Thermodynamics

Name : Roll No. : Topic :

Mohammed Asif

Ph : 9391326657, 64606657

THERMODYNAMICS The branch of physics which deals with the heat energy and other forms of energies is called Thermodynamics. P, V, T are thermodynamic variables (These are inter convertible) → JOULES LAW: - The amount of work performed is directly proportional to the amount of heat produced W∝H W = JH

∴ W = JH mgh = J× m s ∆ θ d r a g g I

C

E

e d

V

m

x gh h

= JS ∆ θ

P.E →heat

gh is melted ∆θ= JS W=J H

Work done against friction so some ice W=JML µ mg× x = JML M=

µ mgx

JL Bullet → V is stopped ∴ all its K. E is converted into heat I C E V i s energy ∴W = J H M 1 Stopped before coming to rest m v2 = J × ms ∆ θ 2 mass of v2 ice melted is M = JS ∆ θ ∴ v W=JH 1 M v2 = J × mL ∴ v2 2 ∆θ = 2 JS Mv2 =m 2 JL

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→

ZEROTH – LAW OF THERMODYNAMICS: - If 2 systems A and B are separately in thermal equilibrium with a 3rd system C, then A and B are also in thermal equilibrium with each other. Internal Energy: - U = Uk + U p

D

I A

A

→

H

E

A D IA B A T IC

C T

W A L L

R

M

A

L

W

A

L

L

B

INDICATOR DIAGRAM: - A graphical representation of the state of system, with the help of two thermodynomical variable is called as Indicator – Diagram.

P P

A E

P

O

∴

P

( P1 V , 1 )

1

X

P

A

N

S

I O

( P2 , V2 )

V

1

V

2

V

( P2 , V2 ) C

N

P

B

2

B 2

O

M

P R

E

S

S

IO

N

A ( P1 V , 1 )

1

V

2

V

1

AREA UNDER P – V – diagram is numerically equal to work done ∴

W = P∆V

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1) P P

2

P

1

W = Area of Rectangle = P2 − P1 × V2 − V1

(

V

V

1

) (

)

V

2

2) P P

2

P

1

W = Area of Triangle 1 = P2 − P1 × V2 − V1 2

(

V

V

1

) (

)

V

2

3) P P

W = Area of the circle = π r1 r2

r1

2

r2 P

=π

1

V

V

1

( P2 − P1 ) 2

 v2 − v1  ×   2 

V

2

4) P P

r1

2

W =− π

r2 P

2

 v2 − v1  ×   2 

1

V

→

( P2 − P1 )

1

V

2

V

WORK DONE IN NON – CYCLIC PROCESS: Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com 3

1) B

P

W = Area under A B C D E A

C

A

E

D

V

2) P

C

W = Area under A B C A

B A V

3)

W = Area under A B C D E FA

THERMODYNAMICS 1.

Two moles of an ideal monoatomic gas are confined within a cylinder by a massless spring loaded with a frictionless piston of negligible mass and of Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com 4

cross – sectional area 4 X 10-3 m2 . The spring is initially in its relaxed state. Now the gas is heated by a heater for sometime. During this time the gas expands and does 50 J of work in moving the piston through a distance of 0.1m. The temperature of the gas increases by 50 K. Calculate the spring constant and the heat supplied by the heater? 2.

Two moles of an ideal monoatomic gas are confined within a cylinder by a massless and frictionless spring loaded piston of cross – sectional area 4 X 10-3 m2 . The spring is initially in its relaxed state. Now the gas is heated by an electric heater, Placed inside the cylinder, for sometime. During this time the gas expands and does 50 J of work in moving the piston through a distance of 0.10 m. The temperature of the gas increses by 50 K. Calculate the spring constant and the heat supplied by the heater?

3.

An ideal gas with adiabatic exponent r is heated at constant pressure. It absorbs Q amount of heat. Determine the fractions of heat absorbed in raising the internal energy and performing the work.

4.

A piston can freely move inside a horizontal cylinder closed from both ends. Initially the piston separates the inside space of the cylinder into two equal parts each of volume V0 , in which an ideal gas is contained under the same pressure P0 and at the same temperature. What work has to be performed in order to increase isothermally the volume of one part of gas n times compared to that of the other by slowly moving the piston?

5.

Consider the cyclic process ABCA, shown in figure performed on a sample of 2 mole of an ideal gas. A total of 1200 J of heat is withdrawn from the sample in the process. Find the work done by the gas during the part BC?

6.

Figure shows the variation in the internal energy U with the volume V of 2.0 moles of an ideal gas in a cyclic process abcda. The temperature of the gas at b and c are 500 K and 300 K respectively. Calculate the heat absorbed by the gas during the process.

7.

A gas expands in a piston – cylinder device from volume V1 to V2 the process being described by P =

a + b, v

a and b are constants find work done in the

process. 8.

An ideal gas is taken through the cycle A → B → C → A , If the net heat supplied to the gas in the cycle is 5 J, the work done by the gas in the process C → A is ___

9.

A thermodynamic system is taken through the closed cyclic P Q R S p process. The net work done by the system is ____

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SOLUTIONS 1.

