Gaseous State and Thermodynamics
Short Description
Dear sir...
Description
Vidyamandir Classes
Gaseous State and Thermodynamics
Gaseous State and Thermodynamics FUNDAMENTAL ASSUMPTIONS OF KINETIC THEORY OF GASES
Section - 1
1.
All gases consist of molecules which are considered as rigid, perfectly elastic, solid spheres. These molecules are extremely small i.e., the volume occupied by the molecules of the gas is taken negligible as compared to the volume of the containing vessel.
2.
The molecules are in a continuous random motion. They are moving in all directions with all possible velocities. There is no particular direction of motion of the particles.
3.
The molecules during their motion collide with one another and also with the walls of the vessel. At each collision, their velocities change both in magnitude and direction; however in the steady state, molecular density remains uniform throughout the gas and does not change with time.
4.
No forces of attraction or repulsion are exerted on a gas molecule by other molecules or by the container except during collisions. This means, the energy of the gas is purely kinetic.
5.
The collisions among the molecules or with the walls of container are purely elastic. A collision is said to perfectly elastic if the energy of the molecules remains same before and after the collision.
6.
Molecules travel in a straight line with a uniform velocity in between the collisions. The average distance travelled by the molecules between the successive collisions is called as mean-free path of a molecule. The time spent during a collision is assumed to be negligibly small as compared with the time taken by the molecule in traversing a mean-free path.
Note : A gas which satisfies all of above set of assumptions of kinetic theory of gases under all conditions of temperature and pressure is called as an ideal gas or perfect gas. A real gas shows appreciable deviations from the perfect gas. At low temperature and high pressure, gases show deviation from the ideal behaviour.
Evidences In Support of Molecular Motion Diffusion : The phenomenon of diffusion presents a remarkable evidence of molecular motion of matter. The particles of one substance diffuse into another against gravity due to the molecular motion. H2 diffuses into CO2 when a jar filled with H2 is held mouth to mouth over a jar filled with CO2, though H2 is lighter than CO2.
Evaporation : The phenomenon of evaporation is related with the motion of molecules. The molecules of liquid are quite free to move about. At any temperature, all the molecules don’t move with the same velocity, some move faster than the others. Evaporation involves the escape of fast moving molecules (having speed greater than average speed). This results in lowering of average speed, which in turn results in lowering of temperature, and thus producing a cooling effect. Self Study Course for IITJEE with Online Support
Section 1
1
Gaseous State and Thermodynamics
Vidyamandir Classes
Brownian Motion : Botanist Robert Brown observed the irregular motion of plane spores floating in water. Such motions are called as Brownian Motion. Here the motion of each particle is due to inequality of the impact of surrounding molecules on it. For example: motion of smoke particles in air is a case of Brownian motion. This supports the view that the molecules of a substance are in constant motion.
Ideal gas equation : For n moles of an ideal gas at pressure P, absolute temperature T occupying a volume V, the following equation holds: PV = nRT (R : Ideal gas constant) The gas equation can also be written in the following form. PM = dRT where M is the molecular weight and d is the density of the gas at pressure P and absolute temperature T.
Pressure Exerted by a Gas Consider a container of volume V occupied by a gas. Let m = mass of the gas in the container N = number of molecules in the container If m0 is the mass of one molecule, m = m0N If N0 is the Avogadro number and M is the molecular weight of the gas, M = m0N0 If the molecules are moving speeds c1, c2, c3, ................., cN ; then the root mean square speed of the gas is defined as :
c2
= crms =
2 c12 +c 22 + c32 + .........+cN
N
The pressure exerted by the gas is : P
1 m0 N 2 crms 3 V
or
P
m N 1 2 crms where = density of gas = 0 3 V
Root Mean Square Velocity (crms) crms can be calculated with any of the following results.
crms =
2
3P 3 RT 3kT M m0
Section 1
Self Study Course for IITJEE with Online Support
Gaseous State and Thermodynamics
Vidyamandir Classes
For the same gas at two different temperatures T1 and T2, crms will be different.
crms 1 T 1 crms 2 T2
(as Crms T)
For two gases at different temperatures T1 and T2 ; c rms (1)
=
c rms (2)
T1 M 2 T 2M 1
Average Speed : (cav) cav = cav =
c1 c2 c3 ..... cN N 8RT pM
Kinetic Energy of Molecules Degrees of Freedom : The degrees of freedom of a dynamical system are defined as the total number of independent quantities required in order to describe the position or motion of the system completely.
Mono-atomic gas molecule : The molecules of a mono-atomic gas (He, Ne, Ar) consist of one atom only. It is capable of only translatory motion in free space which has three compo nent s namely: X, Y and Z directions. Hence it will have three degrees of freedom per molecule.
Di-atomic gas molecule : Each molecule of a diatomic gas (H2, N2, O2, Cl2, etc.) consists of two atoms. In the given diagram, a diatomic molecule A2 (A-A) is shown. A2 is capable of translatory motion in space and thus possess three degrees of freedom. Beside this the molecule may rotate about YY’ axis (perpendicular to the molecular axis and the plane of the paper) or ZZ’ axis (perpendicular to the molecular axis and the plane of the paper). Rotation about molecular axis (XX’) is neglected (because the size of atom is negligible). Thus due to rotational motion, the molecules have two degrees of freedom. Hence a diatomic molecule has five degrees of freedom. Note : A triatomic molecule (CO2, H2O, SO2 etc) has six degrees of freedom (three of them for translatory and three for rotational motion). Self Study Course for IITJEE with Online Support
Section 1
3
Gaseous State and Thermodynamics
Vidyamandir Classes
Kinetic Energy of Translation per molecule : The average kinetic energy or kinetic energy per molecule due to translatory motion is also known as mean KE of translation. 1 1 1 2 m0 c12 m0 c22 ..... m0 cN 1 2 2 2 2 3 1 m c2 (KE) mean = m0 crms 0 rms N 2 23 =
3 PV 2 N
3 kT 2
as PV = NKT
3 (KE)mean = kT 2
KE of translation per mole = (KE)mean N0 =
3 3 kN 0T RT 2 2
3 3 KE of translation in n moles of a gas = n RT = PV 2 2
Equipartition of Energy The Law of equipartition of energy states that for a dynamical system in thermal equilibrium, the total energy of the system is equally divided amongst the various degrees of freedom. The share of each degree of freedom is 1/2 kT.
For monoatomic gas : Total energy per molecule = 3/2 kT Total energy per mole = 3/2 RT
For diatomic gas : Total energy per molecule = 5/2 kT Total energy per mole = 5/2 RT
For triatomic gas : Total energy per molecule = 3kT Total energy per mole = 3RT
4
Section 1
Self Study Course for IITJEE with Online Support
Gaseous State and Thermodynamics
Vidyamandir Classes
Illustration - 1 Calculate the root mean square speed of argon atoms at room temperature (25°C). At what temperature will the root mean square speed will be half that value ? SOLUTION :
Crms 1 T 1 Crms 2 T2
MAr = 0.040 Kg and R = 8.3 J/mol/K
3 RT 3 8.3 273 25 M 0.040
Crms =
= 430.7 m/s
1 T1 2 298 T1 = 74.5 K
For the same gas :
= 198.5 C
Illustration - 2
Calculate the pressure of Hydrogen in a cylinder of capacity 10 lt given that the total kinetic energy of translation is 7.5 × 103 J. What is the total kinetic energy of molecules in the cylinder ? (1 atm pressure = 105 N/m2)
SOLUTION :
The total KE of translation in n moles for a diatomic gas is :
5 Total KE for diatomic gas = PV 2 5 KE = × 5 × 105 × 10 × 10–3 J 2 = 12500 J
3 3 nRT PV 2 2 3 Hence 7.5 × 103 J = P (10 × 10–3m3 ) 2
Illustration - 3
P = 5 × 105 N/m2
An Oxygen storage tank has a capacity of 0.05 m3. The gas pressure is 100 atm. at 27°C.
