CLS Aipmt 17 18 XIII Phy Study Package 3 SET 1 Chapter 12

Chapter 12

Thermodynamics Solutions

SECTION - A

Objective Type Questions 1.

In ther thermo modyn dynam amics ics the Ze Zero roth th law law is is relat related ed to to (1) Work done

(2)

Thermal equilibrium

(3)

Entropy

(4)

Diffusion

(3)

W = = 0

(4)

Both (1) & (3)

(4)) (4

C V =

(4)

All of these

Sol. Answer (2) Zeroth law related to thermal equilibrium. 2.

For a cyc ycli lic c pro roce cess ss (1)) U = (1 = 0

(2)

Q = 0

Sol. Answer (1) Since initial and final points are at same, temperature so U = = 0 3.

Select Sel ect the the incorr incorrect ect rela relation tion.. (Where (Where sym symbols bols have have their their usua usuall meaning meanings) s) (1)) C P = (1

 R  – 1

(2)) (2

C P – C V = R

Pf Vf – Pi Vi

(3)) (3

= U =

(3)

Volume

1 – 

R

 – 1

Sol. Answer (3)

U  4.

 Pi Vi is the correct relation. 1

Pf Vf

Inte In tern rnal al ene energ rgy y of a no nonn-id idea eall gas gas depe depend nds s on (1) Temperature

(2)

Pressure

Sol. Answer (4) Depends on Temperature, Pressure, Volume. 5.

For an an adiabati adiabatic c expansion expansion of an ideal ideal gas gas the fract fractiona ionall change change in its its pressure pressure is is equal equal to (1)) –  (1

V

(2)) (2

dV

dV

–

(3)) (3

 V

dV V

(4)) (4

dV

– 

V

Sol. Answer (4) ∵

PV  = constant

Then,

P P

So, P  V – 

 dV    V 

  

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180

6.

Thermodynami cs

Solutions of Assignment

Which Whic h of the the followi following ng laws laws of therm thermody odynam namics ics defin defines es intern internal al energy energy? ? (1) Zeroth law

(2)

Second law

(3)

First law

(4)

Third law

Sol. Answer (3) Internal energy is defined in first law ∵

Q = U + W

So, U = Q – W 7.

Select Sele ct the the correct correct statem statement ent for for work, work, heat heat and and change change in intern internal al energy energy.. (1)) Heat supplied (1 supplied and work work done depend depend on initial initial and final final states states (2)) Change in internal energy (2 energy depends on the initial initial and final final states only (3)) Heat and work depend (3 depend on the the path between between the the two points (4) Al Alll of of the these se

Sol. Answer (4) All statements are correct. 8.

Morning breakfa breakfast st gives gives 5000 cal cal to a 60 kg person. person. The efficiency efficiency of person person is 30%. 30%. The The height upto which which the person can climb up by using energy obtained from breakfast is (1) 5 m

(2)

10.5 m

(3)

15 m

(4)

16.5 m

Sol. Answer (2) ∵

W =  JQ So, mgh =  JQ

h

9.

JQ mgh

 30  4.2  5000 5000    4.2  100   = 10.5 m 60  10

Select Sele ct the the incorrec incorrectt statem statement ent about about the the specif specific ic heats heats of a gaseous gaseous syste system. m. (1)) Specif (1 pecific ic heat at no exchange exchange conditi condition, on, C A = 0 (3)) Specif (3 pecific ic heat at constan constantt pressure, pressure,

C P



 R   1

(2)

Spec ecif ific ic he heat at at co cons nsttan antt te tempe pera rattur ure, e, C T = 

(4)

Spec pecifi ific c heat heat at con const stant ant vol volume ume,,

C V



R



Sol. Answer (4) The correct value of

C V



R

  1

10. Work done in the the cyclic cyclic process process show shown n in figur figure e is P

3P 0

P 0

(0, 0)

(1) 4P 0V 0

(2)

–4P 0V 0


V 0

V

3V 0 (3)) (3



22 7

P0V 0

(4)

–13P 0V 0


Thermodynamics 181


Sol. Answer (3) Cyclic process is anticlockwise then Work done = –(Area of P -V graph) graph) W = = – R 1R 2

 3P0  P0   3V0  V0        2 2

  

 11.

