NAT (Numerical Answer Type)
Q39.
A diatomic gas at S T P is expanded to thirty-two times its volume under adiabatic conditions. the resulting temperature in ………….. 0 K ( answer must be in two decimal points)
Q40.
The quantity of heat required to raise the temperature of one gram molecule through one degree for a mono atomic gas at constant volume is …………….. R (where R is gas constant )
Q41.
A gas expands adiabatically at constant pressure such that its temperature T ∝
1
V
. The
⎛ Cp ⎞ ⎟⎟ of the gas is ……………… value of γ ⎜⎜ = ⎝ CV ⎠ Q42.
5 ⎞ ⎛ One mole of diatomic gas ⎜ CV = R ⎟ and one mole of a monatomic gas 2 ⎠ ⎝
3 ⎞ ⎛ ⎜ CV = R ⎟ 2 ⎠ ⎝
are mixed. The value of γ for the mixture is, (where γ is the ratio of two specific heats of the gas………….. Q43.
For a ideal gas having 3 translational and 2 rotational degrees of freedom at constant temperature T , the internal energy is ……… k BT
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fiziks Q44.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES The minimum attainable pressure of ideal gas in the process T = T0 + αV 2 where T0 and
are positive constants, and V is the volume of one mole of gas is β R α T0 the value of β is given by ………. Q45.
The volume of one mole of an ideal gas with the adiabatic exponent γ is varied according to the law V = a / T , where a is a constant. Find the amount of heat obtained by the gas in this process if the gas temperature increased by ΔT .
Q46.
Ten grams of ice at 0°C is added to a beaker containing 30 grams of water at 25°C. What is the final temperature of the system is …………. 0C when it comes to thermal equilibrium? (The specific heat of water is 1 cal/gm/°C and latent heat of melting of ice is 80 cal/gm)
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Solutions MCQ (Multiple Choice Questions) Ans. 1: (d) Ans. 2: (b) Solution: dQ = dW sign dQ is positive so dW is positive the gas will do positive work . Ans. 3: (a) Solution: Change in internal energy in both the path is same and P
area under the path A is more than path B so work done in
A
the path A is more than path B .From first law of thermodynamics ΔQ1 > ΔQ2
B V
Ans. 4: (b) Solution: Internal energy is point function Ans. 5: (a) Solution: work done is area under the curve so from the figure continuously increases Ans. 6: (c) Solution: From first law of thermodynamics sign of dQ = dU + dW is not dependent only on
change on temperature rather it can be compensated by sign of internal energy and work done .but work done is positive if volume expand . Ans. 7: (c) Solution: This is isochoric process so work done is zero
(d) Nothing can be said about the relation between ΔW1 and ΔW2 . Ans. 8: (c) Solution: The process is isobaric, so work done is nRdT so change in temperature in B is more
than A so ΔW1 < ΔW2 Ans. 9: (c) Solution: After some time the system will be in mechanical equilibrium, so pressure will remain
constant.
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Ans. 10: (b) Solution: If there is no exchange of heat between a system and its surrounding the process is
known as adiabatic. The gas equation for adiabatic change is given as PV γ = constant
since, volume ∝
γ
⇒
P1 = kd1
since,
d2 = 32 d1
1 γ , P ∝ (density ) density
P2 ⎛ d 2 ⎞ =⎜ ⎟ P1 ⎜⎝ d1 ⎟⎠
γ
P2 = kd 2
( )
p2 7/5 = (32) = 2 5 p1
7/5
γ
= 128
Ans. 11: (c) Solution: Here,
P=
AT − BT 2 V
where, A, B are constants. The work done dW is given as
dW = PdV
(i)
PV = AT − BT 2
Now,
⇒
PdV = d (AT − BT 2 )
⇒
PdV = AdT − 2 BT .dT
(ii)
by equation (i) and (ii)
dW = AdT − 2 BT .dT 2T
work done =
∫ ( A − 2 BT )dT
[
= AT − BT 2
]
2T
T
[
= A(2T − T ) − B (2T ) − T 2 2
]
T
= AT − 3BT 2
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Ans. 12: (c) Solution: The amount of heat required to increase the temperature of unit mass is called specific
heat. Q = mcΔT
where m = mass of substance
ΔT = increment in temperature c = α + βT 2
and
(i) T
The mean specific heat c =
c =α +
1 0 1 cdT = ∫ T0 0 T0
T0
2 ∫ (α + β T )dT = 0
1 T0
⎡ β T03 ⎤ + α T ⎢ 0 ⎥ 3 ⎦ ⎣
BT02 3
By question c = c
⇒α +
BT 2 BT02 Τ = α + BT 2 ⇒ 0 = BT 2 ⇒ T = 0 3 3 3
Ans. 13: (b) Solution: P(V − b ) = nRT
Work done = dW = PdV W = ∫ PdV = nRT ∫
2V
V
Since,
where P =
nRT V −b
W =∫
2V
V
nRT dV V −b
dV = nRT ⎡⎣ log ( 2V − b ) − log (V − b ) ⎤⎦ V −b
n=2
⎛ 2V − b ⎞ ⎟⎟ W = 2 RT log⎜⎜ ⎝ (V − b ) ⎠
Ans. 14: (d) Solution: Slope of adiabatic curve = γ × slope of isothermal curve. This is shown by every
curve given in question. Hence, all curve represent the ideal gas.
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Ans. 15: (c) Solution: We have
⇒
T = a + bV 2
PV = a + bV 2 R
aR V 2 Rb P= + V V
⇒
,P =
dP =0 dV
To find minimum pressure put
⇒
d ⎡ aR ⎤ + VbR ⎥ = 0 ⎢ dV ⎣ V ⎦
⇒
−
aR + Rb = 0 V2
⇒V2 =
Again differential Eq. (i) twice, we get
⇒
aR + VbR V
a b
d 2P is +ve for V 2 , 2 dV
P is minimum at V 2 .
By Eqs. (i) and (ii), we get
⇒
Pmin =
aR a/b
+ bR a / b
= 2 R ab
Ans. 16: (a) Solution: The change in internal energy dU is independent of the path, i.e., if initial and final
states of change are same then dU will be same. Ans. 17: (a) Solution: The work done is given as
50
A
= PdV
= area of the triangle + area of the rectangle ⎡1 ⎤ = ⎢ × (50 − 20 ) × (5 − 2 )⎥ + (20 × 3) ⎣2 ⎦ =
1 × 30 × 3 + 20 × 3 = 105 J 2
p 20
B
C
2
V
5
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Ans. 18: (d) Solution: The first law of thermodynamics is given as
δ Q = dU + δ W Here, δQ = 1.5 × 10 3 = 1500 J
δW = 500 J so,
dU = δ Q − δ W = 1000 J
Ans. 19: (b) Ans. 20: (c) Solution: We know that energy associated with each degree of freedom of one mole gas is given =
1 RT 2
Hence if there is n degree of freedom, then U =
n RT where R → universal gas 2
constant. We know that specific heat at constant volume is given as CV =
dU n ⇒ CV = R dT 2
By Mayor’s formula C P − CV = R = R +
γ =
(i) n ⎛ n⎞ R ⇒ C P = ⎜1 + ⎟ R 2 ⎝ 2⎠
CP 2 ⇒ γ = 1+ Cv n
Ans. 21: (c) Solution:
P
P
V isothermal curve
V adiabatic curve
The ideal gas equation is written as PV = RT For isothermal T is constant differentiate above equation w.r.t. V we get Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES d (PV ) = 0 ⇒ dP = −⎛⎜ P ⎞⎟ ⇒ slope of isothermal curve = − P dV V dV ⎝V ⎠
In case adiabatic expansion gas equation is PV γ = constant . differentiating this w.r.t. V we get V γ dP + γPV γ −1 dV = 0 dP ⎛P⎞ ⎛P⎞ = −γ ⎜ ⎟ ⇒ slope of adiabatic curve = −γ ⎜ ⎟ dV ⎝V ⎠ ⎝V ⎠
dividing equation (ii), we get
(ii)
Slope of isothermal curve 1 = Slope of adiabatic curve γ
Ans. 22: (d) Solution: Specific heat: The rate of change of internal energy w.r.t. temperature at constant
volume is called specific heat at constant volume.
⇒
⎛ dU ⎞ CV = ⎜ ⎟ ⎝ dT ⎠V
by first law of thermodynamics
(i)
ΔQ = dU + ΔW
(ii)
For adiabatic ΔQ = 0 so above equation (ii) becomes
dU + ΔW = 0 ⇒
CV (dT ) + ΔW = 0 ⇒
CV (T2 − T1 ) + ΔW = 0
⇒ ΔW = CV (T1 − T2 )
Ans. 23: (a) Solution: The energy associated with each degree of freedom at temperature T is given as E=
1 k BT , k B is Boltzmann’s constant 2
Here, total number of degrees of freedom= number of degree of rotational freedom + 1 translational freedom 3 + 2 = 5 thus, total energy = 5 × k BT 2
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Ans. 24: (a) Solution: P = P0 e − βV ⇒
⇒T =
RT = P0 e − βV V
P0 − βV Ve R
….(i)
For maximum temperature, This gives V =
1
β
(
)
P dT = 0 ⇒ 0 e − βV − β Ve − βV = 0 dV R
. Putting this value of V in equation (i), gives T =
P0 eβ R
Ans. 25: (b) Solution: Force equation of dr element. dF ⎛ dm ⎞ 2 ⎛ A =⎜ ⎟rω and dm = ⎜ 2 A ⎝ A ⎠ ⎝ rω where A is area of cross section
dF = ( dm ) rω 2 , dP =
⎞ ⎟ dP ⎠
Also we know
ω
⎛ dm ⎞ P ( Adr ) = ⎜ ⎟ RT ⎝M ⎠ PA(dr ) =
M, T
RT ⎛ A ⎜ M ⎝ rω 2
⎞ ⎟ dP ⎠
r
P
0
P0
Mω 2 ∫ rdr = RT ∫
S
P0
r This end is open in air
dP P 2 2
Mω r Mω 2 r 2 P ⇒ P = P0 e 2 RT = RT ln P0 2
Ans. 26: (b) Solution: Suppose number of moles of gas = n
Directional kinetic energy of gas =
1 (nM )v 2 2
When vessel sudden stop, then after long time this directional kinetic energy of gas is converted into random kinetic energy when thermodynamic equilibrium will be achieved and then
1 (nM )v 2 = nCV ΔT 2
⇒ ΔT =
(γ − 1) Mv 2 2R
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Ans. 27: (a) Solution: Suppose initial temperature is T1 and final is (T1 − ΔT ) then
Ui =
P1V ; γ −1
ΔU =
Uf =
P2V γ −1
V0 (P2 − P1 ) γ −1
PV = nRT1 ⇒ P1 = 1
nRT1 PV also nR = 1 0 V T1
P2 =
nRT2 V0
ΔU =
V0 ⎛ nRT2 nRT1 ⎞ nR (ΔT ) − ⎜ ⎟= V ⎠ γ −1 γ −1⎝ V
ΔU =
P1V0 ΔT T1 (γ − 1)
ΔU =
P0V0 ΔT T0 (γ − 1)
here
P1 = P0 T1 = To
ΔU = increase in potential energy ΔQ = ΔU + ΔW
ΔQ = ΔU =
since vessel is sealed then ΔW = 0
PV 0 0 ΔT T0 ( γ − 1)
Ans. 28: (b) Solution: During path AC dU = δ Q − δ W = 150 − 100 = 50 J
Hence internal energy is point function dU will same in all path In path ABC , δ Q = dU + δ W = 50 + 30 = 80 J . Ans. 29: (b) Solution: Heat exchange is perfect differential in isochoric process. ΔQ = dU
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Ans. 30: (b) Solution: ΔW is a perfect differential if the process is adiabatic
ΔW = −d U
MSQ (Multiple Select Questions) Ans. 31: (b) and (c) Solution: If volume increase then workdone is positive and from PV = nRT if volume and
pressure increase then temperature will increase. Ans. 32: (a) and (b) Solution: from PV = nRT if pressure and volume of initial point and final point is same then
temperature is same ,so internal energy will also same Ans. 33: (a) and (c) Solution: ΔU1 = ΔU 2 = 0 because internal energy is path independent so ΔQ − ΔW = 0 Ans. 34: (a) and (d) Solution: δ Q = dU + δ W and dU = nCV dT , dU is negative so temperature decreases.
If − dU = δ W so δ Q = 0 Ans. 35: (b), (c) and (d) Solution: The ratio of C p and CV is known as coefficient of adiabatic expansion. Hence,
γ =
Cp CV
(a) so, for helium γ =
(b) CV =
5 = 3
1 1+
,
Cp
5 C p + CV 3
=
3 8
R 3R 5⎞ ⎛ = for ⎜ γ = ⎟ γ −1 2 3⎠ ⎝
(c) for hydrogen γ =
7 C p − CV = 5 C p + Cv
dividing by CV the numerator and denominator, we get Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 7 −1 C p − CV C p / CV − 1 C p − CV 1 γ −1 5 = = = ⇒ = 7 γ +1 C p + CV C p / CV + 1 C p + CV 6 +1 5 γ R 7R (d) CP = = γ −1 2 Ans. 36: (a), (b) and (c) Solution: The internal energy ΔU is a state function. Hence, it does not depend on path whereas
both Q and W depend on path. Thus, ΔQ is same for every path. Ans. 37: (a) and (d) Solution:
vessel 1 V1 , P1 , T1
2 ?
V2 + V2 P T
V2 , P2 , T2
When value is opened and thermodynamics equilibrium is attained then, number of moles will be constant. Then
P1V1 P2V2 P(V1 + V2 ) + = RT1 RT2 RT P1V1 P2V2 P(V1 + V2 ) + = T1 T2 T
(i)
Also we know, in whole system: ΔQ = 0 = ΔU + ΔW ΔQ = 0 because vessel is insulated. And also
ΔW = 0 because gas does not work on atmosphere because vessel closed then
ΔU system = 0 then n1CV ΔT1 = n2 CV ΔT2
⇒
n1 ΔT1 = n2 ΔT2
P1V1 (T − T1 ) + P2V2 (T − T2 ) = 0 T1 T2 T=
T1T2 (P2V2 + P1V1 ) P1V1T2 + P2V2T1
put in (i) P =
(ii)
P1V1 + P2V2 V1 + V2
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Ans. 38: (a) and (c) Solution:
n=2 T0 P0
Initial (P0 ; T0 ;V ) Here
V = constant P Pf = 0 2
⎛P ⎞ isobaric⎛ P ⎞ → ⎜ 0 ; T1 ;V ⎟ → ⎜ 0 ; T0 ;V ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠
isochoric
PV = nRT P0V = 2 RT0 V=
2 RT0 P0
Isochoric process:
P1 T1′ = P2 T2′ P0 T = 0 P0 / 2 T1
T0 2
⇒
T1 =
⇒
V1 = 2V
Isobaric process:
V1′ T1′ = V2′ T2′ V T0 / 2 = V2 T0
In whole system: from initial to final position: ΔQ = ΔU + ΔW
here
ΔU = 0 final temperature is zero. ΔW = 0 +
P0 [ΔV ] = P0V = P0 2 RT0 2 2 2 P0
ΔW = RT0 and ΔQ = RT0
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES NAT (Numerical Answer Type) Ans. 39:
68.25 K
Solution: For diatomic gas γ =
5 2
∴ Adiabatic equation TV γ −1 = constant
⇒
⎛V T2 = T1 ⎜⎜ 1 ⎝ V2
⎞ ⎟⎟ ⎠
γ −1
here, V1 = V , V2 = 32V , T1 = 273K T2 =
273
⇒
(32 )
γ −1
T2 =
273
(32)
5 −1 2
=
273
(32)
3 2
= 68.25 K
Ans. 40: (a) Solution: The quantity of heat required to increase the temperature at constant volume through
0°C for per degree of freedom
So, Ans. 41:
( Q )V
= nCV dT
CV =
1 R 2
(Q )V
⎛1 ⎞ = 1 × ⎜ R ⎟ × 1 = 0 .5 R ⎝2 ⎠
n = 1, dT = 1
1.5
Solution: We know that for adiabatic expansion TV γ −1 = constant
Given that T ∝
1 V
⇒ TV 1/ 2 = constant
by equation (i) and (ii),
γ −1 =
3 1 1 ⇒ γ = 1 + , γ = = 1.50 2 2 2
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fiziks Ans. 42:
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 1.5
Solution: cV =
n1cV1 + n2 cV2 n1 + n2
n1 = 1, cV1 =
Ans. 43:
and cP =
5R 7R , cP1 = 2 2
n1cP1 + n2 cP2
γ=
n1 + n2
n1 = 1, cV1 =
cp cV
3R 5R , cP1 = 2 2
2.5
Solution: The energy associated with each degree of freedom at temperature T is given as E=
1 k B T is k B Boltzmann’s constant 2
Here, total number of degrees of freedom= number of degree of rotational freedom + 1 translational freedom 3 + 2 = 5thus, total energy = 5 × k B T 2
Ans. 44:
2 T = T0 + αV 2
Solution:
We know PV = nRT = RT If V is increasing, T will increase and hence P will increase. Hence calculation of Pmin : T=
PV put in (i) R
[
]
PV = T0 + αV 2 , P = R T0V −1 + αV R T dP = 0 , R ⎡⎣ −T0V −2 + α ⎤⎦ = 0 ⇒ V = 0 α dV
Put the value of V =
T0
α
[
in P = R T0V −1 + αV
put in (ii)
]
Pmin = 2 R αT0
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fiziks Ans. 45:
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 2
Solution: ⇒
we know
V =
a T
T=
a V
number of mole n = 1
δ Q = dU + δW ΔU = ∫ dU = CV ΔT =
R ΔT γ −1
(i)
δ W = PdV
ΔW = ∫ PdV
(ii)
⇒
PV = nRT = RT RT aR put in (ii) = 2 V V V V2 aR ⎡1 1 2 1⎤ dV = − aR ΔW = ∫ = aR ⎢ − ⎥ 2 V1 V V V1 ⎣V1 V2 ⎦ P=
⎡a a ⎤ ΔW = ⎢ − ⎥ R = − R[ΔT ] ⎣V1 V2 ⎦
ΔQ = ΔU + ΔW =
⎛1− γ +1⎞ RΔT ⎟⎟ − RΔT = RΔT ⎜⎜ γ −1 ⎝ γ −1 ⎠
ΔQ =
Ans. 46:
R ΔT ( 2 − γ )
γ −1
0
Solution: The amount of heat required to melt the ice of mass 10gm at 00C is Q = m × L = 10 × 80 = 800Cal . Where L is the latent heat of melting of ice and m is the
mass of the ice. The amount of heat available in water of mass 30gm at 250C is
Q = m × Cv × T = 30 ×1× 25 = 750Cal Since the heat available is less than the heat required to melt the ice therefore ice will not melt as a result the temperature of the system will be at 00C only Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES
Chapter – 4 Second Law of Thermodynamics and Entropy 4.1
Second Law of Thermodynamics Statement: It is impossible to transfer total heat into work in a cyclic process in the absence of other effect (by Lord Kelvin). In another way it is also stated that it is impossible for heat to be transferred by a cyclic process from a body to one warmer than itself without producing other changes at same time.