Determine : (a) (c)
the total KE of translation of molecules, the total KE.
(b)
the average KE of translation of molecules and
SOLUTION : (a) Total KE of translation = 3/2 PV 5
(c) The total KE (for diatomic gas) = 5/2 nRT 5
= 3/2 (100) x 10 x 0.05 = 7.5 × 10 J (b) Average KE = 3/2 kT = 3/2 (1.38 × 1023) (300) = 6.21 × 1021 J
Self Study Course for IITJEE with Online Support
= 5/2 PV KE = 5/2 (100) × 105 (0.05) = 1.25 × 106 J.
Section 1
5
Gaseous State and Thermodynamics
Vidyamandir Classes
Illustration - 4 Find the number of Hydrogen molecules in 1 cm3 if the pressure is 200 mm Hg and the root mean square velocity of hydrogen is 2400 m/s in these conditions. SOLUTION :
200 P = 200 mm Hg × 1.01 × 105 N/M2 760
Substituting these values P
V = 1 cm3 = 106 m3
N
M m0 = N N0 = 6.023 × 1023 0
N 3
1 m0 N 2 crms 3 V
3 PV 2 m0 crms
200 1.01 105 1 10 6 6.023 10 23 760 0.002 2400 2400
= 4.17 × 1018 molecules.
Illustration - 5 What is the energy of the rotational motion of the molecules contained in 1 Kg of nitrogen at a temperature of 7°C ? SOLUTION : n = moles = 1000/28
2 Rotational Energy = nRT 2
T = 273 + 7 = 280 K =
For a diatomic gas, 2 degrees of freedom are for rotational motion.
1000 8.3 280 J = 8.3 104 J 28
Illustration - 6 A vessel A with a capacity of V1 = 3 lt contains gas at a pressure of P1 = 2 atm and a vessel B with a capacity of V2 = 4 lt contains the same gas at a pressure of 1 atm. The temperature is 0°C in A and 27°C in B. What will be the pressure in the vessels if they are connected by a tube ? SOLUTION : The vessels are initially at temperatures : T1 = 273 K
and
T2 = 300 K
Let n1 and n2 be the moles in A and B respectively. Initial total energy = Final total energy
x x x n1 RT 1 n 2 RT 2 n1 n 2 RT 2 2 2
where x = 3 or 5 according to whether the gas is monoatomic and diatomic.
T = final temperature = 283.2K
Final pressure = P =
n1 n2 RT V1 V2
4 283.2 6 P = 1.43 atm. 272 300 7
Note : It may be useful to remember the following : PV P V P 1 1 2 2 V1 V2
6
Section 1
and
nT n T T= 11 2 2 n1 n2 Self Study Course for IITJEE with Online Support
Vidyamandir Classes
Gaseous State and Thermodynamics
Section - 2
GAS LAWS Before moving ahead let us recapitulate important gas laws.
Boyle’s Law : At a certain temperature, the pressure of a given quantity of any gas varies inversely with the volume.
1 (for given n and T) V PV = constant P
If P1 is the pressure when volume is V1 and P2 is the pressure when the volume is V2 (T is constant), then: P1V1 = P2V2
Charles’ Law : At a constant pressure the volume of a given quantity of any gas varies directly with the absolute temperature. V T (for given n and P) V/T = constant If V1 and V2 are volumes of a gas at T1 and T2 and the pressure is kept constant, then:
V1 V2 T1 T2
Avogadro Law : It states that equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules. If volume V1 of one gas contains N1 molecules and volume V2 of another gas contains N2 molecules at same pressure and temperature conditions, then:
V1 V2 N1 N2
Dalton’s Law of partial pressures : The total pressure of a mixture of non-reacting gases is equal to the sum of partial pressures of each gas. The partial pressure of a gas is defined as the pressure it would exert if it were alone in the container at same temperature. If PT is the total pressure, p1, p2, p3, ......... are the partial pressures of a number of gases in the container, then according to Dalton’s Law of partial pressures: PT = p1 + p2 + p3 + ........... If n1, n2, n3, be the respective number of moles of the gases :
PT
RT n1 2 ..... V
Self Study Course for IITJEE with Online Support
Section 2
7
Gaseous State and Thermodynamics
Vidyamandir Classes
Boltzmann Constant (k) : In the gas equation PV = nRT, the values of R can be taken as: R = 8.314 J/mol/K = 0.082 l-atm/mol/K = 2 cal/mol/K If N0 is the Avogadro number then, k = R/N0 is known as Boltzmann constant. k = 1.38 × 10–23 J per molecule per degree kelvin .
Gas equation in terms of k : R PV = nRT = n N0 N T = n N0 kT 0 PV = NkT
N = nN0 = number of molecules
Illustration - 7
10 gm of oxygen are subjected to a pressure of 3 atm. at a temperature of 10°C. Heating at a constant pressure, the gas is expended to 10 lt. Find : (a) the volume of the gas before expansion (b) the temperature of gas after expansion (c) the density of gas before expansion (d) the density of gas after expansion. SOLUTION : (a) V1
nRT1 10 0.0821 283 = 2.42 L P1 32 3
(b) As pressure is constant :
V1 V2 T1 T2
V T2 2 T1 = 1169.4 K V1
(c) density = d1 =
mass 10 = 4.13 g/L V1 2.42
(d) density = d2 =
mass 10 = 1 g/L V2 10
Illustration - 8 A vertical cylinderical tank 1 m high has its top end closed by a tightly fitted frictionless piston of negligible weight. The air inside the cylinder is at an absolute pressure of 1 atm. The piston is depressed by pouring mercury on it very slowly. How far will the piston descend before mercury spills over the top of cylinder ? The temperature of the air inside the cylinder is maintained constant. SOLUTION :
Let
A = area of cross-section of cylinder.
As the temperature is constant, Use 8
P1 V1 = P2 V2 for air inside the cylinder. Section 2
P1 = initial air pressure = 1 atm = 76 cm Hg P2 = final air pressure = (76 + x) cm Hg V1 = initial volume = 100 A V2 = final volume = (100 x) A 76 (100 A) = (76 + x) (100 x)A 7600 = 7600 76x + 100x x2 x = 24 cm The piston descends by 24 cm. Self Study Course for IITJEE with Online Support
Gaseous State and Thermodynamics
Vidyamandir Classes
Illustration - 9
An air bubble is released from the bottom of a lake at a depth of 11 m. The temperature at the bottom is 4°C and that at the surface is 12°C. What is the ratio of the bubble’s radius at the surface to its radius at bottom ? Assume density of water = 1000 Kg/m3, atmospheric pressure = 75 cm of Hg and density of mercury = 13600 Kg/m3.
SOLUTION : Let P1 = pressure at the bottom of lake = P2 (pressure at surface) + P (due to water column) Taking cm of Hg as the unit of pressure, P1 = 75 cm Hg Pressure of water column of height h = h rw g P = Hwg
P1 = 75 + 80.88 = 155.88 cm Hg. PV PV Using 1 1 2 2 for the air inside the bubble T1 T2
where V1,V2, are the bottom and surface ; T1 and T2 are temperatures at bottom and surface 155.88 V1 75V2 273 4 273 12
h h P w ( ) g w cm of Hg. Hg Hg Hg =
V2 2.138, V1
r2 1/ 3 2.138 1.288 r1
1100 1 cm Hg = 80.88 cm Hg 13.6
Illustration - 10 A vessel containing 1 gm of oxygen at a pressure of 10 atm and a temperature of 47°C. It is found that because of a leak, the pressure drops to 5/8 th of its original value and the temperature falls to 27°C. Find the volume of the vessel and the mass of oxygen that has leaked out. SOLUTION : The pressure, temperature and number of moles of oxygen in the vessel change due to leak while the volume remains fixed. Hence using PV = nRT we have :
P1 P 2 n1T1 n2T2 320 5 1 1 n2 moles 300 8 32 48
Self Study Course for IITJEE with Online Support
volume of vessel = V
n1 RT1 P1
mass leaking out = 1 n2 (32) = 1 2/3 = 1/3 gm.