L = initial position of piston when spring is relaxed. When the gas is heated, it expands and pushes the piston up by a distance x, (say). The spring is compressed. F=kx F kx PS = = A A

( pressure exerted on the gas by

k P L

the spring)

∴ ∴

a s

kx A

Increase in volume dv = A . dx. Work done is given as x x kx  W = ∫ P.dv = ∫  Po + . A. dx = P A  o ∫o dx + k A o 1 w = Po A x + kx2 2

x

∫ x . dx o

∴ 50 = 1.013 × 105 × 4× 10−3 × 0.1 + ∴ k = 1896 N / m Po

x G

∴ P = Po + PS = Po +

0

M

1 k. ( 0.1) 2 2

= 0.76 m of Hg = 0.76 X 9.8 X 13600 = 1.013 × 105 N / m2

W = 50 J we use the A = 4 X 10-3 m2 X = 0.1 m

For the heat energy Q supplied by the heater, first law of thermodynamics  for monoatomic gas C = 3 R  Q ∆U + w   V 2   Now ∆U = n CV ∆T 3 3 ∆V = n R ∆ T = × 2 × 8.31 × 50 2 2 = 1246. 5 J ∴ Q = 1246 . 5 + 50 = 1296 . 5 J

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2.

F = P1 A + k x w = ∫ F.dx =

∫( P

1

)

A + k x dx

1 w = P1 A x+ k x2 2

1 × k ( 0.1) 2 2 Solving we get k = 2000 N/m dQ = du = dw 3 = n R T + 50 2

(

)

50 = 105 × 4 × 10−3 × 0.1 +

H

e a t e r

3 × 2 × 8.3× 50+ 50 2 = 1245 + 50 dQ = 1295 Joule =

3.

At constant pressure heat absorbed by system is Q = n Cp ∆T And ∆U = n CV ∆T ∆U n CV ∆T C 1 = = V = Q n/ CV ∆T Cp r From first law of thermodynamics, we know w = Q − ∆U w Q − ∆U ∆U 1 = =1− = 1− Fraction Q Q Q r

∴ fraction

4.

Let volume of chamber change by ∆V. According to the problem, the final volume of left chamber is n – times final volume of right chamber. ∴ Vo ∆V = n vo − ∆ V

(

)

n − 1 ∆V =   Vo  n + 1 As piston is moved slowly ∴ K . E is zero. By W. E – T, we can write Wgas in right + Wgas in left + Wext = ∆ k .E chamber

chamber

(

Agent

∴ Wext Agent = − wgas ( R ) + wgas ( L )

P

o

Vo T

L

o

P

o

Vo To

R

)

We know that in isothermal process,  Vf  W = nR T ln    Vi   Vo + ∆ V  WL = Po Vo l n    Vo  Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com 7

 2n  = Po Vo l n   n + 1     Vo − ∆ V  WR = Po Vo l n    Vo   2n  = Po Vo l n   n + 1     2n   2n  Wext agent = − Po Vo l n   − PoVo l n    n + 1  n + 1  n + 1 = PoVo l n   4n   

Wext agent 5.

2

In the Cyclic process ∆U = 0 ∴ Q = ∆U + W ∴ W = Q = ∆ U = − 1200− 0 W = − 1200 J From C to A, ∴ WCA 0

T

∆V = 0

C

5 0 0

k

3 0 0

k

B A

O

For the whole cycle WAB + WBC + WCA = W

V

Now work done from A to B In the process V ∝ T so P =

∴ W + W = − 1200 J AB BC ∴ 3324 + WBC = - 1200 J WBC = − 4524 J

constant ∴ PV = n R T P ∆ V=n R ∆ T ∴ WAB = P ∆ V =nR ∆T

(

)

= 2 × 8.31× 500− 300 = 33245 6.

In the process a to b and c to d ∴ ∆ T = 0 ( or ) As ∆ U = 0 Vf

W=

∫ Pdv

wehavePV = n R T

T = constan t

U

a

b 5 0 0

k

d

c

k

Vi

P= vf

∴W =

∫

vi

nR T v

.dv

nR T V

V

0

2 V0

3 0 0

V

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vf

vf  vf  dv = n R T  l n V = n R T  l n  vi v vi   vi 2 vo ∴ Wab = n R Tb l n = 2R × 500l n2 = 1000R l n 2 vo v 1 Wcb = n R Tc l n o = 2R × 300l n = − 600R l n 2 2vo 2 There is no volume charges from b to c and from d to a, so Wbc = Wda = 0 ∴ W = Wab + Wbc + Wcd + Wda W = nR T

∫

= 1000 R l n 2 + 0 – 600 R l n 2 +0 = 400 R l n 2 ∴ from FL OT Q = ∆U +W = 0+ 400R l n 2 ∴ Q = 400R l n 2

7.

W=

v2

∫ P.dv

v2

=

v2

a



∫  v + b .dv

v1

v2

= al n v + bv v1

(

) (

= al nv2 + bv2 − al nv1 + bv1

)

v  w = al n  2  + b( v2 − v1 )  v1  8.

from

A→B

wAB = P  vB − vA  = 10 ( 2 − 1) = 10 J from B → C , V = constan t ∴ WBC = 0 Given Q = 5 J ∴ W = WAB + WBC + WCA 5 = 10 + 0 + wCA

(

2 V

( 3 m)

B

C

1

A P

( N2 ) / m

1 0

)

WCA = − 5 J Q it is anti clock wize 9.

W = Area Q MNR – Area PMNS = 100 × 103 ( 300 − 200) × 10−6 − 300× 103 ( 300 − 200) × 10−6 = ( 10 − 30) J = − 20 J

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P 2 0 0

k p

1 0 0

k p Q

S

P

R M

1 0 0

→

N 3 0 0 c c

Vc c

FLOT: - It is an extension of the principle of conservation of energy dQ = dU + dw In other words Suppose a heat Q is given to a system. This heat is partly used by system in doing work against its surroundings and partly its internal energy gets increased and form energy conservation principle. Heat added to system Q > 0. Q = ∆ U + ∆W Heat removed from system Q < 0. Work done by the system W > 0. Work done on the system W < 0.

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Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com 12