4.2
Heat Engines A machine that can convert heat into work is said to be heat engine. It is a system that performs the conservation of heat or thermal energy to mechanical work. Hot reservoir T1
Q1 W = Q1 – Q2
Q2 Cold reservoir T2 Schematic of operation at heat engine 4.2.1 Heat Reservoir It is an effectively infinite pool of thermal energy of a given, constant temperature. Ideal its heat capacity is large enough that when it is in thermal contact with another system, its temperature remains constant. All heat engines have mainly three essential components. 1. A source: This is a hot region which is a part of the surrounding from which energy flows by heat transfer. Popularly it is known as hot reservoir. Example is a nuclear reactor, furnace. Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 2. The working agent: This under goes change at state as part of a continuous cycle e.g. steam-water. 3. A sink: This is a respectively cold region which is part or the surrounding into which heat is rejected by heat transfer e.g. cooler. 4.2.2 Efficiency of Heat Engine (η):
η=
work done by the system heat given into the system
Hence the process is cyclic dU = 0.
dQ = dW Heat given to system is Q1 and heat rejected by system is Q2.
η=
Q1 − Q2 Q1
4.2.3 Carnot Cycle: It is theoretical thermodynamics cycle proposed by Nicolas, Leonard Sadi Carnot. It can be shown that it is most efficient cycle for converting a given amount of thermal energy into work. A system undergoing a Carnot cycle is called a Carnot heat engine.
(P1 ,V1 )
Stages of the Carnot cycle:
1
P
(P2 ,V2 ) 2
(P4 ,V4 )
T1
4
(P3 ,V3 )
3
T2
V
1. Reversible isothermal expansion of the gas at hot temperature, T1 (isothermal heat addition). During this step (1 to 2) the gas is allowed to expand and it does work on the surrounding. The gas expansion is propelled by absorption of quantity Q1 at heat from the high temperature T1. Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 2. Reversible adiabatic expansion of the gas. For this step (2 to 3) on figure, the gas continues to expand, working on surrounding. The gas expansion causes it to cool to the cold temperature T2. 3. Reversible isothermal compression of the gas at the “cold” temperature T2 (isothermal heat rejection). The process is shown by 3 → 4. Now the surroundings do work on the gas, causing quantity Q2 of heat to flow out of the gas to the low temperature reservoir. 4. Adiabatic compression of the gas (4 to 1). During this step, the surrounding do work on the gas, compressing it and cause the temperature to rise to T1. At this point the gas in the same state as at start at step 1. Efficiency of Carnot engine:
η=
W Q1
W is the work done during cycle and Q1 is heat given to system.
W = W1 + W2 + W3 + W4 W1 = work done during process 1 to 2 for isothermal process = nRT1 ln
V2 V1
change in internal energy during the process dU = 0 ⎛V ⎞ Q1 = W1 = nRT1 ln ⎜ 2 ⎟ (From first law of thermodynamics) ⎝ V1 ⎠ W2 is work done during process 2 to 3 in adiabatic process.
W2 =
dQ = 0
nR (T2 − T1 ) 1− γ
W3 = work done during process 3 to 4 isothermal compression.
⎛V ⎞ W3 = nRT2 ln ⎜ 4 ⎟ ⎝ V3 ⎠ Q2 = W3
dU = 0
(isothermal process)
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES W4 , the work done during the adiabatic process 4 to1 , is given by
W4 =
nR (T1 − T2 )
1− γ
W2 = −W4
⎛V ⎞ ⎛V ⎞ W = nRT1 ln ⎜ 2 ⎟ + nRT2 ln ⎜ 4 ⎟ ⎝ V1 ⎠ ⎝ V3 ⎠ ⎛V ⎞ Q1 = nRT1 ln⎜⎜ 2 ⎟⎟ ⎝ V1 ⎠ ⎛V ⎞ ln ⎜ 4 ⎟ V W T η = = 1+ 2 ⎝ 3 ⎠ Q1 T1 ⎛ V2 ⎞ ln ⎜ ⎟ ⎝ V1 ⎠
….(i)
(P1 ,V1 )
From the figure, γ −1 TV = T2V3γ −1 1 2
….(ii)
γ −1 T2V4γ −1 = TV 1 1
….(iii)
Dividing equation (ii) by (iii) gives
⎛ V2 ⎞ ⎜ ⎟ ⎝ V1 ⎠ ⇒
γ −1
⎛V ⎞ =⎜ 3 ⎟ ⎝ V4 ⎠
1
P
(P2 ,V2 ) 2
(P4 ,V4 )
T1
4
(P3 ,V3 )
γ −1
3
T2
V
V2 V3 = V1 V4
….(iv)
From equations (i) and (iv)
η = 1−
T2 T1
….(v)
For Carnot cycle,
Q2 T2 = Q1 T1
This gives η = 1 −
Q2 Q1
….(vi)
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Example: An ideal gas engine operator in a cycle which when represent on a P-V diagram is a
rectangle. If we call P1 , P2 and the lower and higher pressures respectively and V1 , V2 as lower and higher volume respectively. (a) Calculate the work done in complete cycle (b) indicate in which parts of the cycle heat is absorbed and in which part librated. (c) Calculate the quantity of heat following into the gas in one cycle (d) show that efficiency of the engine is
η=
γ P2
γ −1 +
P2 − P1
V1 V2 − V1
Solution: work done drawing AB = P2 (V2 − V1 )
Heat absorbed = nC p ΔT
= nR =
γ γ −1
γ γ −1
(T2 − T1 ) =
P
( P2 ,V1 ) A
γ γ −1
( P2 ,V2 ) B
( PV 2 2 − PV 2 1)
P2 (V2 − V1 )
D
( P1 ,V1 )
( P1 ,V2 )
C
V
V2 > V1 so heat absorbed in the process.
In process B-C isochoric process W B −c = 0
dQ = dU nCv ΔT
n V RΔT = 2 ( P1 − P2 ) γ −1 γ −1 P1 < P2 heat rejected
In process C-D isobaric process WC − D = P1 (V1 − V2 )
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Heat exchange during the process
γ
QC → D =
γ −1
P1 (V1 − V2 )
V1 < V2 so heat is rejected
In process D → A isochoric process WD→ A = 0
Heat exchange during the process nCv ΔT =
nRΔT V1 ( P2 − P1 ) heat absorbed. = γ −1 γ −1
(a) Work done = P2 (V2-V1) + P1(V1 – V2) W = (P2 – P1) (V2 –V1) (b) Heat absorbed
γ γ −1
P2 (V2 − V1 ) +
V1 (P − P ) γ −1 2 1
(c) Heat flowing into the cycle
γ P2 (V2 − V1 ) V2 γ γ + P1 (V1 − V2 ) + V1 ( P2 − P1 ) ( P1 − P2 ) + γ −1 γ −1 γ −1 γ −1 = =
γ γ −1 γ γ −1
(V2 − V1 )( P2 − P1 ) +
1 ( P − P ) ⋅ (V2 − V1 ) γ −1 1 2
(V2 − V1 )( P2 − P1 ) −
1 ( P − P )(V − V ) γ −1 2 1 2 1
⎡ γ 1 ⎤ − ⎥ = (P2 − P1 )(V2 − V1 ) ⎣ γ − 1 γ − 1⎦
( P2 − P1 )(V2 − V1 ) ⎢ dQ = dW (d) Efficiency =
Therefore, η =
W Qabsorbed
=
( P2 − P1 )(V2 − V1 ) γ P2 (V2 − V1 ) V1 ( P2 − P1 ) + γ −1
γ −1
γ −1 γ
P2 V1 + P2 − P1 V2 − V1
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 4.3 Entropy
Entropy is an extensive thermodynamic property that is measure of a system’s thermal energy per unit temperature that is unavailable for doing useful work. Thermodynamic entropy is a non-conserved state function. For Isolated systems entropy never decreases. In statistical mechanics, entropy is a measure of the number of ways in which a system may be arranged, often taken to be a measure of ‘disorder’ (the higher the entropy the higher the disorder). The infinitesimal change in the entropy ( dS ) of a system is the infinitesimal transfer of heat energy (δQ) to a closed system driving a reversible process, divides by temperature (T) of the system.
ΔS = ∫
δQ T
It has unit Joule/Kelvin or dS = ∫
δQ T
Law of thermodynamics and entropy: According to first law of thermodynamics
δ Q = dU + PdV From definition of Entropy
ΔS =
δQ T
⇒ δ Q = T ΔS ⇒ TdS
TdS = dU + pdV
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 4.3.1 Inequality of Clausius
Consider an irreversible cyclic engine working between T1 and T2. If reversible engine is operating between same temperature then from Carnot theorem. Efficiency of irreversible (ηir) will always smaller than efficiency of reversible engine(η r ).
η irr . < η rev. Q1irr − Q2irr Q1rev − Q2rev < Q1irr Q1rev
1−
Q2irr Q2rev < 1 − Q1irr Q1rev
1−
Q2irr T < 1− 2 irr T1 Q1
Q2irr Q1irr > T2 T1 F0 irreversible cyclic Engine
Q1irr Q2irr − G.M .
So
T1 + T2 > T1T2 So 2
ΔS > 0
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Example: Compute the change in entropy when ice melt into steam. It is given that L1 is latent
heat of fusion, c is specific heat at water and L2 latent heat at vaporization. Solution: Assume T1 be the Kelvin temperature at which ice melts into water and T2 the Kelvin
temperature at which water is boiled to steam. ΔS1 is entropy change when ice is converted into water
ΔS1 =
mL1 T1
ΔS 2 Entropy change when water is heated from T1 to T2 T2
dT T T1
ΔS 2 = mc ∫
ΔS 2 = mc ln
T2 T1
ΔS 3 Entropy change when water change into vapors
ΔS 3 =
mL2 T2
Total change in entropy ΔS = ΔS1 + ΔS 2 + ΔS3 =
mL1 T mL + mc ln 2 + 2 T1 T1 T2
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES MCQ (Multiple Choice Questions)
Q1.
Two Carnot engines A and B are operating between the same source and the same sink. Engine A uses an ideal gas as the working fluid while engine B uses Van der Waals’ gas as the working fluid. Which one of the following is correct? (a) The efficiency of engine A is less than that of engine B (b) The efficiency of engine A is equal to that of engine B (c) The efficiency of engine A is more than that of engine B (d) No comparison can be made
Q2.
A heat engine converts a given quantity of heat into work with maximum efficiency during which one of the following processes?
Q3.
(a) Isobaric process
(b) Isochoric process
(c) Isoenthalpic process
(d) Isothermal process
If heat Q is added reversibly to a system at temperature T and heat Q ′ is taken away from it reversibly at temperature T ′, then which one of the following is correct?
Q4.
(a)
Q Q′ − =0 T T′
(b)
Q Q′ >0 − T T′
(c)
Q Q′ 0 (c) During the process work done is zero but change in entropy of gas as well as universe ΔS > 0 (d) During the process work done is not zero but change in entropy of gas ΔS > 0 as well as universe ΔS > 0 Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Q11.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Each of the two isolated vessels, A and B of fixed volumes, contains N molecules of a
perfect monatomic gas at a pressure P . The temperatures of A and B are T1 and T2 , respectively. The two vessels are brought into thermal contact. At equilibrium, the change in entropy is (a)
⎡ Τ 2 + Τ22 ⎤ 3 Νk B ln ⎢ 1 ⎥ 2 ⎣ 4Τ1Τ2 ⎦
(b)
⎡ (Τ1 + Τ2 )2 ⎤ 3 (c) Νk B ln ⎢ ⎥ 2 ⎣⎢ 4Τ1 Τ2 ⎦⎥
⎛Τ 3 Νk B ln⎜⎜ 2 2 ⎝ Τ1
⎞ ⎟⎟ ⎠
(d) 2NkB
MSQ (Multiple Select Questions)
Q12.
In diagram 1 Carnot cycle is represented in PV diagram while in Diagram II 1 Carnot cycle is represented in TS diagram
Diagram I (P − V )
Diagram II
(T − S )
T1 1 P
A
4
2
T
D
T2 3 V
B C
S
Which one of the following is correct? (a) 1 and A is isothermal expansion and heat is given into the system (b) B is adiabatic compression (c) In process 3 heat is rejected by system (d) Work done during 4 and B are same in magnitude and opposite to sign
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fiziks Q13.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Consider the following statements regarding the characteristics of entropy are correct
(a) Entropy is a measure of disorder. (b) Entropy changes during a reversible adiabatic process. (c) Entropy of a system decreases in all irreversible processes. (d) Change in entropy for a complete reversible thermodynamic cycle is zero Q14.
Consider the following statements regarding transition of a system from one thermodynamic state another which of the following is correct (a) The heat absorbed by it along any reversible path independent of the path. (b) The change of entropy of the system in a reversible : process is independent of the path. (c) The change of entropy of the system in a irreversible process is also independent of the path. (d) The heat absorbed by it along any irrereversible path independent of the path.
Q15.
The temperature entropy diagrams of two engines A and B working between the same temperature T1 and T2 of the source and the sink respectively are shown in the given
T1
B A
Temperature
Temperature
figures.
T2 S1
S2
Entropy The efficiency of A :
T1
T2 S1
S2
Entropy
(a) is less than that of B (b) is equal to that of B (c) is greater than that of B (d) and B cannot be compared on the basis of data given in the diagrams
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fiziks Q16.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Which of the the following statements are correct
(a) The entropy change during a reversible adiabatic process is zero (b) Entropy is a state function (c) The entropy of a thermally isolated system never decreases (d) The entropy change during an reversible adiabatic process is zero. Q17.
An ideal gas is expanded adiabatically from ( P1 , V1 ) to ( p2 , V2 ) . Then it is compressed isobarically to ( P2 , V1 ) . Finally the pressure is increased to P1 at constant volume V1 then P P1
which of the following is correct?
A adiabatic
(a) the P − V indicator diagram is given by fig (b) the work done in the cycle is W =
P2 1 ( PV 2 2 − PV 1 1 ) + P2 (V2 − V1 ) 1− γ
C
V1
(c) heat will absorb in isochoric process
B V2 V
V2 −1 V1 (d) the efficiency of the cycle is η = 1 − γ p1 −1 p2 Q18.
Consider an arbitrary heat engine which operates between reservoirs, each of which has the same finite temperature-independent heat capacity c . The reservoirs have initial temperatures T1 and T2 , where T2 > T1 , and the engine operates until both reservoirs have the same final temperature T3 . Then which of the following statements are correct (a) The change of entropy is given by ln
T32 . T1T2
(b) In general T3 ≤ T1T2 (c) In general T3 ≥ T1T2 (d) The maximum work done is given by Wmax = c
(
T1 − T2
)
2
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fiziks Q19.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Which of the statements are correct?
(a) A mole of an ideal gas undergoes a reversible isothermal expansion from volume V1 to 2V1 .Then change in entropy of the gas is RT ln 2 and there is no change in entropy of the universe . (b) A mole of an ideal gas undergoes a reversible isothermal expansion from volume V1 to 2V1 .Then change in entropy of the gas is RT ln 2 and change in entropy of the universe is RT ln 2 (c) A mole of an ideal gas undergoes free isothermal expansion from volume V1 to 2V1 .Then change in entropy of the gas is RT ln 2 and there is no change in entropy
of the universe (d) A mole of an ideal gas undergoes free isothermal expansion from volume V1 to 2V1 .Then change in entropy of the gas is RT ln 2 and
change in entropy of the
universe is RT ln 2 Q20.
A body of constant heat capacity CP and a temperature Ti is put into contact with a reservoir at temperature T f . Equilibrium between the body and the reservoir is established at established at constant pressure. Assume T f > Ti . Then (a) Change of entropy of the body is C p ln
Tf Ti
(b) The change of entropy of the heat source is
C p (Ti − T f
)
Tf
(c) If T f > Ti then the change in entropy of universe ΔS > 0 (d) If T f < Ti then change of entropy of universe is ΔS < 0
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fiziks Q21.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES n mole of an ideal gas is originally confined to a volume V1 in an insulated container of
volume V1 + V2 . The remainder of the container is evacuated. The partition is then removed and the gas expands to fill the entire container. If the initial temperature of the Insulated container gas was T (a) Then which of the following statement is correct (b) The temperature remain constant T
V1
⎛ V +V ⎞ (c) The work done during the process is nRT ln ⎜ 1 2 ⎟ ⎝ V1 ⎠
V2
⎛ V +V ⎞ (d) The change entropy of gas in the process is nR ln ⎜ 1 2 ⎟ ⎝ V1 ⎠ ⎛ V +V ⎞ The change in entropy of universe during the process is nR ln ⎜ 1 2 ⎟ ⎝ V1 ⎠
NAT (Numerical Answer Type Questions)
Q22.
A Carnot engine has an efficiency of efficiency becomes
Q23.
1 . On reducing the sink temperature by 65 °C , the 6
1 . The source temperature is given by …………. 0 K 3
A Carnot engine whose low-temperature reservoir is at 27 0 C has an efficiency 37.5%.The high-temperature reservoir is …………. 0C
Q24.
In the given T − S diagram, the efficiency is given by …….. % (Answer must be in two decimal point).
200 100
Temperature (K) B
A
500
C
1000
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fiziks Q25.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES An ideal gas is confined to a cylinder by a piston. The piston is slowly pushed in so that
the gas temperature remains at 20 0 C . During the compression, 730 J work is done on the gas. the entropy change by the gas is …….. J / K Q26.
10 g of ice at 0°C is slowly melted to water at 0°C. The latent heat of melting is 80 cal/g. The change in entropy is nearly ………. cal / K
Q27.
If a capacitor of 1μF charged to a potential of 300V is charged, a resistor kept at room temperature, then the entropy change of the universe in……..
Q28.
×10 −4 J / K
One kg of H 2O at 0o C is brought in contact with a heat reservoir at 100o C . When the water has reached 100 o C Then the change in entropy of the universe is ………… J / K (specific heat of water CH 2O = 4.18 J / g )
Q29.
A reversible engine cycle is shown in the following T -S diagram. The efficiency of the engine is ……………….. %
2T1
T
T1
S1
2S1 3S 1 S
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Solution MCQ (Multiple Choice Questions) Ans. 1: (b) Solution: The efficiency of a Carnot working between temperature limits T1 and T2 is given as
η = 1−
T2 T1
T1 → absolute temperature of source T2 → absolute temperature of sink Since, efficiency does not depend on -working substance hence the efficiency of both engines A and B are same Ans. 2: (c) Solution: The first law of thermodynamics dQ = dU + PdV
The maximum efficiency can be obtained, if process is isoenthalpic. Ans. 3: (a) Solution: For a reversible process the change in entropy is zero.