V
1 0.082 320 = 0.082 L 32 10
Section 2
9
Gaseous State and Thermodynamics
Vidyamandir Classes
Illustration - 11 A bulb of 500 cm3 is joined by a narrow tube of another bulb B of 250 cm3 and the whole system is initially filled with air at S.T.P. and sealed. If the temperature of the bulb A is now raised to 100°C and that of B is kept constant, find : (a) the new pressure in the system. (b) mass of air which is transferred from one bulb to another during heating. (density of air at S.T.P. = 1.29 g/l) SOLUTION :
(a) Let P be the final pressure of air in the bulbs. As the bulbs are connected, pressure will be equal in both bulbs. Let TA and TB be the final temperatures of the bulbs.
TA = 373 K,
TB = 273 K
If P0 , T0 are the S.T.P. conditions, then : Initial moles in the system = Final moles in the system
P0VA P0VB PVA PVB RT0 RT0 RTA RTB Using P0 = 1 atm, T0 = 273 K, 1 500 250 P 500 250 373 273 273 P = 1.22 atm.
(b) M = molecular weight of air M = (dSTP) (22.4) = 1.29 22.4 = 28.896 gm. Mass transferred = moles M PV PV = B 0 B M RTB RT0 =
250 103 (1.22 1) 28.896 0.082 273
mass transferred = 0.0709 gm. 10
Section 2
Self Study Course for IITJEE with Online Support
Gaseous State and Thermodynamics
Vidyamandir Classes
Illustration - 12 A thin tube sealed at both ends, is 100 cm long. It lies horizontally, the middle 0.1 m containing mercury and the two ends containing air at standard atmospheric pressure. If the tube is turned to a vertical position, by what amount will the mercury be displaced ? SOLUTION :
Let P0, V0 be the pressure and volume of the air columns initially. Let P1, P2 be the final pressures in cm of Hg and V1, V2 be the final volumes. Applying Boyle’s Law on both air columns, we get : P0 V0 = P1 V1
and
PV P (45) P1 0 0 0 V1 45 x
and
PV P (45) P2 0 0 0 V2 45 x
P0 V0 = P2 V2
Balancing forces on Hg columns in vertical position, we get : P (below Hg column) = P(above Hg column ) + 10 cm P2 = P1 + 10 (if P1, P2 are in cm of Hg) Substituting to get :
45 P0 45 P0 + 10 = 45 + x 45 x Using P0 = 76 cm Hg, we get : x2 + 684 x 2025 = 0 On solving, x = 2.95 cm. NOW ATTEMPT IN-CHAPTER EXERCISE-A BEFORE PROCEEDING AHEAD IN THIS EBOOK
Self Study Course for IITJEE with Online Support
Section 2
11
Gaseous State and Thermodynamics
Vidyamandir Classes
THERMODYNAMICS
Section - 3
Temperature : It may be defined as that thermal property which attains the same value for all bodies placed in thermal contact with each other.
Heat Energy : The transfer or flow of internal energy caused by the difference of temperatures between two bodies is called as Heat Energy.
Zeroth Law of thermodynamics : It two bodies A and B are in thermal equilibrium with another body C, then they are also in thermal equilibrium with each other.
Thermodynamical system : It is a collection of an extremely large numbers of atoms or molecules so that they together have some pressure (P), volume (V) and temperature (T). Examples: a gas, vapour, steam, vapour in contact with the liquid.
Work Done and Internal Energy Work Done By a Gas During Expansion Let us consider an ideal gas enclosed in a perfectly insulated cylinder fitted with a non-conducting and frictionless piston. Let P be the pressure exerted by the gas and V be the volume of the gas at any particular instant. Let A be the area of crossection of the piston. The force exerted by the gas on the piston = P A If the piston moves through an infinitesimal distance dx, this force can be assumed constant, and the work done (dW) is given as : dW = (PA) dx = P dV (A dx = dV = infinitesimal change in volume) W = PdV (a)
If volume is kept constant the piston is not displaced.
(b)
W =0
If pressure is kept constant, V2
W
PdV P dV P V2 V1 V1
12
Section 3
V2 V1
Self Study Course for IITJEE with Online Support
Vidyamandir Classes
W = PV for constant pressure and we also
Gaseous State and Thermodynamics have W = PV2 – PV1 = nRT2 – nRT1
W = nRT for constant pressure
Work Done From P-V Diagram An ideal gas contained in a cylinder fitted with a massless and rictionless piston can be considered as a thermodynamical system. Its state can be represented by variables (P, V, T ). If P and V are known, T can be calculated (PV = nRT).
The system’s state at any instant of time can be specified by two variables (P, V). The relation between pressure P and volume V can be studied on a pressure - volume (P-V) graph known as indicator diagram or a P-V diagram. On such a graph, each equilibrium state of a thermodynamical system can be represented by a point whose x-coordinate represents volume (V) and y-coordinate represents pressure (P). A process can be represented by a curve on the P-V diagram. If the gas goes from initial state A (P1,V1) to the final state B (P2, V2), the work done W is given by : W = area under the curve AB above the X-axis (shaded portion)
Internal Energy Internal energy (U) of a system is the total of all kinds of energy possessed by the atoms and other particles that comprise the system. The change in internal energy U depends only on the temperature difference of initial and final states and not on the process between the states.
First Law of Thermodynamics If some heat is supplied to a system capable of doing some work, the quantity of heat absorbed by the system is the equal to the sum of external work done by the system and increase in its internal energy. If Q is the heat gained by the system, W is the work done against external pressure and U is the change in internal energy, then
heat energy given increase in work done to the system internal energy by the system Q = U + W. Note : First Law of thermodynamics is a direct consequence of Law of conservation of energy.
Self Study Course for IITJEE with Online Support
Section 3
13
Gaseous State and Thermodynamics
Vidyamandir Classes
In Thermodynamics : 1.
Work done by a system is taken as positive (+). Work done on the system is taken as negative (–).
2.
Heat gained by a system is taken as positive (+) and the heat lost by the system is taken as negative (–).
3.
Increase in internal energy is taken as positive (+) and decrease in internal energy is taken as negative (–).
Specific Heat of a Gas The specific heat of gas is defined as the amount of heat required to raise the temperature of unit mass of substance through 1K. Its unit in S.I. system is J/Kg/K. When the unit mass is 1 mole, the specific heat is known as molar specific heat and its units are J/mol/K or cal/mol/K. Molar specific heat = molecular weight (specific heat in J/gm/K) In gases, on giving heat to the gas, besides temperature, its pressure and the volume may also change. Thus specific heat of a gas depends upon the conditions under which the heat is given to the gas. Hence we will define two principal types of specific heats of a gas.
Specific Heat at constant volume (Cv) : Cv of a gas is defined as the amount of heat required to raise the temperature of 1 mole of that gas through 1K, provided that the volume of the gas remains constant. It is also known as molar specific heat at constant volume. In general, if n moles of a gas are heated at constant volume to increase the temperature by T, Heat required = Q = n Cv T
Specific Heat at constant pressure (Cp) : Cp of a gas is defined as the amount of heat required to raise the temperature of 1 mole of that gas through 1°K, provided that the pressure of the gas remains constant. It is also known as molar specific heat at constant pressure. In general, if n moles of a gas are heated at constant pressure resulting in a temperature rise by T, Heat required = Q = n Cp T Note : Cp is greater than Cv (Cp > Cv). When we heat the gas at constant volume, the heat is supplied to raise the temperature of gas only. When we heat the gas at constant pressure, the volume of the gas apart from temperature of the gas rises. When the gas expands, it does some external work for increasing the volume. So in this case, the heat is supplied : (a) to raise the temperature of the gas (b) to do the mechanical work for expansion
14
Section 3
Self Study Course for IITJEE with Online Support
Gaseous State and Thermodynamics
Vidyamandir Classes
As a result more amount of heat is required to increase the temperature of the gas through the same amount when the pressure is kept constant. Hence Cp is always greater than the value of Cv. The ratio of
Cp Cv
is denoted by
i.e. =
Cp Cv
( >1)
Specific Heat Relation : Consider n moles of an ideal gas. If its temperature is raised by T at constant volume, then the heat transferred is n Cv T, where Cv is the specific heat at constant volume. W = 0 (as no volume change is there) By first law of thermodynamics : Q = U + 0
n Cv T = U
. . . . (i)
Next, let us heat the same gas at constant pressure. The heat Q required to change the temperature by T will be n Cp T, where Cp is the specific heat at constant pressure. W = P V
where V is the change in volume.