⇒
dS = 0
⇒
Q1 Q′ − =0 T1 T ′
Ans. 4: (d) Solution: The change in entropy of a system is given as dS = ∫
T2
T1
δQ T
If δQ amount of heat is given to water then T1 change in temperature dT is given as
δQ = mcdT dS = ∫
T2
T1
T mcdT = mc log e 2 T1 T
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Ans. 5: (c) Solution:
1
A
AB → isothermal expansion BC → adiabatic expansion
B 2
P 4
CD → isothermal compression
D
DA → adiabatic compression
C
3
V
Ans. 6: (c) Solution: The change in entropy is given as dS =
δQ T
here, δ Q = mL, L is latent heat
= 1000× 80 ⇒ δ Q = 80000 cal δQ = 80000 × 4.2 J
⇒
(ii)
T = 273 K
80000 × 4.2 J / K = 12.3 × 10 2 J / K 273
so, dS = Ans. 7: (d)
Solution: Efficiency η =
W Q1
T
In a) 1 W = (3T0 − T0 )(2 S0 − S0 ) = T0 S0 2 Q1 =
η=
1 (3T0 )(2 S 0 − S0 ) = 1.5T0 S 0 2
W = 66% Q1
3T0
B
T0
A
C
S0
2S 0
(a )
S
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES T In (b) W=
1 (2T0 − T0 )(3S0 − S0 ) = T0 S0 2
Q1 =
1 (2T0 )(3S0 − S0 ) = 2T0 S0 2
B
2T0
T0
W η = = 50% Q1
C
S0
3S 0
( b)
S
T
In (c) W=
A
1 (3T0 − T0 )(2 S0 − S0 ) = T0 S0 2
3T0
B
C
Q1 = (3T0 )(2 S0 − S0 ) = 3T0 S0
η=
T0
W = 33% Q1
S0
S
2S 0 (c)
T
In (d) W=
A
1 (2T0 − T0 )(3S0 − S0 ) = T0 S0 2
2T0
B
C
Q1 = (2T0 )(3S0 − S0 ) = 4T0 S0
T0
W η = = 25% Q1
A S0
(d )
3S 0
S
Ans. 8: (d) Solution: From dS =
We obtain
1 1 ( dU + PdV ) = ( Cv dT + PdV ) and PV = RT T T
ΔS = Cv ln
T2 V + R ln 2 T1 V1
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Ans. 9: (a) Solution: In the cycle, the energy the working substance absorbs from the source of higher
temperature is Qab = C p (Tb − Ta ) .The energy it gives to the source of lower temperature is Qreject = C p (Tc − Td ) . Thus
η = 1−
Qreject Qab
= 1−
P
adiabatics
P1
Tc − Td Tb − Ta
a
P2
b d
V
form the equation of state pV = nRT and the adiabatic equations γ γ γ γ PV PV 2 d = PV 1 a , 2 c = PV 1 b
we have
⎛ P2 ⎞ ⎟ ⎝ P1 ⎠
η = 1− ⎜
c
γ −1 γ
Ans. 10: (c) Solution: After a hole has been opened, the gas flows continuously to the right side and reaches
equilibrium finally. During the process, internal energy of the system E is unchanged. Since E depends on the temperature T only for an ideal gas, the equilibrium temperature is still T0 so from first law of thermodynamics work done is zero but process is irreversible so change in entropy of gas as well as universe is ΔS > 0 Ans. 11: (c) Solution: Final temperature of each vessel at equilibrium is T =
T1 + T2 2
T ⎡ (Τ1 + Τ2 )2 ⎤ CV dT CV dT 3 3K B ΔS = ∫ +∫ = Νk B ln ⎢ for monatomic gas ⎥ where CV = T T 2 2 T1 T2 ⎣⎢ 4Τ1Τ2 ⎦⎥ T
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES MSQ (Multiple Select Questions) Ans. 12: (a), (c) and (d) Solution:
T1 1
P
A
4
2 T2
T
D
3
B C
V
S
Path 1: Expansion of constant temperature and heat is absorbed because dQ = + ve hence
S will increase Corresponding to A. Path 2: Adiabatic expansion means S is constant Corresponding to B. Path 3: Isothermal compression dQ = − ve so heat will reject S is decrease corresponding
to C. Path A and path B are isoentropic process so work done is dependent on points but direction of both the case will opposite . Ans. 13: (a) and (d) Solution: Entropy is function of no of microstate which will measurement of disorderness
For thermodynamic process dS ≥ 0 and for reversible process dS = 0 . Ans. 14: (b) and (c) Solution: The entropy is point function and perfectly differential so it is path independent Ans. 15: (a) Solution: The efficiency is defined as
η=
H 1 − H 2 T1 − T2 = H1 T1
Ans. 16: (a), (b), (c) and (d) Solution: Entropy:
where δQ → amount of used heat in reversible adiabatic process δ Q = 0 It is a state function and is defined as dS =
δQ T
is exact differential so it is state function
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES If this taken in by a system then change in entropy is positive and so entropy of the
system increases. The entropy change during an reversible adiabatic process is zero. Ans. 17: (a), (b) and (d)
(b)The work the system does in the cycle is W = v∫ PdV =
A
p1
Solution: (a) The cycle is shown in the figure
p2
∫ PdV + P (V − V ) 2
1
adiabatic C
B
2
AB
V1
V2
Because AB is adiabatic and an ideal gas has the equations pV = nkT and C p = Cv + R , we get
∫
AB
1 ( PV 2 2 − PV 1 1) 1− γ
PdV = − ∫ Cv dT = −Cv (T2 − T1 ) = AB
(c) During the CA part of the cycle the gas absorbs heat which is isochoric Q = ∫ TdS = ∫ Cv dT = Cv (T1 − T2 ) = CA
CA
1 V1 ( P2 − P1 ) 1− γ
(d) Hence, the efficiency of the engine is
V2 −1 V1 W η = = 1− γ P1 Q −1 P2 Ans. 18: (a), (c) and (d) Solution: (a) The increase in entropy of the total system is
ΔS = ∫
T3
T1
T3 cdT T2 cdT +∫ = c ln 3 ≥ 0 T2 T T T1T2
(c) Thus T32 ≥ T1T2 , or T3 ≥ T1T2 (d) The maximum amount of work can be obtained using a reversible heat engine, for which ΔS = 0 .
(
) (
Wmax = c (T1 + T2 − 2T3min ) = c T1 + T2 − 2 T1T2 = c
T1 − T2
)
2
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Ans. 19: (a) and (d) Solution: In the process of isothermal expansion, the external work done by the system is W =∫
2V1
V1
PdV = RT ∫
2V1
V1
dV = RT ln 2 V
Because the internal energy does not change in this process the work is supplied by the heat absorbed from the external world. Thus the increase of entropy of the gas is ΔS1 =
ΔQ W = = R ln 2 T T
The change in entropy of the heat source ΔS 2 = −ΔS1 thus the total change in entropy of the universe is ΔS = ΔS1 + ΔS 2 = 0 . If it is a free expansion, the internal energy of the system is constant. As its final state is the same as for the isothermal process, the change in entropy of the system is also the same. In this case, the state of the heat source does not change, neither does its entropy. Therefore the change in entropy of the universe is ΔS = R ln 2 . Ans. 20: (a), (b) and (c) Solution: We assume Ti ≠ T f (because the change of entropy must be zero when Ti = T f ). The
change of entropy of the body is
ΔS1 = ∫
Tf
Ti
C p dT T
= C p ln
Tf Ti
The change of entropy of the heat source is ΔS 2 =
ΔQ C p (Ti − T f = Tf Tf
)
Therefore the total entropy change is ⎛T Tf ΔS = ΔS1 + ΔS 2 = C p ⎜ i − 1 + ln ⎜ Tf Ti ⎝
⎞ ⎟⎟ ⎠
when x > 0 and x ≠ 1 , the function f ( x ) = x − 1 − ln x > 0 .
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES ⎛ V +V ⎞ The change in entropy of universe during the process is nR ln ⎜ 1 2 ⎟ ⎝ V1 ⎠ Ans. 21: (a), (c) and (d) Solution: This is a process of adiabatic free expansion of an ideal gas. The internal energy does
not change; thus the temperature does not change, that is, the final temperature is still T . From first law of thermodynamics the work done is zero . hence the process is irreversible and entropy is state function so one can choose path isothermal reversible ⎛ V +V ⎞ process . so change in entropy in isothermal is nR ln ⎜ 1 2 ⎟ which is equal to change ⎝ V1 ⎠
in entropy of gas as well as universe .
NAT (Numerical Answer Type Questions) Ans. 22:
390
Solution: If T2 is Sink temperature and T1 Source temperature
Then, efficiency is given η =
T1 − T2 T1
Case 1:
T 1 = 1− 2 6 T1
Case 2:
If T2 is reduced by 65 °C , then T2′ = T2 − 65
η′ = 1− ⇒
5 ⇒ T2 = T1 6
T2′ T1
T − 65 T 1 1 65 = 1− 2 ⇒ = 1− 2 + 3 3 T1 T1 T1 1 5 65 ⇒ T1 = 65 × 6 = 390 0K = 1− + 3 6 T1
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fiziks Ans. 23:
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 207
Solution: If T is temperature of source, T is the temperature of the smk then efficiency 37.5 300 T η = 1− 2 ⇒ = 1− T1 T1 100 300 37.5 200 − 75 300 × 8 = 1− ⇒ T1 = = 480 = 5 T1 100 200
⇒
t = 480 − 273 = 207 0 C
Ans. 24:
33.33
Solution:
W η= Q1
Temperature 200
1 W = . ( 200 − 100 )(1000 − 500 ) = 25000 2
100
⎛1 ⎞ Q1 = 50,000 + ⎜ × 500 × 100 ⎟ = 75,000 J ⎝2 ⎠
η= Ans. 25:
B
A
500
C
1000
W = 33.33% Q1
−1.85
Solution: If Q amount of heat is taken by or given to (a) system at temperature T then change
in entropy is given as dS =
δQ
T By first law of thermodynamic
δQ = dU + δW According to question, since temperature is constant so no change in internal energy
⇒
dU = 0
and work of 730J is done on the system
⇒
δQ = −730 J
From first law of thermodynamics, δQ = −0 + (− 730) ⇒ δ Q = −730 J and
T = 273 + 20 = 293 K dS =
δQ T
=
− 730 293
= −1.85J / K
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fiziks Ans. 26:
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 2.9
Solution: The entropy is defined as dS =
δQ T
Here, δQ = ML = 10 × 80 = 800 cal and T = 0 + 273 = 273 K So, Ans. 27:
dS =
800 cal/K 273
= 2.9 cal/K
1.5
Solution: The energy stored in capacitor =
So, change in entropy is given dS =
1 CV 2 2
δQ T
⇒
dS =
CV 2 2T
Here, C = 10 −6 F , V = 300Volt , T = 27 + 273 = 300 K 10 −6 × (300) So, change in entropy dS = 2 × 300 Ans. 28:
2
10 −6 × 9 × 10 4 = 2 × 300
= 1.5 × 10 −4 J / K
184
Solution: We assume the process is a reversible process of constant pressure. The change in
entropy of the water is ΔS H 2O = ∫
373
273
mCH 2O
dT ⎛ 373 ⎞ = mCH 2O ln ⎜ ⎟ T ⎝ 273 ⎠
we substitute m = 1kg , and CH 2O = 4.18 J / g into it, and find
ΔS H 2O = 1305 J / K The change in entropy of the heat source is ΔS ha = −
Q T
= −1000 × 4.18 ×
100 = −11121J / K 373
Therefore the change of entropy of the whole system is ΔS = ΔS H 2O + ΔSha = 184 J / K
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fiziks Ans. 29:
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 33
Solution: W =
1 ( 2T1 − T1 ) (3S1 − S1 ) = T1S1 2
Q=
1 ( 2T1 − T1 ) (3S1 − S1 ) + T1 (3S1 − S1 ) = T1S1 + T1 2 S1 = 3T1S1 2
η=
W T1S1 = = .33 = 33% Q 3T1S1
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES
Chapter - 5 Maxwell relation and Thermodynamic Potential 5.1
Maxwell Relations F = F ( x, y ) and if it is perfect differential then dF = Mdx + Ndy
⎛ ∂F ⎞ ⎛ ∂F ⎞ Where M = ⎜ then M and N will satisfy the condition ⎟ ⎟ and N = ⎜ ⎝ ∂x ⎠ y ⎝ ∂y ⎠ x ⎛ ∂M ⎞ ⎛ ∂M ⎞ ⎜ ⎟ =⎜ ⎟ ⎝ ∂y ⎠ x ⎝ ∂x ⎠ y Maxwell relations are relationship between two derivatives of thermodynamic variables,
and energy due to the equivalence of potential second derivative under a change of d 2F d 2F = where F is thermodynamic potential and x and y are two of dx dy dy dx
operation order
its natural independent variables. Maxwell relations are extremely important for two reasons. First they show us that derivative of thermodynamic parameters are not all independent. This can serve as a consistency check in both experiments and in theoretical analysis. Maxwell relations provide a method for expressing some derivative in other ways. This enables as to connect difficult to measure quantities to those which are readily accessible experimentally. The measurement of entropy and chemical potential can not be directly measurable in lab but with the help of Maxwell relation there thermodynamic property can be determine theoretically. For Maxwell relation. Let us Legendre the independent variable as x , and y such that
So
U = U(x,y), S = S(x, y) V = V(x, y) ⎛ ∂U ⎞ ⎛ ∂U ⎞ ⎟⎟ dy dU = ⎜ ⎟ dx + ⎜⎜ ⎝ ∂x ⎠ y ⎝ ∂y ⎠ x ⎛ ∂S ⎞ ⎛ ∂S ⎞ dS = ⎜ ⎟ dx + ⎜⎜ ⎟⎟ dy ⎝ ∂x ⎠ y ⎝ ∂y ⎠ x
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES ⎛ ∂V ⎞ ⎛ ∂V ⎞ ⎟⎟ dx dV = ⎜ ⎟ dx + ⎜⎜ ⎝ ∂x ⎠ y ⎝ ∂y ⎠ x from first law of thermodynamic
dU = TdS − PdV ⎛ ∂U ⎞ ⎛ ∂S ⎞ ⎛ ∂V ⎞ ⎜ ⎟ = T ⎜ ⎟ − P⎜ ⎟ ⎝ ∂x ⎠ y ⎝ ∂x ⎠ y ⎝ ∂x ⎠ y ⎛ ∂U ⎜⎜ ⎝ ∂y
⎞ ⎛ ∂S ⎞ ⎛ ∂V ⎟⎟ = T ⎜⎜ ⎟⎟ − p⎜⎜ ⎠x ⎝ ∂y ⎠ x ⎝ ∂y
⎞ ⎟⎟ ⎠x
Hence U, V,and S are perfect differential. ⎡ ∂ ⎛ ∂U ⎞ ⎤ ⎡ ∂ ⎛ ∂U ⎞ ⎤ ⎟⎟ ⎥ Then ⎢ ⎜ ⎟ ⎥ = ⎢ ⎜⎜ ⎣⎢ ∂y ⎝ ∂x ⎠ y ⎦⎥ x ⎢⎣ ∂x ⎝ ∂y ⎠ x ⎦⎥ y ⎡ ∂ ⎛ ∂V ⎞ ⎤ ⎡ ∂ ⎛ ∂V ⎟ ⎥ = ⎢ ⎜⎜ ⎢ ⎜ ⎢⎣ ∂y ⎝ ∂x ⎠ y ⎥⎦ x ⎢⎣ ∂x ⎝ ∂y
⎞ ⎤ ⎟⎟ ⎥ ⎠ x ⎥⎦ y
⎡ ∂ ⎛ ∂S ⎞ ⎤ ⎡ ∂ ⎛ ∂S ⎞ ⎤ ⎢ ⎜ ⎟ ⎥ = ⎢ ⎜⎜ ⎟⎟ ⎥ ⎢⎣ ∂y ⎝ ∂x ⎠ y ⎥⎦ x ⎢⎣ ∂x ⎝ ∂y ⎠ x ⎥⎦ y ⎡ ∂ ⎛ ∂U ⎞ ⎤ ⎛ ∂T ⎟ ⎥ = ⎜⎜ ⎢ ⎜ ⎢⎣ ∂y ⎝ ∂x ⎠ y ⎥⎦ x ⎝ ∂y
⎡ ∂ ⎛ ∂S ⎞ ⎤ ⎛ ∂P ⎞ ⎛ ∂V ⎞ ⎡ ∂ ⎛ ∂V ⎞ ⎤ ⎞ ⎛ ∂S ⎞ ⎟⎟ ⎜ ⎟ + T ⎢ ⎜ ⎟ ⎥ − ⎜⎜ ⎟⎟ ⎜ ⎟ − P⎢ ⎜ ⎟ ⎥ ---(1) ⎢⎣ ∂y ⎝ ∂x ⎠ y ⎥⎦ x ⎝ ∂y ⎠ x ⎝ ∂x ⎠ y ⎢⎣ ∂y ⎝ ∂x ⎠ y ⎥⎦ x ⎠ x ⎝ ∂x ⎠ y
Similarly ⎡ ∂ ⎛ ∂U ⎞ ⎤ ⎛ ∂T ⎞ ⎛ ∂S ⎞ ⎡ ∂ ⎛ ∂S ⎞ ⎤ ⎛ ∂P ⎞ ⎛ ∂V ⎢ ⎜ ⎟ ⎥ =⎜ ⎟ ⎜ ⎟ + T ⎢ ⎜ ⎟ ⎥ − ⎜ ⎟ ⎜⎜ ⎢⎣ ∂x ⎝ ∂y ⎠ x ⎥⎦ y ⎝ ∂x ⎠ y ⎝ ∂y ⎠ x ⎢⎣ ∂x ⎝ ∂y ⎠ x ⎥⎦ y ⎝ ∂x ⎠ y ⎝ ∂y
⎡ ∂ ⎛ ∂V ⎞ ⎟⎟ − P ⎢ ⎜⎜ ⎢⎣ ∂x ⎝ ∂y ⎠x
⎞ ⎤ ⎟⎟ ⎥ ---(2) ⎠ x ⎥⎦ y
Equating equation (1) and (2) ⎛ ∂P ⎞ ⎛ ∂V ⎜ ⎟ ⎜⎜ ⎝ ∂x ⎠ x ⎝ ∂y
⎞ ⎛ ∂P ⎞ ⎛ ∂V ⎞ ⎛ ∂T ⎞ ⎛ ∂S ⎞ ⎛ ∂T ⎟⎟ − ⎜⎜ ⎟⎟ ⎜ ⎟ ⎜⎜ ⎟⎟ − ⎜⎜ ⎟ =⎜ ⎠ x ⎝ ∂y ⎠ x ⎝ ∂x ⎠ y ⎝ ∂x ⎠ y ⎝ ∂y ⎠ x ⎝ ∂y
⎞ ⎛ ∂S ⎞ ⎟⎟ ⎜ ⎟ ⎠ x ⎝ ∂x ⎠ y
(A)
Maxwell first relation:- put x = T , y = V ⎛ ∂S ⎞ ⎛ ∂P ⎞ ⎜ ⎟ =⎜ ⎟ ⎝ ∂V ⎠ T ⎝ ∂T ⎠V
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Maxwell Second Relation:- put x = T , y = P ⎛ ∂V ⎞ ⎛ ∂S ⎞ ⎟ ⎜ ⎟ = −⎜ ⎝ ∂T ⎠ P ⎝ ∂P ⎠ T
Maxwell Third Relation:- put x = S , y = V ⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎜ ⎟ = −⎜ ⎟ ⎝ ∂S ⎠V ⎝ ∂V ⎠ S
Maxwell Fourth Relation:- put x = S , y = P ⎛ ∂T ⎞ ⎛ ∂V ⎞ ⎜ ⎟ =⎜ ⎟ ⎝ ∂P ⎠ S ⎝ ∂S ⎠ P
Thermodynamic potential is a scalar function used to represent the thermodynamic state of system. The concept of thermodynamic potentials was introduced by Pierre Duhem in 1886. One main thermodynamic potential that has a physical interpretation is the internal energy. It is energy of configuration of a given system of conservative forces. Expression for all other thermodynamic energy potentials are drivable via Legendre transformation. 5.2
Different Types of Thermodynamic Potential and Maxwell Relation
Thermodynamic potentials are different form of energy which can be used in different thermodynamic process .thermodynamic potentials are path independent variables so they are perfect differential If F is unique thermodynamic potential defined by variables x and y as F = F ( x, y ) and if it is perfect differential then dF = Mdx + Ndy ⎛ ∂F ⎞ ⎛ ∂F ⎞ Where M = ⎜ ⎟ then M and N will satisfy the condition ⎟ and N = ⎜ ⎝ ∂x ⎠ y ⎝ ∂y ⎠ x ⎛ ∂M ⎞ ⎛ ∂M ⎞ ⎜ ⎟ =⎜ ⎟ ⎝ ∂y ⎠ x ⎝ ∂x ⎠ y
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 5.2.1 Internal Energy :- U and second from the first laws of thermodynamics dU = TdS - PdV
from Legendre transformation ⎛ ∂U ⎞ ⎜ ⎟ = T, ⎝ ∂S ⎠ V
⎛ ∂U ⎞ ⎜ ⎟ = −P ⎝ ∂V ⎠ S
from
given
relation
one
can
derive
Maxwell
⎛ ∂T ⎞ ⎛ ∂P ⎞ relation ⎜ ⎟ = −⎜ ⎟ ⎝ ∂V ⎠ S ⎝ ∂S ⎠V
5.2.2 Enthalpy (H) the enthalpy is defined as H = U + PV dH = dU + PdV + VdP
from Laws of thermodynamics TdS = dU + PdV dH = TdS + VdP
from Legendre transformation ⎛ ∂H ⎞ ⎜ ⎟ =T ⎝ ∂S ⎠ P
⎛ ∂H ⎞ ⎜ ⎟ =P ⎝ ∂P ⎠ S
The Enthalpy H is Extensive quantity, which can not be measured directly. Thus change in enthalpy is more useful.