By first law of thermodynamics : Q = U + P V
n Cp T = U + P V
. . . . (ii)
From (i) and (ii), we get : n Cp T = n Cv T + P V From gas equation : PV = nR T
n Cp T = n Cv T + nR T Cp Cv = R
Cp = Cv + R
Combining Cp – CV = R and Cp /Cv = , we set :
CV
R and 1
CV
R 1
Cp and Cv for Gases : Consider 1 mole of a gas heated to raise the temperature by 1°C. When the gas is heated at constant volume, all heat is converted to increase in internal energy. The work done is zero, i.e. Cv = U for 1°C (or 1 K) rise in temperature for 1 mole of a gas.
Self Study Course for IITJEE with Online Support
Section 3
15
Gaseous State and Thermodynamics
Vidyamandir Classes
For mono-atomic gas : 3 Internal energy per mole = RT 2 Cv = V =
Cv =
3 R 2
3 3 3 R T 1 RT R 2 2 2
and
Cp =
5 R (as Cp Cv = R) 2
Cp Cv
5 3
For di-atomic gas : 5 Internal energy per mole = RT 2 5 5 5 Cv = U = R T 1 RT R 2 2 2
5 Cv = R 2
7 Cp = R 2
(as Cp Cv = R)
=
Cp Cv
7 2
Illustration - 13 One mole of oxygen is heated at constant pressure from 0°C. What must be the quantity of heat that should be supplied to the gas for the temperature to be doubled ? If the same heat is supplied to the gas at constant volume, what will be the final temperature ? SOLUTION : = 7930.6 J
T 1 = 0 + 273 = 273 K Temperature is to be doubled
If heating is done at constant volume :
T 2 = 2T1 = 546 K
Q = n Cv T
Q at constant pressure = n Cp T
nR Q = 1 ΔT
7 Hence Q = n Cp T = n R T 2
16
Q = 1
Section 3
7 8.3 (546 273) 2
1 8. 3 7930.6 = 1.4 1 (273)
T = 655 K
Self Study Course for IITJEE with Online Support
Gaseous State and Thermodynamics
Vidyamandir Classes
Illustration - 14 When water is boiled under a pressure of 2 atm., the heat of vapourisation is 2.20 × 106 J/Kg and the boiling point is 120°C. At this pressure 1 Kg of water has a volume of 10–3 m3 and 1 Kg of steam has a volume of 0.824 m3. (a) Compute the work done when 1 Kg of steam is formed at this temperature. (b) Compute the increase in the internal energy. SOLUTION : (a) Work done = P (V) (at constant pressure)
W = 2 atm (0.824 0.001) m3
W = 2 1.013 105 N/m2 0.823 m3
(b) U = Q W U = m L W U = 1 kg 2.20 106 J/kg 166.74 kJ = (2200 166.74) = 2033.26 kJ
= 166.74 kJ
Illustration - 15 The temperature of 3 kg of krypton gas is raised from 20°C to 80°C. (a) If this is done at constant volume, compute the heat added, the work done and the change in internal energy. (b) Repeat if the heating process is at constant pressure. (for mono-atomic gas Kr, Cv = 0.0357 cal/g/K, Cp = 0.0595 cal/g/K) SOLUTION : (a) At constant volume : W =0 Q = m Cv T = 3000 0.0357 100 ; T = 80 (20) = 100°C Q = 10.7 kcal U = Q 0 = 10.7 × 103 × 4.18 J = 44.8 kJ
(b)
At constant pressure : Q = m Cp T = 3000 × 0.0595 × 100 = 17.85 kcal W = P (V2 V1) = nR (T2 T1) 3000 W = 8.3 100 J = 30 kJ 83.8 U = n Cv T = 44.8 kJ
[Note : U is same for both processes as it is a State Function] Note : (i) (ii)
It can be verified that Q = U + W. Q is usually measured in calories and U and W are measured in Joules.
Self Study Course for IITJEE with Online Support
Section 3
17
Gaseous State and Thermodynamics
Vidyamandir Classes
Illustration - 16 A system is taken from state i to the state f (refer to the fig.). Along path “i a f ”, it is found that Q = 50 cal, W = 20 cal. Along the path “i b f ”, Q = 36 cal. Calculate : (a) W along the path “i b f ” (b) If W = 13 cal for curved path “ f i ”, what is Q for this path ? (c) Taking Ui = 10 cal, what is Uf ? (d) If Ub = 22 cal, what is Q for the process “ i b” and for the process “b f ” ? SOLUTION : Path “ i a f ” Q = 50 cal ;
W = 20 cal
U = Q W U = 50 20 = 30 cal
Uf Ui = 30 cal.
As internal energy change is a State function, U will be same for any path from i to f. (a)
Path “ i b f ” W = Q U W = 36 (Uf Ui) = 36 (30) = 6 cal
(b)
Path “ f i ” Q = U + W = (Ui Uf) + W = (30) + (13) = 43 cal.
(c)
Uf Ui = 30 cal
(d)
Uf = Ui + 30 = 40 cal
Process “i b” Q = U + W Q = (Ub Ui) + (W)ibf {(W)ib = (W)ibf because (W)bf = 0} Q = (22 10) + 6 = 18 cal
(d)
Process “b f ” Q = U + W Q = (Uf Ub) + 0 = 40 22 = 18 cal.
18
Section 3
Self Study Course for IITJEE with Online Support
Gaseous State and Thermodynamics
Vidyamandir Classes
THERMODYNAMIC PROCESSES
Section - 4
In a thermodynamic process, the initial and final states are important, but the manner in which the system is taken from initial to final state must also be taken into account. Hence it is important to distinguish between the following processes.
Isothermal process : An isothermal process is one in which the temperature remains constant. A gas under going an isothermal process should remain in perfect thermal contact with a constant temperature heat reservoir. The gas obeys Boyle’s Law, during an isothermal process, i.e. PV = constant As the temperature remains constant internal energy of the system remains constant. i.e., U = 0 (for isothermal process) From Ist Law : Q = W
(a)
Isothermal Expansion : If the gas expands, it takes heat from the heat reservoir and does positive work, i.e. in equation Q = W, both Q and W are positive. The process takes place slowly. As the pressure on the piston is reduced, the gas tries to expand. In expansion, it tends to cool down, but the heat reservoir gives heat input to keep temperature constant.
(b)
Isothermal Compression : If the gas is compressed isothermally, it gives heat to the reservoir and the work done is negative, i.e., in the equation Q = W, both Q and W are negative. To do a perfectly isothermal process, pressure is increased in very small steps. As the gas is compressed slowly, its temperature tends to rise, but the reservoir extracts the heat from the system to keep the temperature constant.
Adiabatic process : In an adiabatic process, no heat enters or leaves the system. Such a situation is achieved by heavily insulating the system. From Ist Law of thermodynamics :
Q = U + W
Self Study Course for IITJEE with Online Support
Section 4
19
Gaseous State and Thermodynamics
Vidyamandir Classes
In adiabatic process Q = 0 U+W=0 The gas in this process obeys the relation : PV = constant
P1V1 = P2V2
( = Cp /Cv)
Note : The adiabatic curve in P-V diagram is much steeper than the isothermal curve.
(a)
Adiabatic Expansion :
As the pressure on the piston is reduced, the gas pushes the piston upwards. Due to the expansion, the temperature of the system falls down. The molecules of the gas lose speed after colliding with a receding (moving away) piston. The loss in the speed causes a fall in the temperature. The gas does positive work and its internal energy decreases. In the equation of Ist law : U + W = 0, U is negative and W is positive. (internal energy of the gas is being converted into work done) (b)
Adiabatic Compression :
As the pressure on the piston is increased, it moves downwards compressing the gas. Due to compression, the temperature of the gas rises. The molecules of the gas gain speed after colliding with an incoming piston. The gain in speed causes a rise in temperature. Work is done on the gas and its internal energy increases. In the equation : U + W = 0, U is positive and W is negative.
Isochoric process : A process in which the volume remains constant is known as isochoric process. For an isochoric process, an ideal gas is taken in a container with rigid walls so that neither expansion nor compression occurs. The gas obeys the relation : P T In isochoric process, W = 0. So from Ist Law of thermodynamics : Q = U. (a) If heat is given to the gas, its temperature and pressure rise. Its internal energy also rises. U and Q are both positive.
20
Section 4
Self Study Course for IITJEE with Online Support
Gaseous State and Thermodynamics
Vidyamandir Classes (b)
If heat is taken out from the gas, its temperature and pressure fall. Its internal energy also decreases. Q and U are both negative.
Isobaric process : A process in which the pressure is kept constant is called as isobaric process. For isobaric process, the external pressure on the piston is kept constant. The gas obeys Charles’ Law : V T The work done : W = P (V2 V1) = n R (T2 T1)
(a)
Isobaric Expansion : If heat is given to the gas, isobaric expansion occurs. The volume and the temperature both rise. The gas expands doing positive work. In equation : Q = U + W, all the three terms : Q, U and W are positive in the isobaric expansion.
(b)
Isobaric Compression : If heat is taken out of the gas, isobaric compression occurs. The temperature falls and the gas contracts causing negative work. In the equation : Q = U + W, all three terms : Q, U and W are negative in the isobaric compression.
Expressions for U, W, and Q for Different Processes Internal Energy Change : (U) U = nCv T for every process. It is state function and depends only on the difference of initial and final temperatures. U = n Cv T =
nR T 1
Self Study Course for IITJEE with Online Support
Section 4
21
Gaseous State and Thermodynamics
Vidyamandir Classes
Process
Work Done : (W)
Heat Exchange : (Q)
Isothermal Process :
W 2.303 nRT log10
Adiabatic Process :
nR T1 T2 PV P V W 11 2 2 1 1
Q = 0
Isochoric Process :
W=0
Q = n Cv T (use definition of Cv)
Isobaric Process :
W = P V = P (V2 V1) W = nR (T2 T1)
Q = n Cp T (use definition of Cp)
V2 V1
V Q 2.303 nRT og10 2 V1
Relations for Thermodynamic Processes (State Variables P, V, T) Isochoric process :
Volume remains constant.
P T
P1 P2 T2 T2
P 1/V
P1 V1 = P2 V2
V T
V1 V2 T2 T2
Isothermal process : Temperature remains constant
Isobaric process :
Pressure remains constant.
Adiabatic process : No heat transfer takes place. P1 V1 = P2 V2
T1 V1 1 = T2 V2 1
T1 P11 = T2 P21
Illustration - 17 An ideal gas undergoes an isothermal expansion at 0°C from 0.010 m3 to 0.200 m3. For 5 moles of gas, compute the work done, the heat added, and the change in internal energy. SOLUTION : Let
V1 = 0.01 m3
V2 = 0.2 m3 ;
;
T1 = T2 = 273 K ;
n = 5 moles
V2 W = 2.303 nRT1 log 10 V 1
W = 2.303 5 8.3 273 log10 20 = 33.946 J. As U = 0 for isothermal process, Q = W = 33946 J.
22
Section 4
Self Study Course for IITJEE with Online Support
Gaseous State and Thermodynamics
Vidyamandir Classes
Illustration - 18 A gasoline engine takes in 5 moles of air at 20°C and 1 atm., and compresses it adiabatically to 1/10th of the original volume. Find the final temperature and pressure. Assume air to be diatomic. Also find the work done and change in internal energy. SOLUTION : Let P1 = 1 atm.
n = 5 moles
T1 = 293 K
V2 = V1/10
Using P1 V1 = P2 V2 1
V1 P2 = P1 = 1 (10)1.4 = 25.12 atm. V2 = 25.12 atm.
Using T1V1 1 = T2V2 1 1
V1 T2 = T1 V2
Work done
=
= 293(10)0.4 = 736 K
nR T1 T2 5 8.3 293 737 1 0. 4
= 46 kJ U = Q W = 0 W = 46 kJ
Illustration - 19 How much work is done by an ideal gas is expanding isothermally from an initial volume of 3 L at 20 atm to a final volume of 24 L ? SOLUTION : In isothermal process at temperature T : V2 W = 2.303 nRT log10 V 1 V2 W = 2.303 (P1V1) log10 V (using P1V1 = nRT) 1
= 2.303 (20 3) log10
24 L atm. 3
= 2.303 60 log10 8 (101) J = 1.26 104 J Self Study Course for IITJEE with Online Support
Section 4
23
Gaseous State and Thermodynamics
Vidyamandir Classes
Illustration - 20 In a cylinder, 2.0 moles of an ideal monoatomic gas initially at 1.0 × 106 Pa and 300 K expands until its volume doubles. Compute the work done if the expansion is : (a)
isothermal
(b)
adiabatic
(c)
isobaric
SOLUTION : Let
P1 = 1× 106 Pa = 106 N/m2
;
T1 = 300 K ; n = 2 moles and Final volume = 2 (initial volume)
V2 = 2 V1
(a)
Wisothermal = 2.303 nRT log (V2 /V1) = 2.303 × 2 × 8.3 × 300 log10 2 = 3452 J
(b)
W(adiabatic) =
nR T1 T2 1
For adiabatic process: T1V1 –1 = T2 V2 –1 5 / 31 V2 1 1 300 T2 = T1 = 2 V1
T2 = 189 K nR T1 T2 W=
1
2 8.3 300 189 5 1 3
= 2764 J (c)
W(isobaric) = P (V2 V1) P1 V1 = nRT1
24
nRT1
V 1=
V1 = 4.98 × 10–3 m3
and
W = P (V2 V1) = P (2V1 V1) = P V1
W = 106 × 4.98 × 103 J = 4980 J.
Section 4
P1
=
2 x 8.3 x 300
106
Self Study Course for IITJEE with Online Support
Gaseous State and Thermodynamics
Vidyamandir Classes
Illustration - 21 2.0 moles of an ideal gas are carried round a cycle as shown. If the process b c is isothermal and Cv = 3 cal/mol/K. Determine: (a) work done (b) change in internal energy (c) heat supplied to the system during processes a b b c and c a. SOLUTION : (a) Process a b is clearly isobaric at P = 4 atm.
Process b c :
W = P (Vb Va) = 4 (16.4 8.2) = 32.8L atm
Isothermal, T = 0
Process c a : U for complete cyclic process is zero because it depends on initial and final States i.e. temperature difference only.
Wa b = 32.8 × 101 J = 3313 J
Process b c is isothermal (given) Vc W = 2.303 nRT log10 V b
1200 + 0 + (U)ca = 0 8.2 16.4
(Q)ab = (U)ab + Wab = (1200 + 3313/4.18) cal
Process c a is isochoric (volume is constant)
= 1992.5 cal
Wc a = 0 J
(Q)ab = 1992.5 cal
(b) Process a b U = n Cv T PbVb PaVa U = nCv nR nR
=
(U)ca = 1200 cal.