ΔH is positive in endothermic reaction and negative in exothermic reaction. ⎛ ∂T ⎞ ⎛ ∂V ⎞ From above relation one can derive Maxwell relation ⎜ ⎟ =⎜ ⎟ ⎝ ∂P ⎠ S ⎝ ∂S ⎠ P
ΔH of a system is equal to sum of non-mechanical work done on it and the heat supplied to it. 5.2.3 Helmholtz Free Energy (F) the Helmholtz free energy is defined F = U − TS dF = dU − TdS − SdT
From laws of thermodynamics dU = TdS − PdV dF = TdS − PdV − TdS − SdT dF = − PdV − SdT
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES From Legendre transformation ⎛ ∂F ⎞ ⎛ ∂F ⎞ ⎟ = −S ⎜ ⎟ = − P, ⎜ ⎝ ∂T ⎠V ⎝ ∂V ⎠ T ⎛ ∂P ⎞ ⎛ ∂S ⎞ From above relation one can derive Maxwell relation ⎜ ⎟ =⎜ ⎟ ⎝ ∂T ⎠V ⎝ ∂V ⎠T
The free Energy F, which is available energy for work in reversible isothermal process. 5.2.4 Gibbs Energy: ‘G’ is defined as G = H – TS. G = U + PV − TS dG = dU + PdV − VdP − TdS − SdT TdS − PdV + PdV + VdP − TdS − SdT dG = VdP − SdT
from Legendre transformation ⎛ ∂G ⎞ ⎟ =V ⎜ ⎝ ∂P ⎠ T
⎛ ∂G ⎞ and ⎜ ⎟ = −S ⎝ ∂T ⎠ P
⎛ ∂V ⎞ ⎛ ∂S ⎞ From above relation one can derive Maxwell relation ⎜ ⎟ = −⎜ ⎟ ⎝ ∂T ⎠ P ⎝ ∂P ⎠T
Gibbs free energy is popularly as free enthalpy. The Gibbs free energy is Maximum amount of nonexpanding work that can be exacted from a closed system. The maximum will activated when the system is in reversible process. Gibbs free energy is also treated as chemical potential. In thermodynamics, chemical potential, as partial molar free energy, is a form of potential energy that can be absorbed or relived during a chemical reaction. The chemical potential of a species in the minute can be defined the slope of the energy at system with respect to a change in the no of moles.
μ=
dG dN
V
where μ is chemical potential, G is Gibbs energy and N is no of molecules Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Example: Prove that internal energy U is given by ⎛ ∂(F /T ) ⎞ (a) U = ⎜ ⎜ ∂ (1/ T ) ⎟⎟ ⎝ ⎠V
⎡ ∂ (G / T ) ⎤ (b) H = ⎢ ⎥ ⎣ ∂ (1 / T ) ⎦ P
Solution: (a) F = U –TS =U = F + TS ⎛ ∂F ⎞ S = −⎜ ⎟ ⎝ ∂T ⎠V ⎛ − ∂F ⎞ ⎛ ∂F ⎞ ⎛ ∂ (F / T ) ⎞ 2 ⎛ ∂ (F / T ) ⎞ U = F +T⎜ ⎟ = F −T⎜ ⎟ = −T ⎜ ⎟ =⎜ ⎟ ⎝ ∂T ⎠V ⎝ ∂T ⎠V ⎝ ∂T ⎠V ⎝ ∂1 / T ⎠V ⎡ ∂ (G / T ) ⎤ Solution: (b) H = ⎢ ⎥ ⎣ ∂ (1 / T ) ⎦ P
G = H – TS ⎛ ∂G ⎞ S = −⎜ ⎟ ⎝ ∂T ⎠ p ⎛ ∂G ⎞ 2⎛ ∂ H = G −T⎜ ⎟ = −T ⎜⎜ ⎝ ∂T ⎠ P ⎝ ∂T
⎛G⎞ ⎞ ⎜ ⎟ ⎟⎟ ⎝ T ⎠P ⎠
⎡ ∂ (G / T )⎤ =⎢ ⎥ ⎣ ∂ (1 / T ) ⎦ P
5.3
Application of Maxwell Relation
5.3.1 First T − dS equation
Let T, and V are independent variable S = S (T, V) ⎛ ∂S ⎞ ⎛ ∂S ⎞ dS = ⎜ ⎟ dT + ⎜ ⎟ dV ⎝ ∂T ⎠V ⎝ ∂V ⎠ T ⎛ ∂S ⎞ ⎛ ∂S ⎞ TdS = T ⎜ ⎟ dT + T ⎜ ⎟ dV ⎝ ∂T ⎠V ⎝ ∂V ⎠ T
⎛ ∂S ⎞ ⎛ ∂P ⎞ put ⎜ ⎟ =⎜ ⎟ ⎝ ∂V ⎠ T ⎝ ∂T ⎠ v
⎛ ∂P ⎞ T ⋅ dS = C v dT + T ⎜ ⎟ dV . ⎝ ∂T ⎠V
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 5.3.2 Second T − dS Equation
Let T and P are independent variable S = S (T, P). ⎛ ∂S ⎞ ⎛ ∂S ⎞ TdS = T ⎜ ⎟ dT + T ⎜ ⎟ dP ⎝ ∂T ⎠ P ⎝ ∂P ⎠ T
From Maxwell relation ⎛ ∂S ⎞ ⎛ ∂V ⎞ ⎜ ⎟ = −⎜ ⎟ ⎝ ∂P ⎠ ⎝ ∂T ⎠ P ⎛ ∂V ⎞ TdS = C P dT − T ⎜ ⎟ dP ⎝ ∂T ⎠ P
5.3.3 Third T-dS Equation:
Let P,V are independent variable S = S (P,V) ⎛ ∂S ⎞ ⎛ ∂S ⎞ dS = ⎜ ⎟ dP + ⎜ ⎟ dV ⎝ ∂P ⎠ P ⎝ ∂V ⎠V ⎛ ∂S ⎞ ⎛ ∂S ⎞ TdS = T ⎜ ⎟ dP + T ⎜ ⎟ dV ⎝ ∂P ⎠V ⎝ ∂V ⎠V ⎛ ∂S ⎞ ⎛ ∂T ⎞ ⎛ ∂S ⎞ ⎛ ∂T ⎞ = T⎜ ⎟ ⎜ ⎟ dP + T ⎜ ⎟ ⎜ ⎟ dV ⎝ ∂T ⎠V ⎝ ∂P ⎠V ⎝ ∂T ⎠ P ⎝ ∂V ⎠ P ⎛ ∂T ⎞ ⎛ ∂T ⎞ = CV ⎜ ⎟ dP + C P ⎜ ⎟ dV ⎝ ∂P ⎠V ⎝ ∂V ⎠ P
5.3.4 The First Energy Equation
Let T and V are independent variable U = U(T, V) ⎛ ∂U ⎞ ⎛ ∂U ⎞ dU = ⎜ ⎟ dT + ⎜ ⎟ dV ⎝ ∂T ⎠V ⎝ ∂V ⎠ T
From first law of thermodynamics.
dU = TdS − PdV ⎛ ∂U ⎞ ⎛ ∂S ⎞ ⎜ ⎟ = T⎜ ⎟ −P ⎝ ∂V ⎠ T ⎝ ∂V ⎠ T
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES ⎛ ∂S ⎞ ⎛ ∂P ⎞ Using Maxwell relation T ⎜ ⎟ =T⎜ ⎟ ⎝ ∂V ⎠T ⎝ ∂T ⎠V ⎛ ∂U ⎞ ⎛ ∂P ⎞ ⎜ ⎟ = T⎜ ⎟ −P ⎝ ∂V ⎠ T ⎝ ∂T ⎠V ⎛ ∂U ⎞ ⎛ ∂U ⎞ dU = ⎜ ⎟ dT + ⎜ ⎟ dV ⎝ ∂T ⎠V ⎝ ∂V ⎠ T ⎛ ∂U ⎞ = C v dT + ⎜ ⎟ dV ⎝ ∂V ⎠ T ⎛ ∂U ⎞ ⎛ ∂P ⎞ ⎜ ⎟ = T⎜ ⎟ −P ⎝ ∂V ⎠ T ⎝ ∂T ⎠V
⎞ ⎛ ⎛ ∂P ⎞ dU = C v dT + ⎜⎜ T ⎜ ⎟ − P ⎟⎟dV ⎠ ⎝ ⎝ ∂T ⎠V
5.3.5 Second Energy Equation
dU = TdS − PdV ⎛ ∂U ⎞ ⎛ ∂S ⎞ ⎛ ∂V ⎞ ⎜ ⎟ = T ⎜ ⎟ − P⎜ ⎟ ⎝ ∂P ⎠ T ⎝ ∂P ⎠ T ⎝ ∂P ⎠ T
Using Maxwell relation ⎛ ∂S ⎞ ⎛ ∂V ⎞ ⎜ ⎟ = −⎜ ⎟ ⎝ ∂P ⎠ T ⎝ ∂T ⎠ P ⎛ ∂U ⎞ ⎛ ∂V ⎞ ⎛ ∂V ⎞ ⎜ ⎟ = −T ⎜ ⎟ − P⎜ ⎟ This is popularly known as second energy ⎝ ∂P ⎠ T ⎝ ∂T ⎠ P ⎝ ∂P ⎠ T
equation Application of second energy equation If U is function of independent variable of T and P. U = U (T , P ) ⎛ ∂U ⎞ ⎛ ∂U ⎞ dU = ⎜ ⎟ dP ⎟ dT + ⎜ ⎝ ∂P ⎠ T ⎝ ∂T ⎠ P
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES ⎡ ⎛ ∂V ⎞ ⎛ ∂V ⎞ ⎤ ⎛ ∂U ⎞ dU = ⎜ ⎟ ⎥ dP ⎟ + P⎜ ⎟ dT − ⎢T ⎜ T P ∂ ∂ ⎠T ⎦ ⎠ ⎝ ⎝ ⎝ ∂T ⎠ P P ⎣ Example: From relation dU = TdS − PdV ⎛ ∂P ⎞ ⎛ ∂T ⎞ Derive Maxwell relation ⎜ ⎟ = −⎜ ⎟ ⎝ ∂S ⎠V ⎝ ∂V ⎠ S
Solution:
⎛ ∂U ⎞ ⎛ ∂U ⎞ ⎟ =T ⎟ = − P, ⎜ ⎜ ⎝ ∂S ⎠V ⎝ ∂V ⎠ S
dU = TdS − PdV
Hence U is exact differential ⎡ ∂ ⎡ ∂ ⎛ ∂U ⎞ ⎤ ⎟ ⎥ =⎢ ⎢ ⎜ ⎣ ∂S ⎝ ∂V ⎠ S ⎦ v ⎣ ∂V
⎛ ∂U ⎞ ⎤ ⎟ ⎥ ⎜ ⎝ ∂S ⎠V ⎦ S
⎛ ∂T ⎞ ⎛ ∂P ⎞ −⎜ ⎟ = ⎜ ⎟ ⎝ ∂S ⎠V ⎝ ∂V ⎠ S
Example: A real gas which obey van der Waal’s equation of state are kept in container which
has temperature T0 and volume V0. if volume of container changes to V such that temperature of gas become T what is change in entropy? Solution: Assume CV is specific heat of constant volume
For van der Waal’s gas a ⎛ ⎜P + 2 V ⎝
⎞ ⎟(V − b ) = RT ⎠
From first T − dS equation ⎛ ∂P ⎞ TdS = CV dT + ⎜ ⎟ dV ⎝ ∂T ⎠V R ⎛ ∂P ⎞ ⎟ = ⎜ ⎝ ∂T ⎠V V − b T
dS = CV
V
dT R ∫T T + V∫ (V − b)dV 0 0
⎛T ⎞ ⎛ V −b ⎞ S = CV ln ⎜ ⎟ + R ln ⎜ ⎟ + S0 ⎝ T0 ⎠ ⎝ V0 − b ⎠
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES where S 0 is integration constant Example: For Vander wall gases, prove that a ⎛ ∂U ⎞ ⎟ = 2 where U is internal energy. ⎜ ⎝ ∂V ⎠ T V
Solution: From first energy equation ⎛ ∂P ⎞ ⎛ ∂U ⎞ ⎟ −P ⎟ = T⎜ ⎜ ⎝ ∂T ⎠V ⎝ ∂V ⎠ T a ⎛ ⎜P + 2 V ⎝
P=
-----(i)
⎞ ⎟(V − b ) = RT ⎠
RT a − 2 V −b V
R ⎛ ∂P ⎞ put the value of ⎟ = ⎜ ⎝ ∂T ⎠V (V − b )
⎛ ∂P ⎞ ⎜ ⎟ in equation (i) ⎝ ∂T ⎠V
a ⎛ ∂U ⎞ ⎟ = 2 ⎜ ⎝ ∂V ⎠ T V
Example: Prove that ⎛ ∂2P ⎞ ⎛ ∂C ⎞ (a) ⎜ V ⎟ = T ⎜⎜ 2 ⎟⎟ ⎝ ∂V ⎠ T ⎝ ∂T ⎠V ⎛ ∂ 2V ⎛ ∂C ⎞ (b) ⎜ P ⎟ = −T ⎜⎜ 2 ⎝ ∂P ⎠ T ⎝ ∂T
⎞ ⎟⎟ ⎠P
⎛ ∂S ⎞ Solution: (a) we know CV = T ⎜ ⎟ ⎝ ∂T ⎠v ⎡ ∂ 2 ⎛ ∂S ⎞ ⎤ ⎡ ∂ 2 ⎛ ∂S ⎞ ⎤ ⎛ ∂CV ⎞ ⎟ =T⎢ ⎜ ⎟ ⎥ =T⎢ ⎜ ⎜ ⎟ ⎥ ⎝ ∂V ⎠ T ⎣ ∂T ⎝ ∂V ⎠ T ⎦ V ⎣ ∂V ⎝ ∂T ⎠V ⎦ T
using Maxwell relation ⎡ ∂ 2 ⎛ ∂S ⎞ ⎤ ⎡ ∂2 ⎛ ∂P ⎞ ⎛ ∂S ⎞ = T ⎟ one can get T ⎢ ⎜ ⎟ =⎜ ⎜ ⎟ ⎥ ⎢ ⎝ ∂V ⎠ T ⎝ ∂T ⎠V ⎣ ∂T ⎣ ∂T ⎝ ∂V ⎠T ⎦V
⎛ ∂P ⎞⎤ ⎟⎥ ⎜ ⎝ ∂T ⎠⎦ V
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 2 ⎛∂ P⎞ ⎛ ∂CV ⎞ ⎜ ⎟ = T ⎜⎜ 2 ⎟⎟ ⎝ ∂V ⎠ T ⎝ ∂T ⎠V
⎛ ∂S ⎞ (b) C p = T ⎜ ⎟ ⎝ ∂T ⎠ p ⎡ ∂ ⎛ ∂S ⎞ ⎤ ⎡ ∂ ⎛ ∂S ⎞ ⎤ ⎛ ∂C P ⎞ ⎜ ⎟ =T⎢ ⎜ ⎟ ⎥ =T⎢ ⎜ ⎟ ⎥ ⎝ ∂P ⎠ T ⎣ ∂P ⎝ ∂T ⎠ P ⎦ T ⎣ ∂T ⎝ ∂P ⎠ T ⎦ P
Use Maxwell relation ⎛ ∂V ⎞ ⎛ ∂S ⎞ ⎟ ⎜ ⎟ = −⎜ ⎝ ∂T ⎠ P ⎝ ∂P ⎠ T ⎛ ∂ 2V ⎛ ∂C P ⎞ ⎜ ⎟ = −T ⎜⎜ 2 ⎝ ∂P ⎠ T ⎝ ∂T
⎞ ⎟⎟ ⎠P
Example: If α p is thermal expansivity at constant pressure and KT isothermal compressibility
then prove that ⎛ ∂S ⎞ (i) ⎜ ⎟ = −Vα p ⎝ ∂P ⎠ T
αp ⎛ ∂P ⎞ (ii) ⎜ ⎟ = ⎝ ∂T ⎠V K T (iii) C P − CV =
TVα 2 P KT
Solution: From Maxwell relation ⎛ ∂V ⎞ ⎛ ∂S ⎞ ⎟ ⎜ ⎟ = −⎜ ⎝ ∂T ⎠ P ⎝ ∂P ⎠ T
α=
1 ⎛ ∂V ⎞ ⎟ ⎜ V ⎝ ∂T ⎠ P
⎡ 1 ⎛ ∂V ⎞ ⎤ ⎛ ∂S ⎞ ⎛ ∂S ⎞ ⎟ ⎥ , ⎜ ⎟ = −α V ⎜ ⎟ = −V ⎢ ⎜ ⎝ ∂P ⎠ T ⎣V ⎝ ∂T ⎠ P ⎦ ⎝ ∂P ⎠ T ⎛ ∂P ⎞ ⎛ ∂S ⎞ (ii) ⎜ ⎟ =⎜ ⎟ ⎝ ∂T ⎠V ⎝ ∂V ⎠T Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES ⎛ ∂P ⎞ ⎛ ∂V ⎞ ⎛ ∂T ⎞ ⎟ = −1 ⎟ ⎜ ⎜ ⎟ ⎜ ⎝ ∂T ⎠V ⎝ ∂P ⎠ T ⎝ ∂V ⎠ P
−1 ⎛ ∂P ⎞ ⎜ ⎟ = ⎝ ∂T ⎠V ⎛ ∂V ⎞ ⎛ ∂T ⎞ ⎟ ⎜ ⎟ ⎜ ⎝ ∂P ⎠ T ⎝ ∂V ⎠ P
1 ⎛ ∂V ⎞ ⎟ ⎜ V ⎝ ∂T ⎠ P = 1 ⎛ ∂V ⎞ ⎟ ⎜ V ⎝ ∂P ⎠ T −
−α ⎛ ∂P ⎞ ⎜ ⎟ = ⎝ ∂T ⎠V K T ⎛ ∂S ⎞ ⎛ ∂S ⎞ dS = ⎜ ⎟ dV ⎟ dT + ⎜ ⎝ ∂V ⎠ T ⎝ ∂T ⎠V
dS =
CV ⎛ ∂S ⎞ dT + ⎜ ⎟ dV T ⎝ ∂V ⎠ T
CV ⎛ ∂S ⎞ ⎛ ∂V ⎞ ⎛ ∂S ⎞ +⎜ ⎟ ⎟ ⎜ ⎜ ⎟ = T ⎝ ∂V ⎠ T ⎝ ∂T ⎠ P ⎝ ∂T ⎠ P
C P CV ⎛ ∂S ⎞ = +V αP⎜ ⎟ T T ⎝ ∂V ⎠ T
Use Maxwell relation C P CV ⎛ ∂P ⎞ = + Vα P ⎜ ⎟ T T ⎝ ∂T ⎠V C P − CV =
⎛α C P CV = + V α P ⎜⎜ P T T ⎝ KT
⎞ ⎟⎟ ⎠
T V α p2 KT
Example: Prove that ⎛ ∂P ⎞ ⎛ ∂V ⎞ (a) CP − CV = T ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠V ⎝ ∂T ⎠ P
(b) For the Vander Waal’s gas prove that 2a ⎞ ⎛ CP − CV = R ⎜ 1 + ⎟ ⎝ RTV ⎠
⎛ ∂Q ⎞ ⎛ ∂Q ⎞ Solution: (a) CP − CV = ⎜ ⎟ −⎜ ⎟ ⎝ ∂T ⎠ P ⎝ ∂T ⎠V ⎛ ∂S ⎞ ⎛ ∂S ⎞ = T⎜ ⎟ −T⎜ ⎟ ⎝ ∂T ⎠ P ⎝ ∂T ⎠V
---- (A)
S = S (T , V )
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES ⎛ ∂S ⎞ ⎛ ∂S ⎞ ⎛ ∂S ⎞ ⎛ ∂S ⎞ ⎛ ∂V ⎞ ⎛ ∂S ⎞ dS = ⎜ ⎜ ⎟ =⎜ ⎟ +⎜ ⎟ dT + ⎜ ⎟ ⎜ ⎟ dV ⎟ ⎝ ∂T ⎠ P ⎝ ∂T ⎠V ⎝ ∂V ⎠ T ⎝ ∂T ⎠ P ⎝ ∂T ⎠V ⎝ ∂V ⎠T ⎛ ∂S ⎞ Put the value of ⎜ ⎟ in equation (A) ⎝ ∂T ⎠ P ⎛ ∂S ⎞ ⎛ ∂S ⎞ ⎛ ∂V ⎞ ⎛ ∂S ⎞ ⎛ ∂S ⎞ ⎛ ∂V ⎞ CP − CV = T ⎜ ⎟ +T⎜ ⎟ ⎜ ⎟ −T ⎜ ⎟ ⇒ CP − CV = T ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠V ⎝ ∂ V ⎠ T ⎝ ∂T ⎠ P ⎝ ∂T ⎠V ⎝ ∂ V ⎠ T ⎝ ∂T ⎠ P ⎛ ∂P ⎞ =T⎜ ⎟ ⎝ ∂T ⎠V
⎡⎛ ∂S ⎞ ⎛ ∂V ⎞ ⎛ ∂P ⎞ ⎤ ⎜ ⎟ ∵ ⎢⎜ ⎟ =⎜ ⎟ ⎥ ⎝ ∂T ⎠ P ⎣⎝ ∂ V ⎠ T ⎝ ∂T ⎠V ⎦
(b) For van der Waal’s gas a ⎞ ⎛ ⎜ P + 2 ⎟ (V − b ) = RT V ⎠ ⎝
a ⎞ RT ⎛ ⎜P+ 2 ⎟ = V ⎠ (V − b ) ⎝
---- (B) → differentiate w.r.t. to T
R ⎛ ∂P ⎞ ⎜ ⎟ = ⎝ ∂T ⎠V V − b
Differentiate (B) with respect to V −
2a ⎛ ∂V ⎞ RT ⎛ ∂V ⎞ R ⎟ =− ⎟ + 2 ⎜ 3 ⎜ V ⎝ ∂T ⎠ P (V − b ) ⎝ ∂T ⎠ P V − b
⎛ ∂V ⎞ ⎜ ⎟ = ⎝ ∂T ⎠ P
R V −b RT
(V − b )
2
−
2a V3
⎛ ∂P ⎞ ⎛ ∂V ⎞ Substituting the value ⎜ ⎟ ⎜ ⎟ in equation ⎝ ∂T ⎠V ⎝ ∂T ⎠ P ⎛ ∂S ⎞ ⎛ ∂V ⎞ CP − CV = T ⎜ ⎟ ⎜ ⎟ ⎝ ∂V ⎠T ⎝ ∂T ⎠ P
R ⎛ R ⎞ ⎜ ⎟ V − b) ⎝ V − b ⎠ ( CP − CV = T RT 2a − 3 2 (V − b ) V
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES R = 2 ⎛ 2a ⎞ (V − b ) 1− ⎜ 3 ⎟ ⎝ V ⎠ RT −1
R 2a ⎞ 2a ⎞ ⎛ ⎛ CP − CV = , CP − CV = R ⎜ 1 + = R ⎜1 − ⎟ ⎟ 2 RTV ⎠ RTV ⎠ ⎛ 2a ⎞ V ⎝ ⎝ 1− ⎜ 3 ⎟ ⎝ V ⎠ RT ⎛ ∂P ⎞ ⎛ ∂V ⎞ Example: From CP − CV = T ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠V ⎝ ∂T ⎠ P
Prove C P − CV = TEα 2V where E is bulk modulus of elasticity and α is coefficient of volume expansion. ⎛ ∂P ⎞ ⎛ ∂V ⎞ Solution: Let CP − CV = T ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠V ⎝ ∂T ⎠ P P = P(T , V ) ⎛ ∂P ⎞ ⎛ ∂P ⎞ dP = ⎜ ⎟ dT + ⎜ ⎟ dV ⎝ ∂T ⎠V ⎝ ∂V ⎠ T
For constant pressure dP = 0 ⎛ ∂P ⎞ ⎛ ∂P ⎞ ⎜ ⎟ dT = −⎜ ⎟ dV ⎝ ∂T ⎠V ⎝ ∂V ⎠T ⎛ ∂P ⎞ ⎛ ∂P ⎞ ⎛ ∂V ⎞ ⎜ ⎟ = −⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠V ⎝ ∂V ⎠ T ⎝ ∂T ⎠ P 2
⎛ ∂P ⎞ ⎛ ∂V ⎞ CP − CV = −T ⎜ ⎟ ⎜ ⎟ ⎝ ∂V ⎠T ⎝ ∂T ⎠ P 1 ⎛ ∂V ⎞ ⎛ ∂P ⎞ E = −V ⎜ ⎟ and α = ⎜ ⎟ V ⎝ ∂T ⎠ P ⎝ ∂V ⎠ T 2
V ⎛ ∂P ⎞ V 2 ⎛ ∂V ⎞ ⎛E⎞ 2 2 CP − CV = −T ⎜ ⎟ ⎜ ⎟ = −T ⎜ ⎟ V α V ⎝ ∂V ⎠T V ⎝ ∂T ⎠ ⎝V ⎠ C P − CV = −TVEα 2
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Example: If Helmholtz free energy for radiation is given by
F =−
8π 5 K B4 T 4 V 45h 3 c 3
(a) What is radiation pressure (b) If S entropy of system prove that specific heat at constant volume is given by CV = 3S Solution: (a) dF = − SdT − PdV
8π 5 K B4T 4 ⎛ ∂F ⎞ P = −⎜ ⎟ = 45h 3c 3 ⎝ ∂V ⎠ T (b)
32π 5 K B4 ⎛ ∂F ⎞ S = −⎜ VT 3 ⎟ = 3 3 ∂ T 45 h c ⎝ ⎠V ⎛ 32π 5 K B4 ⎛ ∂S ⎞ CV = T ⎜ ⎟ = T ⋅ 3 ⋅ ⎜⎜ 3 3 ⎝ ∂T ⎠V ⎝ 45h c
⎞ 2 ⎛ 32π 2 K B4 ⎟⎟VT = 3 ⋅ ⎜⎜ 3 3 ⎠ ⎝ 45h c
⎞ 3 ⎟⎟VT ⎠
CV = 3S Example: The internal energy E of a system is given by E =
bS 3 where b is constant and other VN
symbols have their used meaning. (a) Find the temperature of system (b) Find Pressure of system Solution: From first law of thermodynamics
bS 3 VN
TdS = dU + PdV
U =E=
(a)
⎛ ∂U ⎞ T =⎜ ⎟ ⎝ ∂S ⎠V
T=
(b)
⎛ ∂U ⎞ P = −⎜ ⎟ ⎝ ∂V ⎠ S
⎛ bS 3 ⎞ P = −⎜⎜ − 2 ⎟⎟ ⎝ V N⎠
dU = TdS − PdV
3bS 2 VN bS 3 P= 2 V N
Example: Consider an Ideal gas where entropy is given by
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES n⎡ U V⎤ S = ⎢σ + 5 R ln + 2 R ln ⎥ 2⎣ n n⎦
where n = number of moles, R = universal gas constant, U = internal energy V = volume and σ = constant
(a) Calculate specific heat at constant pressure and volume (b) Prove that internal energy is given by U =
5 PV 2
Solution: (a) From first law of thermodynamics
TdS = dU − PdV , dS =
n 5R 1 = 2U T
U=
1 P dU − dV T T
1 ⎛ ∂S ⎞ ⎜ ⎟ = ⎝ ∂U ⎠V T
5 nRT 2
7 ⎛ ∂U ⎞ 5 CV = ⎜ ⎟ = nR ⇒ CP = CV + R ⇒ CP = nR 2 ⎝ ∂T ⎠ 2 (b)
⇒
U 5 nRT = V 2 V
PV = nRT
⇒
V=
nRT P
U 5 = P V 2
⇒
U=
5 PV 2
U=
5 nRT 2
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 3R Example: Using the equation of state PV = nRT and the specific heat per mole C v = for 2
monatomic ideal gas (a) Find Entropy of given system. (b) Find free energy of given system dU = nCV dT ,
p=
nRT V
CV =
3R 2
TdS = dU + PdV dS = S=
dU p + dV T T
or ds = nCV
dT dV + nR T V
3 NR ln T + N ln V + S0 where S0 is constant 2
(b) F = U − TS =
3nRT ⎛ 3nRT ⎞ −⎜ ln T − nRT ln V ⎟ + F0 where F0 = T ⋅ S 0 is again constant 2 ⎝ 2 ⎠
Example: From electromagnetic theory Maxwell found that the pressure P from an isotropic
radiation equal to
1U 1 the energy density i.e. P = where V is volume of the cavity 3V 3
using the first energy equation prove that Energy density u is proportional to T 4 . ⎛ ∂U ⎞ ⎛ ∂S ⎞ ⎛ ∂P ⎞ Solution: ⎜ ⎟ =T⎜ ⎟ −P =T⎜ ⎟ −P ⎝ ∂V ⎠T ⎝ ∂V ⎠T ⎝ ∂T ⎠T 1 U u= P= u 3 V 1 ⎛ ∂U ⎞ ⎛ ∂U ⎞ T ⎛ ∂U ⎞ U ⎛ ∂P ⎞ =⎜ ⎟ = ⎜ ⎟ ⎜ ⎟ = ⎜ ⎟ − ⎝ ∂T ⎠V 3 ⎝ ∂T ⎠V ⎝ ∂V ⎠T 3V ⎝ ∂T ⎠V 3V T ⎛ du ⎞ 1 ⎛ ∂U ⎞ u= ⎜ ⎜ ⎟ =u ⎟ − u 3 ⎝ dT ⎠ v 3 ⎝ ∂V ⎠T
T
du du dT = 4u = =4 dT u T
u ∝T 4 =u = α T 4 where α is a constant. Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES MCQ (Multiple Choice Questions)
Q1.
Which of the following is not a Maxwell’s thermodynamic relation? ⎛ ∂S ⎞ ⎛ ∂P ⎞ (a) ⎜ ⎟ =⎜ ⎟ ⎝ ∂V ⎠ T ⎝ ∂T ⎠V ⎛ ∂T ⎞ ⎛ ∂V ⎞ (c) ⎜ ⎟ = ⎜ ⎟ ⎝ ∂P ⎠V ⎝ ∂S ⎠ P
Q2.
⎛ ∂T ⎞ ⎛ ∂P ⎞ (b) ⎜ ⎟ = −⎜ ⎟ ⎝ ∂V ⎠ S ⎝ ∂S ⎠V ⎛ ∂V ⎞ ⎛ ∂S ⎞ (d) ⎜ ⎟ =⎜ ⎟ ⎝ ∂P ⎠ T ⎝ ∂T ⎠ P
Which one of the following thermo dynamical relations is used for certain adiabatic changes, such as the sudden compression of a liquid or sudden -stretching of a rod? ⎛ ∂S ⎞ ⎛ ∂V ⎞ (a) ⎜ ⎟ = −⎜ ⎟ ⎝ ∂P ⎠ T ⎝ ∂T ⎠ P ⎛ ∂P ⎞ ⎛ ∂S ⎞ (c) ⎜ ⎟ = −⎜ ⎟ ⎝ ∂T ⎠ P ⎝ ∂V ⎠ T
Q3.
⎛ ∂S ⎞ ⎛ ∂P ⎞ (b) ⎜ ⎟ = −⎜ ⎟ ⎝ ∂T ⎠ P ⎝ ∂V ⎠ T ⎛ ∂V ⎞ ⎛ ∂S ⎞ (d) ⎜ ⎟ = −⎜ ⎟ ⎝ ∂P ⎠ T ⎝ ∂T ⎠ P
Which one of the Maxwell’s thermodynamic relations given below leads to Clausius-Clapeyron equation?
Q4.
⎛ ∂P ⎞ ⎛ ∂T ⎞ (a) ⎜ ⎟ = −⎜ ⎟ ⎝ ∂V ⎠ S ⎝ ∂S ⎠V
⎛ ∂V ⎞ ⎛ ∂T ⎞ (b) ⎜ ⎟ = −⎜ ⎟ ⎝ ∂P ⎠ S ⎝ ∂S ⎠ P
⎛ ∂S ⎞ ⎛ ∂P ⎞ (c) ⎜ ⎟ =⎜ ⎟ ⎝ ∂V ⎠ T ⎝ ∂T ⎠V
⎛ ∂S ⎞ ⎛ ∂V ⎞ (d) ⎜ ⎟ = −⎜ ⎟ ⎝ ∂P ⎠ T ⎝ ∂T ⎠ P
Which of the following is correct if α is volume expansivity and other variables have usual meaning in thermodynamics .
Q5.
⎛ ∂C ⎞ (a) ⎜ P ⎟ = −TV α 2 ⎝ ∂P ⎠T
⎛ ∂C ⎞ (b) ⎜ P ⎟ = TV α 2 ⎝ ∂P ⎠T
TV ⎛ ∂C ⎞ (c) ⎜ P ⎟ = − 2 α ⎝ ∂P ⎠T
TV ⎛ ∂C ⎞ (d) ⎜ P ⎟ = 2 ⎝ ∂P ⎠T α
Which of the following can be derived by S = S (T ,V ) ? ⎛ ∂P ⎞ (a) TdS = CV dT + T ⎜ ⎟ dV ⎝ ∂T ⎠V
⎛ ∂V ⎞ (b) TdS = CP dT − T ⎜ ⎟ dV ⎝ ∂T ⎠ P
⎛ ∂P ⎞ (c) TdS = CV dT − T ⎜ ⎟ dV ⎝ ∂T ⎠V
⎛ ∂V ⎞ (d) TdS = CP dT + T ⎜ ⎟ dV ⎝ ∂T ⎠ P
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fiziks Q6.
Q7.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Which of the following can be derived by S = S (T , P) ? ⎛ ∂P ⎞ (a) TdS = CV dT + T ⎜ ⎟ dV ⎝ ∂T ⎠V
⎛ ∂V ⎞ (b) TdS = CP dT − T ⎜ ⎟ dV ⎝ ∂T ⎠ P
⎛ ∂P ⎞ (c) TdS = CV dT − T ⎜ ⎟ dV ⎝ ∂T ⎠V
⎛ ∂V ⎞ (d) TdS = CP dT + T ⎜ ⎟ dV ⎝ ∂T ⎠ P
For an isolated thermodynamically system, P, V , T ,U , S and F represent the pressure, volume, temperature, internal energy, entropy, and Helmholtz free energy respectively. Then the following relation is true
Q8.
⎛ ∂F ⎞ (a) ⎜ ⎟ = −S ⎝ ∂T ⎠V
⎛ ∂F ⎞ (b) ⎜ ⎟ = −S ⎝ ∂T ⎠ P
⎛ ∂U ⎞ (c) ⎜ ⎟ =T ⎝ ∂S ⎠V
⎛ ∂U ⎞ (d) ⎜ ⎟ = −P ⎝ ∂V ⎠V
Which of the following thermodynamic relation will give the Maxwell relation ⎛ ∂S ⎞ ⎛ ∂V ⎞ ⎜ ⎟ = −⎜ ⎟ ? ⎝ ∂P ⎠T ⎝ ∂T ⎠ P
Q9.
Q10.
(a) dU = TdS − PdV
(b) dH = TdS + VdP
(c) dF = − SdT − PdV
(d) dG = − SdT + VdP )
Which of the following is not an exact differential? (a) dQ where Q heat absorbed
(b) dU where U is internal energy
(c) dS where S is entropy
(d) dF where S is entropy
Which among the following sets of Maxwell relations is correct? (U-internal energy, Henthalpy, A-Helmholtz free energy and G-Gibbs free energy)? ⎛ ∂U ⎞ ⎛ ∂U ⎞ ⎟ and P = ⎜ ⎟ ⎝ ∂V ⎠ S ⎝ ∂S ⎠V
(a) T = ⎜
⎛ ∂G ⎞ ⎛ ∂G ⎞ ⎟ and V = ⎜ ⎟ ∂ V ⎝ ⎠T ⎝ ∂P ⎠ S
(c) P = −⎜
⎛ ∂H ⎞ ⎛ ∂H ⎞ ⎟ and T = ⎜ ⎟ ⎝ ∂P ⎠ S ⎝ ∂S ⎠ P
(b) V = ⎜
⎛ ∂A ⎞ ⎛ ∂A ⎞ ⎟ and S = ⎜ ⎟ ∂ S ⎝ ⎠T ⎝ ∂P ⎠V
(d) P = −⎜
Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Q11.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES When a system is held at constant temperature and pressure in a state of equilibrium, then
it attains a minimum value of:
Q12.
(a) internal energy
(b) enthalpy
(c) Helmholtz energy
(d) Gibb’s free energy
Given that H = the enthalpy of a system T = absolute temperature and
S = entropy G = H − TS is the Gibbs function for the system In the case of a reversible, isotherm and isobaric process: (a) G = constant (b) G > 0 and changes with T (c) G < 0 and changes with S (d) G changes with both T and S Q13.
The Gibb’s function G in thermodynamics is defined is
G = H − TS where H is the enthalpy, T is the temperature and S is the entropy. In an isothermal, isobaric, reversible process, G :
Q14.
(a) remains constant, but not zero
(b) varies linearly
(c) varies non-linearly
(d) is zero
The internal energy E of a system is given by E =
bS 3 , where b is a constant and other VN
symbols have their usual meaning. The temperature of this system is equal to (a)
bS 2 VN
(b)
3bS 2 VN
(c)
bS 3 V 2N
⎛S⎞ (d) ⎜ ⎟ ⎝N⎠
2
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fiziks Q15.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES The free energy of gas of N particles in a volume V and at a temperature T is
[
F = Nk B T ln a0V (k BT )
5/ 2
]
/ N , where a 0 is a constant and k B denotes the Boltzmann
constant. The internal energy of the gas is (a)
3 Nk B T 2
(b)
[
(c) Nk B T ln a 0V (k B T ) Q16.