(c) Using first law of thermodynamics:
Wb c = 4592.5J
U = 0
(U)ab + (U)bc + (U)ca = 0
Vc W = 2.303 (Pb Vb) log10 V b
W = 2.303 (4 16.4 101) log10
(Q)bc = (U)bc + Wbc = (0 4592.5/4.18) cal = 1098.7 cal PV = nRT
Cv 3 PbVb PaVa R 0.0821
(Q)bc = 1098.7 cal. (Q)ca = (U)ca + Wca = 1200 + 0 = 1200 cal
Q = 1200 cal.
(4 16.4 4 8.2) = 1200 cal.
Self Study Course for IITJEE with Online Support
Section 4
25
Gaseous State and Thermodynamics
Vidyamandir Classes
Illustration - 22 Figure shows three processes for an ideal gas. The temperature at “a” is 600K, pressure 16 atm and volume 1L.The volume at “b” is 4 L. Out of two processes, ab and ac, one is adiabatic and other is isothermal. The ratio of specific heats of the gas is 1.5. (a) Which of the processes “ab” and “ac” is adiabatic ? Why ? (b) Compute the pressure of the gas at “b” and “c”. (c) Compute the temperature at “b” and “c”. (d) Compute the volume at “c”. SOLUTION : (a)
The process ab is adiabatic because it steeper than ac.
(b)
Pa = 16 atm Vb = 4 L Va = 1 L Ta = 600K Process “ab” (adiabatic)
Pa Va = Pb Vb
1.5 V 1 Pb Pa a 16 4 Vb
Pb = 2 atm.
As bc is isobaric, Pb = Pc = 2 atm. (c)
Tc = Ta = 600K because ac is isothermal. Process “ ab ” (adiabatic) Tb can be calculated using Ta Va 1 = Tb Vb 1 or
PaVa PbVb Ta Tb 1
Hence Tb = Ta Va Vb (d)
0.5
300 K
Process “ac” (isothermal)
26
1 600 4
Pa Va = Pc Vc P V 16 1 Vc = a a = =8 L 2 Pc
Section 4
Self Study Course for IITJEE with Online Support
Gaseous State and Thermodynamics
Vidyamandir Classes
Illustration - 23 6 grams of hydrogen gas at a temperature of 273K were isothermally expanded to five times its initial volume and then isochorically heated so that the pressure in the final state becomes equal to that in the initial state. Find the total amount of heat absorbed by the gas during the entire process. SOLUTION : Represent the processes on the PV diagram by the curve 1 to 2 (isothermal) and the line 2 to 3 (isochoric). P1 = P3 For the process 1 to 2 (isothermal) T2 = T1 = 273 ;
V2 = 5 V1
Q = U + W = 0 + W
Q = 2.303 nRT1 log10
Q = 2.303
V2 V1
6 8.3 273 log10 5 2 = 10942.4 J
State 1 and state 3 are at the same pressure.
V1 V3 T1 T3
V3 T3 = V T1 = 5 273 = 1365 K 1
Process 2 to 3 (isochoric) Q = U + W Q = n Cv T + 0
6 5 R T3 T2 2 2
Q =
Q = 7.5 8.3 (1365 273) J = 67977 J
Thus total heat absorbed = Q 1 2 + Q2 3 = 78919.4 J
Self Study Course for IITJEE with Online Support
Section 4
27
Gaseous State and Thermodynamics
Vidyamandir Classes
Illustration - 24 One mole of a perfect gas, initially at a pressure and temperature of 105 N/m2 and 300K respectively expands isothermally until its volume is doubled and then adiabatically until its volume is again doubled. Find the final pressure and temperature of the gas. Find the total work done during the isothermal and adiabatic processes. Given = 1.4. Also draw the P-V diagram for the process. SOLUTION : Let (P1 V1 T1) be the initial pressure, volume and the temperature of the gas. For isothermal expansion (1 to 2) We have :
V2 = 2V1
Using,
P1 V1 = P2 V2
P2 = P1(V1/V2) = P1/2 = 0.5 × 105 N/m2.
and
T2 = T1 = 300 K
For adiabatic expansion (2 to 3) Using
P2 V2 = P3 V3
and
Hence final pressure and temperatures are Work Done :
1.9 × 104 N/m2 and 227.35 K
V3 = 2 V2
V2 P3 = P2 = 0.5 105 (0.5)1.4 V3 4
W = W1 2 + W2 3
W = 2.303 nRT1
2
= 1.9 10 N/m Also,
T2 V2
1
= T3 V3 -1
v2 T3 = T 2 v3
log10
1
1 = 300 4
0.4
= 227.35 K
V2 nR T2 T3 V1 1
W = R (2.303 300 log10 2) +
R 300 227.35 0.4
= 3233.56 J
NOW ATTEMPT IN-CHAPTER EXERCISE-B BEFORE PROCEEDING AHEAD IN THIS EBOOK
28
Section 4
Self Study Course for IITJEE with Online Support
Gaseous State and Thermodynamics
Vidyamandir Classes
SUBJECTIVE SOLVED EXAMPLES Example - 1 A flask of volume 2l, provided with a stopcock contains oxygen at 300 K and atmospheric pressure. The system is heated to a temperature of 400 K, with the stopcock open to the atmosphere. The stopcock is then closed and the flask is then cooled to its original temperature. (a) What is the final pressure of oxygen in the flask ? (b) How many grams of oxygen remain in the flask ? SOLUTION :
Let V = 2 L (remains constant) When the flask is heated open to the atmosphere its pressure remains the same at P atmosphere . Hence using PV = nRT, we apply : n1T1 = n2T2 for the heating process. n1(300) = n2(400) 3n1 = 4n2 For the cooling process(taking place at constant volume and given moles)
P2 P 3 T2 T3
P3
P3 T3 T2
P 3 P3 at 300 atm. 400 4
Loss of grams of O2 = (n1 n2) 32 3 = n1 n1 32 = 8n1 4
P V = 8 at RT1
8 1 2 = 0.6504 gm 0.082 300
Self Study Course for IITJEE with Online Support
Subjective Solved Examples
29
Gaseous State and Thermodynamics
Vidyamandir Classes
A tube closed at one end contains a mercury column 15 cm long inside it. The length of air Example - 2 column trapped between the closed end and mercury is 40 cm when the tube is horizontal. Find the length of air column if: (a) the tube is made vertical with open end downwards. (b) the tube is kept inclined at 60° with the vertical with open end downwards.(Patm = 75 cm Hg) SOLUTION :
Let A be the crosssectional area and x be the length of Hg column. Let P1, P2 be the pressures in air column in the vertical and inclined positions respectively. Let y1, y2 be the lengths of the corresponding air columns. Applying Boyle’s Law :
P0 V0 = P1 V1 = P2 V2
Where P0 = atm. pressure, V0 = 40 A, V1 = y1 A, V2 = y2 A Balancing forces on Hg columns : P1 + x = P and P2 + x cos 60° = P0 P0 V0 = P1 V1 75 (40 A) = (75 15) y1 y1 = 50 cm. P0 V0 = P2 V2 75 (40 A) = (75 15 cos 60°) (y2 A) y2 = 44. 4 cm
30
Subjective Solved Examples
Self Study Course for IITJEE with Online Support
Gaseous State and Thermodynamics
Vidyamandir Classes
Example - 3 An open glass tube is immersed in Hg in such a way that a length of 8 cm extends above the Hg level. The open end of the tube is then closed and the tube is raised vertically up by 44 cm. What will be the length of the air column above Hg in the tube now ? SOLUTION :
Let A be the area of crosssection of the tube. P0 = atm. pressure = 76 cm Hg x = length of the air column in the final position. P1 = air pressure (final) As the tube is raised, the volume of the air column trapped between the closed end and Hg increases and hence pressure drops. Due to drop in pressure, Hg rises in the tube. Height of Hg column = (8 + 44) x = 52 x Due to hydrostatic conditions :
P1 + (52 x) = Patm.