5/ 2
]
/N −
5 Nk B T 2
[
3 Nk B T 2
(d) Nk BT ln a0V / (k B T )
5/ 2
]
The entropy S of a thermodynamic system as a function of energy E is given by the following graph
S
C B A →E
The temperatures of the phases A, B and C , denoted by T A , TB and TC , respectively, satisfy the following inequalities: (a) TC > TB > T A Q17.
(b) T A > TC > TB
(c) TB > TC > T A
(d) TB > T A > TC
The entropy of an ideal paramagnet in a magnetic field is given approximately by S = S0 − cU 2 where U is energy of the spin system and c is constant then which one is
correct plot between internal energy and temperature T where −∞ < T < ∞
U (a)
U
(b)
T T U (c)
U T
(d)
T
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fiziks Q18.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES A certain system is found to have Gibbs free energy given by
⎛ aP ⎞ G(p, T) = RT ln ⎜ 5/2 ⎟ ⎝ (RT) ⎠ Where a and R are constants then specific heat at constant pressure ( c p ) is given by (a) Q19.
3 R 2
5 (b) R 2
7 (c) R 2
(d)
9 R 2
Helmholtz free energy is given by F = −CT 4 where C is constant and T is temperature in Kelvin then which one is correct relation between specific heat at constant volume Cv and entropy S is given by (a) CV = 2S
Q20.
(b) CV = 4S
(c) CV = 3S
(d) C v =
3 S 2
If CV is the specific heat of the ideal gas then which of the following is correct of Vander wall gases for same degree of freedom . (a) dU = CV dT (c) dU = CV dT −
Q21.
(b) dU = −
a dV V2
a dV V2
(d) dU = CV dT +
a dV V2
For a Van der Waals gas the equation of the adiabatic curve in the variables T ,V ; (a) T (V − b )R / C P = cons tan t
(b) T (V − b )R / CV = cons tan t
(c) T (V − b )− R / C P = cons tan t
(d) T (V − b )− R / CV = cons tan t
Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES MSQ (Multiple Select Questions)
Q22.
If H is enthalpy and G is Gibbs free energy of the thermodynamic system then which of the following is correct (a) H = G − TS ⎛ ∂ (c) H = −T 2 ⎜ ⎝ ∂T
Q23.
(b) H = G + TS ⎛ G ⎞⎞ ⎜ ⎟⎟ ⎝ T ⎠ ⎠P
⎛ ∂ (d) H = T 2 ⎜ ⎝ ∂T
⎛ G ⎞⎞ ⎜ ⎟⎟ ⎝ T ⎠ ⎠P
Which of the following statements are correct . ⎛ ∂U ⎞ ⎛ ∂P ⎞ (a) The first energy equation is be given by ⎜ ⎟ = T⎜ ⎟ −P ⎝ ∂V ⎠ T ⎝ ∂T ⎠V ⎛ ∂U ⎞ (b) The value of ⎜ ⎟ = 0 for Ideal gas ⎝ ∂V ⎠ T a ⎛ ∂U ⎞ (c) The value of ⎜ ⎟ = 2 for Vander Waal’s gases . ⎝ ∂V ⎠ T V ⎛ ∂U ⎞ ⎛ ∂V ⎞ (d) The second energy equation is given by ⎜ ⎟ = T⎜ ⎟ −P ⎝ ∂P ⎠ T ⎝ ∂T ⎠ P
Q24.
Which of following is correct for heat capacity at constant pressure CP and volume CV ⎡ ⎛ ∂U ⎞ ⎤ ⎛ ∂V ⎞ (a) C p − CV = ⎢ p + ⎜ ⎟ ⎥⎜ ⎟ ⎝ ∂V ⎠T ⎦ ⎝ ∂T ⎠ P ⎣
(b) for van der waal’s gas C p − CV =
R 1 − 2a (1 − b / V ) / VRT 2
(c) van der Waal’s gas behave like a ideal as high temperature (d) for Ideal gas c p − cV = R Q25.
Which of the following is correct? ⎛ ∂P ⎞ ⎛ ∂V ⎞ (a) CP − CV = T ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠V ⎝ ∂T ⎠ P ⎛ ∂P ⎞ ⎛ ∂V ⎞ (c) CP − CV = −T ⎜ ⎟ ⎜ ⎟ ⎝ ∂V ⎠T ⎝ ∂T ⎠
⎛ ∂P ⎞ ⎛ ∂V ⎞ (b) CP − CV = −T ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠V ⎝ ∂T ⎠ P 2
P
⎛ ∂P ⎞ ⎛ ∂V ⎞ (d) CP − CV = T ⎜ ⎟ ⎜ ⎟ ⎝ ∂V ⎠T ⎝ ∂T ⎠
2
P
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fiziks Q26.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES If α is volume expansivity of substance and βT is isothermal compressibility β S is ⎛ ∂P ⎞ ⎛ ∂V ⎞ adiabatic compressibility and it is given that CP − CV = −T ⎜ ⎟ ⎜ ⎟ ⎝ ∂V ⎠T ⎝ ∂T ⎠
2
then which P
one of the following is correct (a) CP − CV = −
T α 2V
βT
T α 2V (c) β s − βT = CP
Q27.
Q28.
(b) CP − CV =
T α 2V
βT
T α 2V (d) β s − βT = − CP
Which of the following is correct if all variable have usual meaning in thermodynamics ⎛ ∂S ⎞ ⎛ ∂P ⎞ (a) ⎜ ⎟ =⎜ ⎟ ⎝ ∂V ⎠T ⎝ ∂T ⎠V
⎛ ∂2 P ⎞ ⎛ ∂C ⎞ (b) ⎜ V ⎟ = T ⎜ 2 ⎟ ⎝ ∂V ⎠T ⎝ ∂T ⎠V
⎛ ∂S ⎞ ⎛ ∂V ⎞ (c) ⎜ ⎟ =⎜ ⎟ ⎝ ∂P ⎠T ⎝ ∂T ⎠V
⎛ ∂ 2V ⎞ ⎛ ∂C ⎞ (d) ⎜ P ⎟ = −T ⎜ 2 ⎟ ⎝ ∂P ⎠T ⎝ ∂T ⎠V
One mole of oxygen is expanded from a volume V1 to V2 at a constant temperature T . If the The gas is assumed to be a Van der Waals gas (a) the increment of the internal energy of the gas is zero ⎛1 1 ⎞ (b) the increment of the internal energy of the gas a⎜⎜ − ⎟⎟ ⎝ V1 V2 ⎠
Q29.
(c) heat exchange during the process is RT ln
⎡1 V2 − b 1⎤ + a⎢ − ⎥ V1 − b ⎣V2 V1 ⎦
(d) heat exchange during the process is RT ln
V2 − b V1 − b
⎛a⎞ The free energy for a photon gas is given by F = −⎜ ⎟VT 4 , where a is a constant. The ⎝3⎠
entropy S and the pressure P of the photon gas are 4 a (b) , P = T 4 aVT 3 3 3 4 4 a 4 (c) S = − aVT 3 (d) P = T 3 3 Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com
(a) S =
Email:
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Solutions MCQ (Multiple Choice Questions) Solution Ans. 1: (d) Ans. 2: (a) Solution: The sudden stretching of wire of compression of liquid is given by ⎛ ∂S ⎞ ⎛ ∂V ⎞ ⎜ ⎟ = −⎜ ⎟ ⎝ ∂P ⎠ T ⎝ ∂T ⎠ P
Ans. 3: (c) Solution: The rate of change of temperature pressure is given by Clausius-Clapeyron equation
which is given as
dP L = , L is latern heat dT T (V2 − V1 )
This can be derived by Maxwell’s first thermodynamical relation given as ⎛ ∂P ⎞ ⎛ ∂S ⎞ ⎟ =⎜ ⎟ ⎜ ⎝ ∂V ⎠ T ⎝ ∂T ⎠V
Ans. 4: (a) ⎛ ∂S ⎞ ⎛ ∂V ⎞ Solution: from Maxwell relation ⎜ ⎟ = − ⎜ ⎟ ⎝ ∂P ⎠T ⎝ ∂T ⎠V ⎛ ∂ ⎛ T ∂S ⎞ ⎞ ⎛ ∂ ⎛ ∂V ⎞ ⎞ ⎛ ∂ ⎛ ∂S ⎞ ⎞ ⎛ ∂ ⎛ ∂V ⎞ ⎞ ⎜ ⎟ ⎟ = −T ⎜ ⎜ ⎟ ⎟ ⎜ ⎟ ⎟ = −⎜ ⎜ ⎟ ⎟ ⇒⎜ ⎜ ⎝ ∂P ⎝ ∂T ⎠ P ⎠T ⎝ ∂T ⎝ ∂T ⎠ P ⎠ P ⎝ ∂T ⎝ ∂P ⎠T ⎠ P ⎝ ∂T ⎝ ∂T ⎠ P ⎠ P ⎛ ∂ 2V ⎞ ⎛ ∂C ⎞ ⇒ ⎜ P ⎟ = −T ⎜ 2 ⎟ ⎝ ∂P ⎠T ⎝ ∂T ⎠V
and α =
1 ⎛ ∂V ⎞ ⎛ ∂CP ⎞ 2 ⎜ ⎟ so ⎜ ⎟ = −TV α V ⎝ ∂T ⎠ P ⎝ ∂P ⎠T
Ans. 5: (a) Solution:
⎛ ∂S ⎞ ⎛ ∂S ⎞ dS = ⎜ ⎟ dT + ⎜ ⎟ dV ⎝ ∂T ⎠V ⎝ ∂V ⎠T
from
Maxwell
relation
⎛ ∂S ⎞ ⎛ ∂P ⎞ ⎜ ⎟ =⎜ ⎟ ⎝ ∂V ⎠T ⎝ ∂T ⎠V
and
⎛ T ∂S ⎞ ⎜ ⎟ = CV ⎝ ∂T ⎠V ⎛ ∂P ⎞ TdS = CV dT + T ⎜ ⎟ dV ⎝ ∂T ⎠V
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Ans. 6: (b) Solution:
⎛ ∂S ⎞ ⎛ ∂S ⎞ dS = ⎜ ⎟ dT + ⎜ ⎟ dP ⎝ ∂T ⎠ p ⎝ ∂P ⎠T
from
Maxwell
relation
⎛ ∂S ⎞ ⎛ ∂V ⎞ ⎜ ⎟ = −⎜ ⎟ and ⎝ ∂P ⎠T ⎝ ∂T ⎠ P
⎛ T ∂S ⎞ ⎜ ⎟ = CP ⎝ ∂T ⎠V ⎛ ∂V ⎞ TdS = CP dT − T ⎜ ⎟ dV ⎝ ∂T ⎠ P
Ans. 7: (a) Solution: dF = − SdT − PdV Ans. 8: (d) Solution: dG = − SdT + VdP ⎛ ∂G ⎞ ⎜ ⎟ = − S and ⎝ ∂T ⎠ P
⎛ ∂ ⎛ ∂G ⎞ ⎞ ⎛ ∂ ⎛ ∂G ⎞ ⎜ ⎟ =V ⇒ ⎜ ⎜ ⎟ ⎟ =⎜ ⎝ ∂P ⎠T ⎝ ∂P ⎝ ∂T ⎠ P ⎠T ⎝ ∂T
⎛ ∂S ⎞ ⎛ ∂V ⎞ ⎛ ∂G ⎞ ⎞ ⎜ ⎟ ⎟ ⇒ ⎜ ∂P ⎟ = − ⎜ ∂T ⎟ ⎝ ⎠T ⎝ ⎠P ⎝ ∂P ⎠T ⎠ p
And. 9: (a) Solution: Heat exchange is path dependent, so it is not perfect differential. Ans. 10: (b) Solution: dH = TdS + VdP
⎛ ∂H ⎞ ⎛ ∂H ⎞ V =⎜ ⎟ and T = ⎜ ⎟ ⎝ ∂P ⎠ S ⎝ ∂S ⎠ P
Ans. 11: (d) Solution: The change in Gibb’s free energy is given as
dG = VdP − SdT at constant P, dP = 0
at constant T, dT = 0
G ⇒ constant
Ans. 12: (a) Solution: Enthalpy H = PV + U
From laws of thermodynamics is given as
TdS = dU + PdV so,
G = U + PV − TS
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES ⇒ dG = dU + d ( PV ) − d (TS )
= dU + PdV + VdP − SdT − TdS by equation (i) dG = VdP − SdT
dG = VdP − S dT
(ii)
for isobaric dP = 0 , for isothermal
dT = 0
dG = 0
so, equation (ii) becomes
⇒ G = constant Ans. 12: (d) Ans. 13: (a) Solution: Gibb’s function is given as
G = H − TS
H = U + PV
And The enthalpy
G = U + PV − TS
dG = dU + d (PV ) − d (TS )
⇒
dG = dU + PdV + VdP − TdS − SdT dG = VdP − SdT According to question the process is isobaric i.e., SdP = 0 isothermal i.e., dT = 0 . So Equation (v) becomes as dG = 0 ⇒ G = c Ans. 14: (b)
3bS 2 ⎛ ∂E ⎞ Solution: TdS = dE + PdV ⇒ dE = TdS − PdV ⇒ ⎜ ⎟ = T ⇒ T = VN ⎝ ∂S ⎠V Ans. 15: (b)
[
Solution: F = − Nk BT ln a0V (k B T )
5/ 2
]
/ N , F = U − TS , U = F + TS
⎛ ∂F ⎞ ⎛ ∂F ⎞ ⎛ ∂F ⎞ dF = − SdT − PdV ⇒ ⎜ ⎟ ⎟ = − S or S = −⎜ ⎟ ⇒ U = F −T⎜ ⎝ ∂T ⎠V ⎝ ∂T ⎠V ⎝ ∂T ⎠V
(
F = − Nk B T ln C T
5/ 2
)
a 0Vk B5 / 2 where C = N
5 C 5 3/ 2 ⎛ ∂F ⎞ ⎛ ∂F ⎞ 5/ 2 5/ 2 − Nk B T T ⇒ T⎜ − Nk B T ⎜ ⎟ = − Nk B ln CT ⎟ = − Nk B T ln CT 5/ 2 2 2 CT ⎝ ∂T ⎠V ⎝ ∂T ⎠V
(
)
(
)
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 5 5 ⎛ ∂F ⎞ ⎛ ∂F ⎞ T⎜ ⎟ = Nk B T . ⎟ = F + Nk B T ⇒ U = F − T ⎜ 2 ⎝ ∂T ⎠V 2 ⎝ ∂T ⎠ V Ans. 16: (c) Solution: Now temperature of phase TA, TB , TC ⎛ dS ⎞ 1 Now ⎜ ⎟= ⎝ dE ⎠ T
Now
C S
dS will be stops then it will be zero for B - phase dE
B A
So TB = ∞
→E
And in C and A phases external energy of C phase is more so TC > TA Now TB > TC > T A Ans. 17: (c) 1 ⎛ ∂U ⎞ Solution: T = ⎜ i.e ⎟ =− 2cU ⎝ ∂S ⎠ V
U=−
1 2cT
Ans. 18: (b) Solution:
⎡ ap ⎤ 5 ⎛ ∂G ⎞ S = −⎜ ⎟ = R − R ln ⎢ 5/2 ⎥ ⎝ ∂T ⎠ p 2 ⎣ (RT) ⎦ ⎛ ∂S ⎞ 5 Cp = T ⎜ ⎟= R ⎝ ∂T ⎠ 2
Ans. 19: (c) ⎛ ∂F ⎞ Solution: S = −⎜ ⎟ = 4CT 3 ⎝ ∂T ⎠V
TdS = CV dT + PdV ⇒ dS = CV ⎛ ∂S ⎞ ⎜ ⎟ = ⎝ ∂T ⎠V T
⇒
CV P dT + dV T CV
⎛ ∂S ⎞ CV = T ⎜ ⎟ ⇒ CV = 3S ⎝ ∂T ⎠V
Ans. 20: (d) ⎛ ∂U ⎞ ⎛ ∂U ⎞ Solution: U = U (T , V ) ⇒ dU = ⎜ ⎟ dT + ⎜ ⎟ dV ⎝ ∂T ⎠V ⎝ ∂V ⎠T
⎛ ∂U ⎞ ⇒ dU = CV dT + ⎜ ⎟ dV ⎝ ∂V ⎠T
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES a a ⎛ ∂U ⎞ ⎜ ⎟ = 2 so dU = CV dT + 2 dV V ⎝ ∂V ⎠T V Ans. 21: (b) a ⎞ a ⎛ Solution: ⎜ P + 2 ⎟(V − b ) = RT and dU = CV dT + 2 dV V V ⎠ ⎝
For adiabatic process: dQ = 0 = dU + pdV − CV dT −
a a ⎛ RT − 2 dV = ⎜ 2 V ⎝V − b V
⎞ ⎟ dV ⎠
⎡ RT ⎤ − CV dT = dV ⎢ ⎣V − b ⎥⎦
−∫ −
CV dT dV =∫ RT V −b
CV ln Tk = ln(V − b ) R
V − b = (Tk )
− CV / R
V − b = T − CV / R × k − CV / R
(V − b )T C
V
T (V − b )
/R
R / CV
= k − CV / R = cons tan t
MSQ (Multiple Select Questions) Ans. 22: (b) and (c) Solution:
⎛ ∂G ⎞ dG = − SdT + VdP ⇒ ⎜ ⎟ = −S ⎝ ∂T ⎠ P
G = H − TS
⎛ ∂G ⎞ H = G + TS so H = G − T ⎜ ⎟ ⎝ ∂T ⎠ P
⎛ ∂ ⎛ G ⎞⎞ ⇒ H = −T 2 ⎜ ⎜ ⎟⎟ ⎝ ∂T ⎝ T ⎠ ⎠ P
Ans. 23: (a), (b) and (c) Solution: The first law of thermodynamics is given as dU + PdV = TdS
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES ⎛ ∂S ⎞ ⎛ ∂P ⎞ ⎛ ∂U ⎞ ⎛ ∂S ⎞ From Maxwell’s second relation ⎜ ⎟ =⎜ ⎟ ⎜ ⎟ = T⎜ ⎟ −P ⎝ ∂V ⎠T ⎝ ∂T ⎠V ⎝ ∂V ⎠ T ⎝ ∂V ⎠ T ⎛ ∂U ⎞ ⎛ ∂P ⎞ ⎜ ⎟ = T⎜ ⎟ −P ⎝ ∂V ⎠ T ⎝ ∂T ⎠V ⎛ ∂U ⎞ For real gas PV = nRT so ⎜ ⎟ =0 ⎝ ∂V ⎠ T a a ⎞ ⎛ ∂U ⎞ ⎛ For Vander wall’s ⎜ P − 2 ⎟ (V − b ) = nRT so ⎜ ⎟ = 2 V ⎠ ⎝ ∂V ⎠ T V ⎝ ⎛ ∂U ⎞ ⎛ ∂S ⎞ ⎛ ∂V ⎞ ⎜ ⎟ =T⎜ ⎟ − P⎜ ⎟ ⎝ ∂P ⎠T ⎝ ∂P ⎠T ⎝ ∂P ⎠T
⎛ ∂S ⎞ ⎛ ∂P ⎞ ⎟ =⎜ ⎟ ⎜ ⎝ ∂V ⎠T ⎝ ∂T ⎠V
Ans. 24: (a), (b), (c) and (d) Solution: (a) From H = U + PV , we obtain, ⎛ ∂H ⎞ ⎛ ∂U ⎞ ⎛ ∂V ⎞ ⎜ ⎟ =⎜ ⎟ + P⎜ ⎟ ⎝ ∂T ⎠ P ⎝ ∂T ⎠ P ⎝ ∂T ⎠ P