P1 = 76 (52 x) = 24 + x
PA = PB
Applying Boyle’s Law on air column : P0 V0 = P1 V1
76 (8A) = (24 + x)(x A) 76 (8) = 24 x + x2
On solving, we get : x = length of air column = 15.42 cm
Self Study Course for IITJEE with Online Support
Subjective Solved Examples
31
Gaseous State and Thermodynamics
Vidyamandir Classes
Example - 4
A barometer tube contains some air above the Hg level inside it. The atmospheric pressure is 76 cm of Hg. The Hg level in the barometer tube is 74 cm high and the tube has a length of 100 cm above Hg level outside the tube as shown. The tube is raised vertically upwards by a distance of 10 cm. What is the height of Hg level inside the tube after it has been raised ? SOLUTION : Let P1 and P2 be the initial and final pressures in the air column above Hg inside the tube. Let x be the height of Hg column after the tube has been raised.
From Hydrostatic conditions : P1 + 74 = Patm
P1 = 76 74 = 2 cm Hg.
P2 + x = Patm
P2 = 76 x.
Applying Boyle’s Law : P1 (26 A) = P2 (110 x) A where A is crosssectional area. 2 (26) = (76 x) (110 x) x2 186 x + 8308 = 0 On solving, we get : x = 74.53 cm.
32
Subjective Solved Examples
Self Study Course for IITJEE with Online Support
Gaseous State and Thermodynamics
Vidyamandir Classes
Example - 5 A cylinder fitted with a movable piston contains hydrogen at a pressure of 3.5 × 105 N/m2 and a temperature 366 K. Hydrogen expands adiabatically until the pressure in the cylinder falls to 0.7 ×105 N/m2. The piston is then fixed and the gas is heated until the temperature becomes 366 K. The pressure in the cylinder is now found to be 1.1 × 105 N/m2. Determine the molar specific heats of hydrogen. SOLUTION : Represent the processes on P-V diagram by curve 1 to 2 (adiabatic) and the line 2 to 3 (isochoric). Hence P1 = 3.5 × 105 N/m2 P2 = 0.7 × 105 N/m2 T1 = 366 K
T2 = ? 5
2
P3 = 1.1 × 10 N/m
T3 = 366 K
Process 2 to 3 (isochoric) P3 P2 T = T 2 3
P2 0.7 T2 = P T3 1.1 366 232.90 K 3
Process 1 to 2 (adiabatic) T1 P11 = T2 P21
T1 1 1 T2 5
(log10 11 log10 7) = ( 1) log10 5
log10 5 = log 5 log 7 - log 11 1.39 10 10 10
Thus Cp =
1
11 1 7 5
R = 29.58 J mol1 K1 1
and
Self Study Course for IITJEE with Online Support
Cv =
R = 21.28 J mol1 K1 1
Subjective Solved Examples
33
Gaseous State and Thermodynamics
Vidyamandir Classes
Example - 6 The rectangular box shown in fig. has a partition which can slide without friction along the length of the box. Initially each of two chambers of the box has one mole of a monoatomic ideal gas ( = 5/3) at a pressure of P0, volume V0 and temperature T0. The chamber on the left is slowly heated by an electric heater. The walls of box and partition are thermally insulated. Heat loss through lead wires of heater is negligible. The gas in the left chamber expands by pushing the partition until the final pressure in both chambers becomes (243/32) P0. Determine : (a) the final temperature of the gas in each chamber, (b) the work done by the gas in the right chamber.
1/
SOLUTION : Let P1, V1, T1 and P2, V2, T2 be the final pressure, volume, temperature in the right and left chambers respectively. It is very important to observe that P1 = P2 because the partition is in equilibrium and V1 + V2 = 2V0 because total volume remains same. The process in the left chamber is none of the four standard processes, so we will use
PV = T
constant. The process in the right chamber is adiabatic. We have P1 = P2
243 P 32 0
3/ 5 P0 8V 32 V1 V0 V0 0 27 243 P1 and T0 V0 1 = T1 V1 1
1
V T1 T0 0 V1
27 T0 8
2/3
9 T0 4
In left chamber : We have : V2 = 2V0 V1 = 2V0
P0V0 P2V2 T0 T2
T2 =
8 46 V0 V0 27 27
P2V2 243 46 207 T0 T0 V0 P0V0 32 27 27
Hence the final temperatures are
207 9 T0 and T0 16 4
Work done : Considering adiabatic compression in right chamber :
W=
P0V0 PV 1 1 1
W=
P0V0 243 8 15 1 P0 V0 1 32 27 8
In right chamber : We have :
34
P0 V0 = P1 V1
Subjective Solved Examples
Self Study Course for IITJEE with Online Support
Gaseous State and Thermodynamics
Vidyamandir Classes
Example - 7 An ideal mono-atomic gas is confined in a cylinder by a spring loaded piston of cross-section 8 × 10– 3 m2. Initially the gas is at 300 K and occupies a volume of 2.4 × 10–3 m3 and the spring is in its relaxed (unstretched, uncompressed) state. The gas is heated by a small electric heater until the piston moves out slowly by 0.1 m. Calculate the final temperature of the gas and the heat supplied in joules by the heater. The force constant of the spring is 8000 N/m, atmospheric pressure is 1 × 105 N/m2. The cylinder and the piston are thermally insulated. The piston is massless and there is no friction between the piston and the cylinder. Neglect heat loss through the lead wires of the heater. The heat capacity of the heater coil is negligible. Assume the spring to be massless. SOLUTION : The process taking place in the cylinder is none of the four standard processes.
Applying PV/T = constant, we get :
Hence, we will use PV/T = constant.
T2
Let (P1, V1, T1) and (P2, V2, T2) denote the initial and the final pressure, volume and temperature respectively. P1 = Patm = 105 N/m2 ; V1 = 2.4 × 10–3 m3
;
Balancing forces on the piston : P2 = Patm +
Here,
x = 0.1 m ;
and
k = 8000 N/m
P1 V1
300 2 105 3.2 103 105 2.4 103
= 800 K Heat supplied = Q = U + W We have U = n Cv T
T1 = 300 K
We have :
T1 P2 V2
kx A
A = 8 × 10–3 m3
P2 = 105 + 105 = 2 × 105 N/m2
and
V2 = V1 + x A = 3.2 × 103 m3
Self Study Course for IITJEE with Online Support
PV 3 = 1 1 R T2 T1 RT1 2 105 2.4 103 3 500 600 J = 300 2
W = work done against atmosphere + work done against spring W = Patm (V) + 1/2 kx2
W = 105 (0.1 × 8 × 10–3) + 0.5 × 8000 × 0.01 = 120 J
Hence Q = U + W = 600 + 120 = 720 J.
Subjective Solved Examples
35
Gaseous State and Thermodynamics
Vidyamandir Classes
Example - 8 Two moles of an ideal monatomic gas, initially at pressure p1 and volume V1, undergo an adiabatic compression until its volume is V2. Then the gas is given heat Q at constant volume V2. (a) Sketch the complete process on a P-V diagram (b) Find the total work done by the gas, the total change in its internal energy and the final temperature of the gas. [Give your answers in terms of P1, V1, V2, Q and R.] SOLUTION : (a) Figure displays the PV diagram of the gas undergone the given two processes.