Let U = U ⎡⎣T , V (T , P ) ⎤⎦ . Let U = U ⎡⎣T , V (T , P ) ⎤⎦ . The above expression becomes ⎛ ∂U ⎞ ⎛ ∂U ⎞ ⎛ ∂U ⎞ ⎛ ∂U ⎞ ⎛ ∂U ⎞ ⎛ ∂V ⎞ dU = ⎜ ⎟ dT + ⎜ ⎟ dV ⇒ ⎜ ⎟ =⎜ ⎟ +⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠V ⎝ ∂V ⎠T ⎝ ∂T ⎠ P ⎝ ∂T ⎠V ⎝ ∂V ⎠T ⎝ ∂T ⎠ p ⎛ ∂H ⎞ ⎛ ∂U ⎞ ⎡ ⎛ ∂U ⎞ ⎤ ⎛ ∂V ⎞ ⎜ ⎟ =⎜ ⎟ + ⎢P + ⎜ ⎟ ⎥⎜ ⎟ ⎝ ∂T ⎠ P ⎝ ∂T ⎠V ⎣ ⎝ ∂V ⎠T ⎦ ⎝ ∂T ⎠ P
Hence
⎡ ⎛ ∂U ⎞ ⎤ ⎛ ∂V ⎞ cP − cV = ⎢ P + ⎜ ⎟ ⎥⎜ ⎟ ⎝ ∂V ⎠T ⎦ ⎝ ∂T ⎠ P ⎣
a ⎞ ⎛ (b) to find c p − cv for a Van der Waals gas ⎜ P + 2 ⎟ (V − b ) = RT V ⎠ ⎝
For the Van der Waals gas, we have R ⎛ ∂V ⎞ ⎜ ⎟ = 2a (V − b ) ⎤ ⎝ ∂T ⎠ P ⎡ RT − ⎢ ⎥ V3 ⎣V − b ⎦
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES R Hence, c p − cv = 2 1 − 2a (1 − b / V ) / VRT (c) and d) When V & T → ∞, cP − cV → R , which is just the result for an ideal gas. Ans. 25: (a) and (c) Solution: S = S (T , V ) ⎛ ∂S ⎞ ⎛ ∂S ⎞ ⎛ ∂S ⎞ ⎛ ∂S ⎞ ⎛ ∂S ⎞ ⎛ ∂V ⎞ dS = ⎜ ⎟ dT + ⎜ ⎟ dV ⇒ T ⎜ ⎟ =T ⎜ ⎟ +⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠V ⎝ ∂V ⎠T ⎝ ∂T ⎠ P ⎝ ∂T ⎠V ⎝ ∂V ⎠T ⎝ ∂T ⎠ P ⎛ ∂P ⎞ ⎛ ∂V ⎞ CP − CV = T ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠V ⎝ ∂T ⎠ P P = P (T , V )
⎛ ∂P ⎞ ⎛ ∂P ⎞ dP = ⎜ ⎟ dT + ⎜ ⎟ dV for constant pressure dP = 0 ⎝ ∂T ⎠V ⎝ ∂V ⎠T ⎛ ∂P ⎞ ⎛ ∂P ⎞ ⎛ ∂V ⎞ ⎛ ∂P ⎞ ⎛ ∂V ⎞ Put ⎜ ⎟ = −⎜ ⎟ ⎜ ⎟ in CP − CV = T ⎜ ⎟ ⎜ ⎟ one will ⎝ ∂T ⎠V ⎝ ∂V ⎠T ⎝ ∂T ⎠ P ⎝ ∂T ⎠V ⎝ ∂T ⎠ P ⎛ ∂P ⎞ ⎛ ∂V ⎞ CP − CV = −T ⎜ ⎟ ⎜ ⎟ ⎝ ∂V ⎠T ⎝ ∂T ⎠
2
P
Ans. 26: (b) and (d) ⎛ ∂P ⎞ ⎛ ∂V ⎞ Solution: CP − CV = −T ⎜ ⎟ ⎜ ⎟ ⎝ ∂V ⎠T ⎝ ∂T ⎠
So CP − CV =
γ −1 =
T α 2V
βT
2
it is known α = P
,we know that
β T α 2V T α 2V ⇒ β s − βT = S CV βT βT CV
1 ⎛ ∂V ⎞ 1 ⎛ ∂V ⎞ ⎟ ⎜ ⎟ and βT = − ⎜ V ⎝ ∂T ⎠ P V ⎝ ∂p ⎠T
C P βT = =γ CV β S
⇒ β s − βT = −
T α 2V CP
Ans. 27: (a), (b) and (d) ⎛ ∂S ⎞ ⎛ ∂P ⎞ Solution: from Maxwell relation ⎜ ⎟ =⎜ ⎟ ⎝ ∂V ⎠T ⎝ ∂T ⎠V
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES ⎛ ∂ ⎛ T ∂S ⎞ ⎞ ⎛ ∂ ⎛ ∂P ⎞ ⎞ ⎛ ∂ ⎛ ∂S ⎞ ⎞ ⎛ ∂ ⎛ ∂P ⎞ ⎞ ⎜ ⎜ ⎟ ⎟ =T ⎜ ⎜ ⎟ ⎟ ⎜ ⎟ ⎟ =⎜ ⎜ ⎟ ⎟ ⇒⎜ ⎝ ∂T ⎝ ∂V ⎠T ⎠V ⎝ ∂T ⎝ ∂T ⎠V ⎠V ⎝ ∂V ⎝ ∂T ⎠V ⎠T ⎝ ∂T ⎝ ∂T ⎠V ⎠V ⎛ ∂2 P ⎞ ⎛ ∂C ⎞ ⇒⎜ V ⎟ =T ⎜ 2 ⎟ ⎝ ∂V ⎠T ⎝ ∂T ⎠V ⎛ ∂S ⎞ ⎛ ∂V ⎞ Similarly another Maxwell relation is given by ⎜ ⎟ = − ⎜ ⎟ ⎝ ∂P ⎠T ⎝ ∂T ⎠V ⎛ ∂ ⎛ ∂S ⎞ ⎞ ⎛ ∂ ⎛ ∂V ⎞ ⎞ ⎛ ∂ ⎛ T ∂S ⎞ ⎞ ⎛ ∂ ⎛ ∂V ⎞ ⎞ ⎜ ⎜ ⎟ ⎟ = −⎜ ⎜ ⎟ ⎟ ⇒⎜ ⎜ ⎟ ⎟ = −T ⎜ ⎜ ⎟ ⎟ ⎝ ∂T ⎝ ∂P ⎠T ⎠ P ⎝ ∂T ⎝ ∂T ⎠ P ⎠ P ⎝ ∂P ⎝ ∂T ⎠ P ⎠T ⎝ ∂T ⎝ ∂T ⎠ P ⎠ P ⎛ ∂ 2V ⎞ ⎛ ∂C ⎞ ⇒ ⎜ P ⎟ = −T ⎜ 2 ⎟ ⎝ ∂P ⎠T ⎝ ∂T ⎠V
Ans. 28: (b) and (d) a ⎛ ∂U ⎞ Solution: (a) for Vander waal’s gas ⎜ ⎟ = 2 ⎝ ∂V ⎠T V ΔU =
⎛1 1 ⎞ a a − = a⎜⎜ − ⎟⎟ V1 V2 ⎝ V1 V2 ⎠
(b) We know a ⎞ ⎛ ⎜ P + 2 ⎟(V − b ) = RT V ⎠ ⎝ V2 RT a and work done W = ∫ PdV P= − V −b V 2 V1
ΔW = dW = Now
2 ⎡1 V2 − b 1⎤ RT a dV − ∫V V − b ∫V V 2 dV = RT ln V1 − b + a ⎢⎣V2 − V1 ⎥⎦ 1 1
V2
ΔQ = ΔW + ΔU = RT ln
V
⎡1 ⎡1 V2 − b 1⎤ 1⎤ + a⎢ − ⎥ + a⎢ − ⎥ V1 − b ⎣V2 V1 ⎦ ⎣V1 V2 ⎦
ΔQ = RT ln
V2 − b V1 − b
Ans. 29: (a) and (b) Solution: dF = − SdT − PdV ⎛ ∂F ⎞ ⎛ ∂F ⎞ ⎜ ⎟ = −S , ⎜ ⎟ = −P ⎝ ∂T ⎠V ⎝ ∂V ⎠T
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Chapter - 6 Phase Transition and Low Temperature Physics 6.1
Third Law of Thermodynamics and Attainable of Low Temperature The third law of thermodynamics is some time stated as follows: It is impossible for any process, no matter how idealized, to reduce the entropy of a system to its zero point value in a finite number of operations. Properties of material at low temperature At T → 0
CP
Heat Capacity
CV
CP = CV = 0 At T → 0 T
S→0 At T → 0 Thermal expansion coefficient α =
1 ⎛ ∂V ⎞ ⎜ ⎟ =0 V ⎝ ∂T ⎠ P
α =0 6.2
Production of Low Temperature: The Joule – Kelvin Expansion: The Joule – Kelvin Expansion is essentially a continuous steady – state flow process in which a compressed gas is made to expand adiabatically irreversibly through a porous plug and do work.
porous plug
p2
wool p1
p1
Constant temperatur Fig: A schematic diagram of the porous-plug experiment for Joule-Kelvin Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Let us assume that we start off with a gas of internal energy U1 and volume V1 . After passing through the porous plug let find internal energy and volume of the gas by U 2 and V2 . No heat enters the system.
So this work has to performed at the expense of internal energy
U 1 + P1V1 = U 2 + P2V2 H1 = H 2 Joule – Kelvin expansion is isenthalpic process H = H (T, P) ⎛ ∂H ⎞ ⎛ ∂H ⎞ dH = ⎜ ⎟ dP ⎟ dT + ⎜ ⎝ ∂P ⎠ T ⎝ ∂T ⎠ P
⎛ ∂H ⎞ CP = ⎜ ⎟ ⎝ ∂T ⎠ P
⎛ ∂H ⎞ dH = C P dT + ⎜ ⎟ dP ⎝ ∂P ⎠ T
dH = TdS + VdP
⎛ ∂S ⎞ ⎛ ∂H ⎞ ⎜ ⎟ = T⎜ ⎟ +V ⎝ ∂P ⎠T ⎝ ∂P ⎠T
⎛ ⎛ ∂V ⎞ ⎞ dH = C P dT + ⎜⎜V − T ⎜ ⎟ ⎟⎟dP ⎝ ∂T ⎠ P ⎠ ⎝
Hence H does not change dH = 0 ⎛ ∂H ⎞ The μ is defined ⎜ ⎟ ⎝ ∂P ⎠ H ⎞ T2 ⎛ ∂ 1 ⎛ ⎛ ∂V ⎞ ⎛ ∂T ⎞ ⎜⎜ T ⎜ ⎜⎜ ⎜ ⎟ =μ= ⎟ − V ⎟⎟ = C P ⎝ ⎝ ∂T ⎠ P ⎝ ∂P ⎠ H ⎠ C P ⎝ ∂T
⎛V ⎞ ⎞ ⎜ ⎟ ⎟⎟ ⎝ T ⎠P ⎠
where μ is known as the Joule – Kelvin Coefficient. The equation defines a curve in the (T, P) plane and is known as the inversion curve when μ = 0.
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H = H5 A
T1 T2
B
P2
H = H4 H = H3 H = H2 Inversion curve H = H1
P1
P
Figure: Curves of constant enthalpy. The bold curve is the inversion curve. Inside it, the gas is cooled on expansion. The temperature change ΔT = T2 − T1 produced in a A series of isoenthalphs, i.e. curve with H (T , P ) = constant. At T = T j μ = 0 known as inversion temperature.
T < Ti and μ = Positive . There is heating effect of gases i.e. temperature and pressure move in same direction. T > Ti μ = Negative . There is cooling effect i.e. temperature and pressure move in
opposite direction. Example: (a) For van der Waals gas. Prove that inversion temperature Ti =
2a where a and b Rb
are parameter used in van der Waals gas. (b) Why Hydrogen and helium shows heating effect as pressure increased at constant enthalphy. Solution: (a)
⎞ ⎛ ∂T ⎞ 1 ⎛ ⎛ ∂V ⎞ ⎜⎜ T ⎜ ⎟⎟ = ⎟ − V ⎟⎟ ⎝ ∂p ⎠ H C p ⎝ ⎝ ∂T ⎠ P ⎠
μ = ⎜⎜
For van der Waals gas a ⎞ ⎛ ⎜ P + 2 ⎟(V − b ) = RT V ⎠ ⎝
R ⎛ ∂V ⎞ ⎟ = ⎜ ⎝ ∂T ⎠ P ⎛ P + a ⎜ V2 ⎝
⎞ ⎟ ⎠
−
2a (V − b ) V3
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES (V − b ) ⎛ ∂V ⎞ ⎜ ⎟ = ⎛ 2a(V − b )2 ⎞ ⎝ ∂T ⎠ P ⎟ T ⎜⎜1 − 3 ⎟ RTV ⎝ ⎠
since b Tc, the gas phase of the matter cannot be converted into the liquid phase, but for T < Tc, the gas phase can in general be converted into the liquid phase. A
Melting Solid
Liqui
C
Critical P
Saturation
O
Triple B
Gas
T
Figure 1: P-T phase of one component 6.3.2 Equilibrium Between Two Phases
Let us consider an isolated system having a matter which is existing in two phases, denoted by 1 and 2, simultaneously in equilibrium with each other (Figure 2). Suppose V1 and V2 are volumes, N1 and N2 the number of particles, E1 and E2 the internal energies, and S1 and S2 the entropies of the two phases, respectively. For each phase, entropy is a function of its volume, number of particles (mass) and internal energy. From these relations, it follows that
B
T1 = T2
thermal equilibrium
P1 = P2
mechanical equilibrium
μ1 = μ 2
chemical equilibrium
A
E2
E1
Hence, when two different phases of the matter are in equilibrium,
their
temperatures,
pressures
and
chemical
potentials must be equal. If the chemical potentials are expressed
Figure 2: Equilibrium of two phases of an isolated one component system
as functions of pressure and temperature, we have
μ1 (P, T ) = μ 2 (P, T ) Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES where P (= P1 = P2) and T (= T1 = T2) are the common pressure and temperature,
respectively, of the two phases in equilibrium. Thus, from above equation , we have G1 (P, T )N 1 = G 2 (P, T )N 2
G1 G2 = N1 N 2 where G1(P, T) and G2(P, T) are the Gibbs free energies, and N1 and N2 the number of particles in the two phases, respectively. Since during the phase transition, the number of particles is not changing (i.e., N1 = N2), we have G1 (P, T ) = G 2 (P, T )
Hence, during the phase transition, the Gibbs free energy does not change. Gibbs energies
G1 and G2 of the two phases 1 and 2, respectively, can be exhibited as shown in figure 4.
G1 > G 2
G1 < G2 P
G1 = G2
1
2
T Figure 3: Phase equilibrium curve (G1 – G2) separating two phases 1 and 2. 6.3.3 Clapeyron-Clausius equation
When the two phases, denoted by 1 and 2, of the given matter are in equilibrium, we have G1 (P, T ) = G 2 (P, T )
where G1 and G2 are Gibbs free energies of the two phases, respectively, and P (= P1 =
P2) and T (= T1 = T2) are the common pressure and temperature, respectively, of the two phases. In the P-T diagram, along the phase-transition line, let us consider a point, where the pressure is P + dP and the temperature is T + dT so that we have G1 (P + dP , T + dT ) = G 2 (P + dP , T + dT )
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Using Taylor series expansion and neglecting the higher order terms, we have
⎛ ∂G ⎞ ⎛ ∂G ⎞ ⎛ ∂G ⎞ ⎛ ∂G ⎞ G1 (P, T ) + ⎜ 1 ⎟ dP + ⎜ 1 ⎟ dT = G2 (P, T ) + ⎜ 2 ⎟ dP + ⎜ 2 ⎟ dT ⎝ ∂T ⎠ P ⎝ ∂P ⎠ T ⎝ ∂T ⎠ P ⎝ ∂P ⎠ T Using above two equation we get V1 dP − V2 dP − S1 dT + S 2 dT = 0
dP S 2 − S1 ΔS = = dT V2 − V1 ΔV
ΔS =
ΔH L = T T
where ΔH (= H 2 − H 1 ) is the change in heat (enthalpy) which is the molar latent heat L. thus, from equation an we have
dP L = dT TΔV for V2 > V1, we have ⎛ ∂G2 ⎞ ⎛ ∂G1 ⎞ ⎟ ⎜ ⎟ ⎜ ⎝ ∂P ⎠ T ⎝ ∂P ⎠ T further, for S2 > S1, we have ⎛ ∂G2 ⎞ ⎛ ∂G1 ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ P ⎝ ∂T ⎠ P
(A)
(B)
6.3.4 Liquid-Vapour Phase Transition
Let us consider a phase transition from a liquid state to a vapor one. If Vi and Vg, respectively, denote the volume in the liquid and gas phases, and Lv is the heat of vaporization (latent heat for the transition from liquid to vapour), the Clapeyro-Clausius equation is
Lv dP = dT T (V g − Vi ) Since in the phase transition, Vg is always greater than Vi and the heat of vaporization Lv is positive and we have
dP >0 dT Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES It shows that the boiling point of a liquid increases with the increase in pressure.