Work done by the gas on heating at constant volume Since the volume is held constant W2 = 0 The total work done by the gas is W = W1 + W2
(b) (i) Total Work Done by the Gas : Work done by the gas in adiabatic compression P1 V1 P2V2
We have : PV 1 1 P2 V2
PV 1 1
Hence,
W1
2
1
V 1 PV 1 1 1 1 W1 1 V2 3P V V W1 1 1 1 1 2 V2
2/3
[For a monatomic gas, = 5/3] 36
Subjective Solved Examples
3 P1 V1 2
V 1 V2
2/3
1
and U2 = Q (given) The total change in internal energy of the gas is 2/3 3 P1 V1 V1 Q 1 U = U1 + U2 = 2 V2 (iii) Final temperature of the Gas : Change in temperature in adiabatic compression Since U = CV T,
PV 1 1 V
V2
2/ 3
U1 = W1
U1
1 For a gas undergoing adiabatic process :
(ii) Total change in Internal Energy : Change in internal energy in adiabatic compression
In an adiabatic process : W1
3 P1 V1 V1 1 2 V2
U1 We get : T = C V
3 P1 V1 T2 – T1 = 2 C V
V 1 V2
2/3
1
Self Study Course for IITJEE with Online Support
Gaseous State and Thermodynamics
Vidyamandir Classes
3 P1 V1 3 2 n R 2
P1 V1 T2 = n R
V 1 V2
2/ 3
1
Q = CV T = CV (T3 – T2) PV V Q Q T2 1 1 1 T3 = CV n R n R V2
V 2 / 3 1 1 V2
2/3
Since n = 2 mol, we get : Q p V T3 1 1 3R 2R
Change in temperature on heating the gas
V1 V2
2/3
Two moles of an ideal monatomic gas is taken through Example - 9 a cycle ABCA as shown in the P–T diagram. During the process AB, pressure and temperature of the gas vary such that PT = constant. If T1 = 300 K, calculate (a) the work done on the gas in the process AB and (b) the heat absorbed or released by the gas in each of the processes. Give answers in terms of the gas constant R. SOLUTION : In the process AB, we will have PV = nRT and PT = K Eliminating P, we get :
. . . . (i)
(b) The change in energy of the gas in the process AB is
nRT 2 V K
dV 2nRT dT K
. . . . (ii)
(a) The expression of work done by the gas is : W P dV TB
K 2nRT W dT T K TA [Using Equations (i) and (ii)]
WAB
3 U AB n CV T 2 R T1 2T2 2
= 900 R Now from the first law of thermodynamics, QAB = UAB + WAB = 1200 R – 900 R = – 2100 R The negative sign implies that the heat is released in the process AB. The process BC takes place at constant pressure. Hence WBC = P V = (2P1) (VC – VB)
T1
The negative sign implies that the work is done on the gas.
2nR dT
2T1
= 2 nT1R = 2 2 (300 K) R = – 1200 R Self Study Course for IITJEE with Online Support
nRT1 nRT1 = 2 P1 P 2 P 1 1 = nRT1 = 2 300 R = 600 R Subjective Solved Examples
37
Gaseous State and Thermodynamics
Vidyamandir Classes V W AC nRT ln A VC
Now, 3 U BC n CV 2 R TC TB 2 = 3 (R) (2T1 – T1) = 3(R) 300 = 900 R
(where T = 2T1) = 2 R 2 300 log10 2 = 1200 R log10 2
The positive sign implies that the heat is absorbed in the process BC.
UCA = 0 QCA = UCA + WCA
The process CA takes place at constant temperature. Hence
= 0 + 1200 R log10 2
Example - 10
A cubical box of side 1 metre contains helium gas (atomic weight 4) at a pressure of 100 N/s . During an observation time of 1 second, an atom travelling with the root-mean-square speed parallel to one of the edges of the cube, was found to make 500 hits with a particular wall, without any collision 2
25
with other atoms.Take R J/mol-Kand 3 k = 1.38 10–23 JK. (a) Evaluate the temperature of the gas. (b) (c) Evaluate the total mass of helium gas in the box.
Evaluate the average kinetic energy per atom.
SOLUTION : (a) We have : Distance travelled by the atom in successive hit to one of the edges of the cube is l = 2m Number of hits with the wall in one second is N = 500/s
1000 2 4 103 kg
mol 1
3 25 / 3 J 1 mol 1
160 K
(b) The average kinetic energy per atom is 3 3 KE kT 1.38 1023 J K 1 160 K 2 2
= 3.31 10–21 J (c) Using the expression
Hence, distance travelled in one second, i.e. root mean square speed of atom, is Vrms = l N = (2 m) (500/s) = 1000 m/s Since the root mean square speed of gaseous atoms is given by vrms 3 RT / M , we get :
v2 M T rms 3R
38
Subjective Solved Examples
m PV nRT RT M
We get : m
PVM RT
100 Pa 1 m3 4 103 kg
mol 1
25 / 3 J K 1 mol 1 160 K
= 3.0 10–4 kg
Self Study Course for IITJEE with Online Support
NOW ATTEMPT OBJECTIVE WORKSHEET BEFORE PROCEEDING AHEAD IN THIS EBOOK
Gaseous State and Thermodynamics
Vidyamandir Classes
THINGS TO REMEMBER
1. (a) If the molecules are moving speeds c1, c2, c3, ................., cN ; then the root mean square speed of the gas is defined as :
c 2 = crms = (b)
3P 3 RT 3kT M m0
The pressure exerted by the gas is P
3.
N
crms can be calculated with any of the following results. crms =
2.
2 c12 +c 22 + c32 + .........+cN
1 m0 N 2 crms 3 V
or
P
m N 1 2 crms where = density of gas 0 V 3
Kinetic energ of a gas : For monoatomic gas : Total energy per molecule = 3/2 kT Total energy per mole = 3/2 RT
For diatomic gas : Total energy per molecule = 5/2 kT Total energy per mole = 5/2 RT
For triatomic gas : Total energy per molecule = 3kT Total energy per mole = 3RT
4. (a) Work done W = PdV
(b) Work Done From P-V Diagram
W = area under the curve AB above the X-axis Self Study Course for IITJEE with Online Support
(shaded portion) Things to Remember
39
Gaseous State and Thermodynamics
5.
Vidyamandir Classes
Internal Energy Internal energy (U) of a system is the total of all kinds of energy possessed by the atoms and other particles that comprise the system. The change in internal energy U depends only on the temperature difference of initial and final states and not on the process between the states. U = Uf – Ui
6.
First Law of Thermodynamics heat energy given increase in work done to the system internal energy by the system Q = U + W.
7.
Combining Cp – CV = R and Cp /Cv = , we set :
CV
8.
Cp and Cv for Gases : (a)
For mono-atomic gas : Cv =
(b)
9.
R R and CV 1 1
3 R 2 Cp Cv
and
Cp =
5 R 2
5 3
For di-atomic gas :
5 Cv = R 2
Cp =
7 R 2
Cp Cv
5 3
Relations for Thermodynamic Processes (State Variables P, V, T) Isochoric process : Volume remains constant.
P T
P1 P2 T2 T2
P 1/V
P1 V1 = P2 V2
Isothermal process : Temperature remains constant
40
Things to Remember
Self Study Course for IITJEE with Online Support
Gaseous State and Thermodynamics
Vidyamandir Classes
Isobaric process : Pressure remains constant.
V
V
1 2 T T 2 2
V T
Adiabatic process : No heat transfer takes place. P1 V1 = P2 V2 T1 V1 1 = T2 V2 1 T1 P11 = T2 P21
10. Expressions for U, W, and Q for Different Processes Internal Energy Change : (U) U = nCv T for every process. It is state function and depends only on the difference of initial and final temperatures. U = n Cv T =
nR T 1
Process
Work Done : (W)
Heat Exchange : (Q)
Isothermal Process :
W 2.303 nRT log10
Adiabatic Process :
nR T1 T2 PV P V W 11 2 2 1 1
Q = 0
Isochoric Process :
W=0
Q = n Cv T (use definition of Cv)
Isobaric Process :
W = P V = P (V2 V1)
Q = n Cp T (use definition of Cp)
V2 V1
Q 2.303 nRT og10
V2 V1
W = nR (T2 T1)
Self Study Course for IITJEE with Online Support
Things to Remember
41
Vidyamandir Classes
My Chapter Notes
Self Study Course for IITJEE with Online Support
Vidyamandir Classes
Illustration - 1
Self Study Course for IITJEE with Online Support
Vidyamandir Classes
Self Study Course for IITJEE with Online Support
Vidyamandir Classes
Self Study Course for IITJEE with Online Support
View more...
Comments