Now, if the vapour pressure is low, i.e., Vg >> Vi, in comparison to Vg, and we have L dP = v dT TVg Using the ideal gas equation, PVg = RT, we have dP Lv P = dT RT 2 ln[P (T )] = −
dP Lv dT = P R T2 Lv 1 +C R T
where C is a constant of integration. At the critical point, we have P = Pc, T = Tc and equation is ln[Pc (Tc )] = −
Lv 1 +C R T
⎡L P(T ) = Pc (Tc ) exp⎢ v ⎣R
⎛ 1 1 ⎞⎤ ⎜⎜ − ⎟⎟⎥ ⎝ Tc T ⎠⎦
Here, we have assumed that the heat of vapourisation Lv is independent of the temperature. However, it depends on the temperature. Suppose it varies as Lv = a − bT , then for an ideal gas at low pressure, we have
dP (a − bT ) dT = P R T2 ln[P(T )] = −
a 1 b − ln (T ) + C RT R
where C is a constant of integration. At the critical point, we have P = Pc , T = Tc and equation is ln[Pc (Tc )] = −
a 1 b − ln(Tc ) + C R Tc R
On subtracting equation from and rearranging, we have ⎡ P(T ) ⎤ a ⎛ 1 1 ⎞ b ⎛ Tc ⎞ ln ⎢ ⎥ = ⎜⎜ − ⎟⎟ + ln⎜ ⎟ ⎣ Pc (Tc ) ⎦ R ⎝ Tc T ⎠ R ⎝ T ⎠ Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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(1) Gibbs free energy is continuous (2) First order derivative with respect to temperature and pressure have finite discontinuity i.e. entropy ( S ) and pressures ( P) have finite discontinuity. (3) Second and more higher order differential is infinite
g
1
2
1
2
2 1
T
T
T
Fig. 4: A schematic representation of first order phase transitions
Example: For a two phase system in equilibrium, p is a function of T only, so that ⎛ ∂P ⎞ ⎛ ∂P ⎞ ⎜ ⎟ =⎜ ⎟ ⎝ ∂T ⎠V ⎝ ∂T ⎠ S
Show that ⎛ dP ⎞ = TV ⎜ ⎟ βS ⎝ dT ⎠
CV
2
Solution: Let us take T and V as independent variables and write S = S (T , V )
so that ⎛ ∂S ⎞ ⎛ ∂S ⎞ dS = ⎜ ⎟ dV ⎟ dT + ⎜ ⎝ ∂V ⎠T ⎝ ∂T ⎠V
For an adiabatic process, it yields ⎛ ∂S ⎞ ⎛ ∂V ⎞ ⎛ ∂S ⎞ ⎟ ⎟ ⎜ ⎟ = −⎜ ⎜ ⎝ ∂V ⎠ T ⎝ ∂T ⎠ S ⎝ ∂T ⎠V
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Using first Maxwell relation, we obtain ⎛ ∂S ⎞ ⎛ ∂P ⎞ ⎛ ∂V ⎞ ⎜ ⎟ = −⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠V ⎝ ∂T ⎠V ⎝ ∂T ⎠ S ⎛ ∂P ⎞ Since CV = T ⎜ ⎟ we can write ⎝ ∂T ⎠V ⎛ ∂P ⎞ ⎛ ∂V ⎞ CV = −T ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠V ⎝ ∂T ⎠ S
⎛ ∂P ⎞ ⎛ ∂V ⎞ ⎛ ∂P ⎞ = −T ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠V ⎝ ∂P ⎠ S ⎝ ∂T ⎠ S
⎛ ∂P ⎞ ⎛ ∂P ⎞ = TV β S ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠V ⎝ ∂T ⎠ S
⎛ dP ⎞ = TV β S ⎜ ⎟ ⎝ dT ⎠
2
where β S is adiabatic compressibility. Example: Calculate under what Pressure water would boil at 120° C. One gram of steam
occupies a volume of 1677 cm3. Latent heat of steam = 540 cal/g, J = 4.2 × 107 erg/cal. atmospheric pressure = 1.0 × 106 dyne/cm3 Solution:
L × dT (V2 − V1 )
dP L = dT T (V2 − V1 )
dP =
V2 = 1677 cm3/g
V2 = 1 cm3/g
L = 4.2 × 107 × 540 erg/g
dT = 20 o k
dP = 0.725
P2 – P1 = .725
P2 = 0.725 + P1 = 1 + 0.725 = 1.725 Example: Liquid helium – 4 has normal boiling point of 4.2 k. However at pressure at 1 mm of
mercury it boils at 1.2 k. Estimate the average latent heat of vaporization of helium in this temperature range. Solution:
dP L L = = dT T (Vg − Ve ) TVg
PVg = RT
Vg =
RT P
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES dP LP dP L = = RT dT RT 2 dT T P P
T
dP L dT ∫P P = R T∫ T 2 0 0
ln
P L⎛ 1 1⎞ = ⎜ − ⎟ P0 R ⎜⎝ T0 T ⎟⎠
P P0 L= 1 1 − T0 T R ln
P0 = 746 mm
T0 = 4.2 k
P1 = 1 mm
T = 1.2
L = 93 J/mol. Example: Liquid helium boils at temperature T0 when its vapour pressure is equal to P0 we now
pump on the vapour and reduce the pressure to much smaller value P. Assume that the Latent heat L is approximately independent at temperature and helium vapour density is much smaller than that of liquid, calculate the approximate temperature Tm of the liquid in equilibrium with its vapour at pressure P. Express your answer in terms of L, T0, P0, Pm and any other required constants. Solution:
dP L = dT TΔV dP LP = dT RT 2 Tm =
ΔV = Vgas − Vliq ≈ Vgas Pm
T
dP L m dT ∫ P = R T∫ T 2 P0 0
T0 RT P 1 + 0 ln 0 L Pm
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Example: In the phase transition from a liquid state to a vapour state. The heat of vapourisation Lv varies with temperature T as Lv = a − bT 1 / 2 . Considering the gas as an ideal one at
low pressure, show that the pressure P (T ) at temperature T in terms of the critical pressure Pc (Tc ) at critical temperature Tc is given by ⎡ P(T ) ⎤ a ⎛ 1 1 ⎞ 1 b⎛ 1 ln ⎢ ⎥ = ⎜⎜ − ⎟⎟ + 2 ⎜⎜ 1 / 2 − 1 / 2 R ⎝T Tc ⎣ Pc (Tc ) ⎦ R ⎝ Tc T ⎠
⎞ ⎟ ⎟ ⎠
Solution: Clapeyron-Clausius equation for the phase transition from liquid to vapour is
Lv dP = dT T (V g − Vi )
where Lv is the heat of vapourisation and Vi and Vg , respectively, denote the volume is the liquid and gas phase. For low pressure, V g >> Vi L dP = v dT TV g
Using Lv = a − bT 1 / 2 , we have dP a − bT 1 / 2 = dT TV g
For an ideal gas equation, PV g = RT , and thus,
(
)
(
)
dP a − bT 1 / 2 P = dT RT 2
dP a − bT 1 / 2 P a dT b dT = − dT = 2 dT R T 2 R T 3/ 2 RT ln[P(T )] = −
a 1 b 1 + +C R T R T 1/ 2
On subtracting equation above and rearranging, we have ⎡ P(T ) ⎤ a ⎛ 1 1 ⎞ 2b ⎛ 1 1 ⎞ ln ⎢ ⎥ = ⎜⎜ − ⎟⎟ + ⎜⎜ 1 / 2 − 1 / 2 ⎟⎟ Tc ⎠ ⎣ Pc (Tc ) ⎦ R ⎝ Tc T ⎠ R ⎝ T
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES ⎛ ∂S ⎞ ⎛ ∂P ⎞ Derive Glausius-Clapeyron equation from Maxwell relation ⎜ ⎟ . ⎟ =⎜ ⎝ ∂V ⎠ T ⎝ ∂T ⎠V ⎛ ∂S ⎞ ⎛ ∂P ⎞ ⎟ ⎜ ⎟ =⎜ ⎝ ∂V ⎠ T ⎝ ∂T ⎠V
⎛ ∂S ⎞ ⎛ ∂P ⎞ T⎜ ⎟ = T⎜ ⎟ ⎝ ∂V ⎠ T ⎝ ∂T ⎠V
⎛ ∂Q ⎞ ⎛ ∂P ⎞ ⎟ ⎜ ⎟ = T⎜ ⎝ ∂V ⎠T ⎝ ∂T ⎠V
δQ = Ldm L ⎛ ∂P ⎞ T⎜ ⎟ = ⎝ ∂T ⎠V Vvap − Vliq
L ⎛ ∂P ⎞ ⎜ ⎟ = ⎝ ∂T ⎠V T (Vvap − Vliq )
(b) Draw Phase diagram for water and explain why water expand after freezing. Phase diagram for water The slope of solid liquid phase is negative. So from Clausius:-
P
solid
liquid
critical point
Clapeyron equation
L ⎛ ∂P ⎞ ⎜ ⎟ = ⎝ ∂T ⎠ sat T (Vliq − Vsolid ) ⎛ ∂P ⎞ ⎜ ⎟ = −ve ⎝ ∂T ⎠
gas
T
Vliq − Vsolid < 0
Vliq < Vsolid
so water expand on freezing 6.3.6 Second Order Phase Transition:
In some cases the state of matter does not change but the arrangement of its constituent particle changes. This kind of phase transition is known as second order phase transition. In the case of second order phase transition, no heat is evolved or absorbed. In second order phase transition 1. Gibbs free energy is continuous 1 1
2
2
g
g
T
p
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changes smoothly
p
T
p
T
3. Second order differential of Gibbs energy with respect to temperature ie specific heat and second order differential of Gibbs energy with respect to pressure ie isothermal and
αT
Cp
isobaric expansivity have finite discontinuity at critical temperature.
T
p
Figure 5: A schematic representation of a second order phase transition
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derivative will be infinite at critical temperature, P 75 50
solid He
25
vapour He
0
1
2
3
4
5
T
Figure: Phase diagram of helium
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES MCQ (Multiple Choice Questions)
Q1.
In phase transition of first order, there is a finitely discontinuity at the transition point of (a) Gibbs free energy G (b) The first-order derivatives of G (c) The second -order-derivatives of G (d) The higher -order-derivatives more than second order of G (where G is the Gibbs function)
Q2.
In phase transition of second order, there is a infinitely discontinuity at the transition point of G is the Gibbs function (a) Gibbs free energy G (b) The first-order derivatives of G (c) The second -order-derivatives of G (d) The higher -order-derivatives more than second order of G (where G is the Gibbs function)
Q3.
Which of the following is finitely discontinuous at transition temperature for first order transition? (a) Gibbs free energy (c) Specific Heat
Q4.
(b) Entropy (d) Volume expansibility
The Clausius-Clapeyron equation indicates that the increase of pressure increases the melting point: (a) in the case of all substances (b) in the case of substances which expand on solidification (c) in the case of substances which contract on solidification (d) in the case of substances which neither expand nor contract on solidification
Q5.
Consider the following statements in respect of first-order phase transition: 1. Clausius-Clapeyron latent heat equation holds well in the first-order phase transition. 2. There is change in entropy and volume in the first-order phase transition. Which of the above statements is/are correct?
(a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Q6.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES The Joule-Thomson expansion produced cooling:
(a) at all initial temperatures and pressures (b) above certain initial temperature (c) above certain initial pressure (d) below certain initial temperature Q7.
After Joule-Thomson expansion, the gas is: (a) always heated (b) heated or cooled depending upon the initial temperature of the gas (c) neither heated nor cooled at any temperature (d) always cooled
Q8.
Match List I and List II and select the correct answer using the codes given below the Lists: List I
List II
A Temperature of inversion in Joule.
1.
⎛ ∂U ⎞ ⎜ ⎟ =0 ⎝ ∂V ⎠T
2.
1 ⎡ 2a ⎛ ∂T ⎞ ⎤ − b⎥ ⎜ ⎟ = ⎢ ⎝ ∂P ⎠ H C P ⎣ RT ⎦
3.
∂ (PV )T ≠ 0 ∂P
4.
⎛ ∂V ⎞ −V ⎟ = 0 ⎜T ⎝ ∂T ⎠
Thomson effect is related to
B For perfect gases . C For a perfect gas Joule-Thomson .
effect vanishes because
D Deviation from Boyle’s law implies . Codes:
Q9.
A
B
C
D
(a)
2
4
1
3
(b)
3
1
4
2
(c)
2
1
4
3
(d)
3
4
1
2
On the inversion curve, the Joule-Thomson coefficient is:
(a) positive (b) zero (c) negative (d) infinite Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES MSQ (Multiple Select Questions)
Q10.
Which of the following is correct statement about first order phase transition. (a) The Gibbs free energy is continuous at transition temperature (b) The entropy is finitely discontinues at transition temperature (c) Pressure is infinitely discontinues at transition temperature (d) Specific heat is infinitely discontinuous at transition temperature
Q11.
Which of the following is incorrect statement about second order phase transition? (a) The Gibbs free energy is continuous at transition temperature (b) The entropy is finitely discontinues at transition temperature (c) Pressure is infinitely discontinues at transition temperature (d) Specific heat is infinitely discontinuous at transition temperature
Q12.
Which of the following is finitely discontinuous at transition temperature for second order transition?
Q13.
(a) Gibbs free energy
(b) Entropy
(c) Specific Heat
(d) Volume expansibility
Which of the following is correct statement for phase transition? (a) Clausius-Clapeyron latent heat equation holds well in the first-order phase transition. (b) There is not change in entropy and volume at critical temperature in the second-order phase transition. (c) The change of phase when water transformed in Ice is first order transition (d) The concept of superconductor and super fluid can be explain by second order transition
Q14.
Which is correctly matched for van der Waal’s gas?
2a Rb 2a = (b) Boyle temperature is given by TBoyle Rb 8a (c) Critical temperature is given by Tcritical = 27bR a (d) Boyle temperature is given TBoyle = Rb Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com (a) Inversion temperature is given by Tinversion =
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fiziks Q15.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Which of the following pairs are correctly matched?
(a) Law of liquefaction of gases → triple point (b) Law of liquefaction of gases → critical temperature (c) Cooling effect of gas → inversion temperature (d) Phase transition → Boyle temperature Q16.
⎛ ∂T ⎞ The Joule-Thomson coefficient μ = ⎜ ⎟ : ⎝ ∂P ⎠ H
(a) less than zero at all temperatures and pressures for ideal gas (b) zero at all temperatures and pressures for ideal gas (c) less than zero at all temperatures and pressures for real gas (d) can have any value depending on pressure and temperature for real gases Q17.
Consider the following statements: A gas can be liquefied by increasing the pressure (a) Above the critical below critical temperature only. (b) Only when the temperature of the enclosed gas in below the critical temperature. (c) Only when the volume of the enclosed gas is below the critical volume. (d) Ideal gas can not be liquefied.
Q18.
Consider the following statements: When a compressed real gas is allowed to pass through a narrow hole, the temperature (a) Falls for some gases if T < Ti (b) Falls for some gases if T > Ti . (c) Rise for some gases if T > Ti . (d) Rise for some gases T < Ti
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Solutions MCQ (Multiple Choice Questions) Ans. 1: (b) ⎛ ∂G ⎞ Solution: In first order phase transition Entropy S = −⎜ ⎟ is discontinuous at transition ⎝ ∂T ⎠ P
point Phase transition and low temperature physics. Ans. 2: (d) Ans. 3: (b) ⎛ ∂G ⎞ ⎛ ∂G ⎞ Solution: for first order phase transition G is continuous S = − ⎜ ⎟ is ⎟ and V = − ⎜ ⎝ ∂T ⎠ P ⎝ ∂p ⎠ S
⎛ ∂ 2G ⎞ finitely discontinuous and CP = −T ⎜ 2 ⎟ is infinitely discontinuous ⎝ ∂T ⎠ P
Ans. 4: (c) Solution: Clausius-Clapeyron equation is given as dP L ⇒ = dT T (V2 − V1 )
If V2 > V1 ,
dP is positive dT
If volume increases the pressure increases with T . Ans. 5: (c) Ans. 6: (d) Solution: Cooling effect is produced when temperature is below a temperature known as
temperature of inversion. Ans. 7: (b) Solution: The drop of temperature dT in Joule.
Thomson effect is given as dT =
P CP M
⎛ 2a ⎞ − b⎟ ⎜ ⎝ RT ⎠
(i)
2a 2a − b = 0 ⇒ Ti = RT Rb Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com
⇒ No change in temperature if
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES This temperature is known as temperature of inversion Equation (i)
⇒
dT > 0 if
2a −b > 0 RT
⇒ T<
2a Rb
⇒ If temperature T of the gas is less than temperature of inversion then cooling effect is observed otherwise heating effect. Ans. 8 : (c) Solution: (i) The temperature of inversion in Joule-Thomson effect is the temperature above of
which all gases show heating effect. Below of this all gases show colling effect and at the temperature of inversion no effect. This is given as 1 ⎛ 2a ⎛ ∂T ⎞ ⎞ − b⎟ ⎜ ⎟ = ⎜ ⎝ ∂P ⎠ H C P ⎝ RT ⎠
(ii) The internal energy of a perfect gas does not depend on volume this depends only on temperature. Thus, ⎛ ∂U ⎞ ⎛ ∂U ⎞ ⎟ =0 ⎟ =⎜ ⎜ ⎝ ∂V ⎠ T ⎝ ∂P ⎠ T
a ⎛ ∂U ⎞ ⎜ ⎟ = 2 ⎝ ∂V ⎠ T V
For real gases
(iii) For a perfect gas Joule-Thomson effect vanishes because
⎛ ∂V ⎞ −V ⎟ = 0 ⎜T ⎝ ∂T ⎠ (iv) The pressure P and PV is related as
PV = A + BP + CP 2 + DP 3 + A > B > C > D … are virial constants.
The Boyle’s law is applied when P is very low hence when PV = A
⇒
∂ ( PV ) = 0 ∂P
So, deviation from Boyle’s law ⇒
∂ (PV )T ≠ 0 ∂P
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Ans. 9: (b)
representing equilibrium of same enthalpy. If temperature is not very high then such curves pass through a maximum, called the inversion point. The locus of all inversion points is called the inversion curve. Since, towards the left of the
temperature
Solution: An isenthalpic curve is the locus of all points
inversioncurve cooling heating T
maximum of either curve of constant enthalpy, the gas shows cooling effect, the region inside the
P
pressure
inversion curve is the cooling region, for this region gas shows cooling effect. On the other hand, the region outside the inversion curve the gases show heating effect and region is called region of heating. The Joule-Thomson coefficient is defined as ⎤ 1 ⎡ ⎛ ∂V ⎞ ⎛ ∂T ⎞ ⎟ −V ⎥ ⎟ = ⎢T ⎜ ⎝ ∂P ⎠ H C P ⎣ ⎝ ∂T ⎠ P ⎦
μ =⎜
On the inversion μ = 0
Thus, Joule-Thomson coefficient is zero.
MSQ (Multiple Select Questions) Ans. 10: (a), (b) and (d) ⎛ ∂G ⎞ ⎛ ∂G ⎞ Solution: For first order phase transition G is continuous S = − ⎜ ⎟ is ⎟ and V = − ⎜ ⎝ ∂T ⎠ P ⎝ ∂p ⎠ S
⎛ ∂ 2G ⎞ finitely discontinuous and CP = −T ⎜ 2 ⎟ is infinitely discontinuous ⎝ ∂T ⎠ P
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Ans. 11: (b), (c) and (d) ⎛ ∂G ⎞ ⎛ ∂G ⎞ Solution: For second order phase transition G is continuous S = − ⎜ ⎟ is ⎟ and V = − ⎜ ⎝ ∂T ⎠ P ⎝ ∂p ⎠ S
⎛ ∂ 2G ⎞ continuous and CP = −T ⎜ 2 ⎟ is finitely discontinuous . ⎝ ∂T ⎠ P
Ans. 12: (c) and (d) ⎛ ∂G ⎞ ⎛ ∂G ⎞ Solution: For second order phase transition G is continuous S = − ⎜ ⎟ is ⎟ and V = − ⎜ ⎝ ∂T ⎠ P ⎝ ∂p ⎠ S
⎛ ∂ 2G ⎞ continuous and CP = −T ⎜ 2 ⎟ is finitely discontinuous ⎝ ∂T ⎠ P
Ans. 13: (a), (b), (c) and (d). Ans. 14: (a), (c) and (d) Solution: Tinversion =
8a 2a a , TBoyle = and Tcritical = Rb Rb 27bR
Ans. 15: (b) and (c) Ans. 16: (b) and (d) Solution: Joule-Thomson coefficient for ideal gas is zero
i.e.,
⎛ ∂T ⎞ ⎜ ⎟ =0 ⎝ ∂P ⎠ H
whereas for real gas it is given as 1 ⎛ 2a ⎞ ⎛ ∂T ⎞ − b⎟ ⎜ ⎟ = ⎠ ⎝ ∂P ⎠ H C P ⎝ RT
μ =⎜
a, b are Vander Waal’s constant. Ans. 17: (b) and (d) Solution: When temperature of a gas is above a fixed temperature, known as critical
temperature, the gas cannot be liquid whatever the pressure is applied. If temperature of the gas is equal to or smaller than the critical temperature, then it can be liquefied. Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Ans. 18: (a) and (c) Solution: The fall of temperature is given as dT =
⇒
Ti =
dP ⎛ 2a ⎞ − b⎟ ⎜ C P ⎝ RT ⎠
2a Rb
Thus, temperature T > Ti then heating effect, if temperature T < Ti then cooling effect.
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