22 7

P0V 0

In following figures figures (a) to to (d), variation variation of volume volume by change of pressure is shown shown in figure. The gas is taken taken along the path ABCDA. Change in internal energy of the gas will be (a)) (a

(b)) (b

P D

P

(c)

D

C

C

A

B V

D C

A

B

B O

V

O

D

P

C A

A

(d)) (d

P

V

O

O

B V

(1)) Posi (1 Positive tive in in all cases cases from from (a) to (d) (d) (2)) Posi (2 Positive tive in cases cases (a), (a), (b) and (c) (c) but zero zero in case (d) (d) (3)) Negative in cases (a), (b) and (c) but zero in case (d) (3 (d) (4)) Zer (4 Zero o in all the the four four case cases s Sol. Answer (4) = 0 in all cases because cyclic process. U = 12. In a thermodynamic thermodynamic process process pressure of a fixed mass of a gas is changed in such a manner manner that the gas releases 20 J of heat when 8 J of work was done on the gas. If the initial internal energy of the gas was 30 J, then the final internal energy will be (1) 2 J

(2)

18 J 18

(3)

42 J 42

(4)

58 J 58

Sol. Answer (2) We know by 1st Law of Thermodynamics

Q = U + W – 20 J = U – – 8 J ∵

U = U final – U initial = – 12 J U =

So, U final = U initial + U = 30 + (–12) = 18 J 13.. A perf 13 perfect ect gas goe goes s from from stat state e A to state B by absorbing 8 × 105 joule and doing 6.5 × 105 joule of external work. If it is taken from same initial state A to final state B in another process in which it absorbs 105 J of heat, then in the second process work done (1) On ga gas s is is 1 10 05 J

(2)

On gas is 0.5 × 105 J

(3) By ga gas s is is 1 10 05 J

(4)

By gas is 0.5 × 105 J



182

Thermodynami cs


Sol. Answer (2)

Q = U + W 8 × 105 = U + + 6.5 × 105 1.5 × 105 J = U Again using Q = U + + W for for the second case U will will stay the same. Now, 105 = 1.5 × 10 5 + W – 0.5 × 105 = W negative sign indicates work is being done on the gas. 14.. Figur 14 Figure e sho shows ws two two pro proces cesse ses s a and b for a given sample of gas. If Q1, Q2 are the amount of heat absorbed by the system in the two cases; and U 1, U 2 are changes in internal energy respectively, then a P b

(0,0)

V

(1)) Q1 = Q2; U 1 = U 2 (1

(2)) (2

Q1 > Q2; U 1 > U 2

(3)) Q1 < Q2; U 1 < U 2 (3

(4)) (4

Q1 > Q2; U 1 = U 2

Sol. Answer (4) ∵

Initial and final states are same.

a

 U 1 = U 2

P b

Area under 'a' > area under 'b' i .e., W 1 > W 2

 Heat absorbed by a > heat absorbed by b

(0,0)

V

Q1 > Q2 15. A gas undergoes undergoes a change change at constant constant temperatur temperature. e. Which of the following following quantities quantities remain remain fixed? (1) Pressure

(2)

Entropy

(3) Heat exchanged with the system

(4)

All the above may change

Sol. Answer (4) When temperature change = 0 then, P 1V 1 = P 2V 2 = constant

Rest may change. 16.. Fol 16 Follow lowing ing fig figure ure sho shows ws P -T graph graph for four processes A, B, C and and D. Select the correct alternative. P

D

C B

A T

(0, 0) (1)) A – Isobaric process (1

(2)

B – Adiabatic process

(3)) C – Isochoric process (3

(4)

D – Isothermal process



Thermodynamics 183


Sol. Answer (3)

P

D

(A)) Temper (A Temperature ature is is constant constant – isotherm isothermal al

C B

(B)) Pres (B Pressure sure is constan constantt – Isobaric Isobaric (C) Pr Pres essu sure re  Temperature – Isochoric process

A

(D)) P 1–  T  = constant – Adiabatic process (D

T

(0, 0)

17. An ideal gas with adiab adiabatic atic expo exponent nent  is is heated at constant pressure. It absorbs Q amount of heat. Fraction of heat absorbed in increasing the temperature is (1))  (1

1

(2)) (2

(3)) (3



1

1

(4)



2 

Sol. Answer (2) Heat absorbed in increasing temperature = U = = Q – W = = nC V T Fraction of heat absorbed =

Heat absorbed Total heat





nCV T nCP T C V C P



1



18. A certain amount amount of an ideal monatomic monatomic gas needs needs 20 J of heat energy to raise raise its temperature temperature by 10°C at constant pressure. The heat needed for the same temperature rise at constant volume will be (1) 30 J

(2)

12 J

(3)

200 J

(4)

215.3 J

Sol. Answer (2)

Q = nC P T 20 = nC P × 10

...(1)

= nC V T U =

U  n

C P



∵

T

 mono  5 / 3

= 12 J U = 19. Two Two cylinders contain contain same amount of ideal monatomic monatomic gas. Same Same amount of heat is given given to two two cylinders. If temperature rise in cylinder A is T 0 then temperature rise in cylinder B will be

4

(1)) (1

3

T 0

(2)

Free Fr ee pi pist ston on

Fix ixed ed pi pist ston on

A

B

Heat

Heat

2T 0


(3)) (3

T 0

2

5

(4)) (4

3

T 0


184

Thermodynami cs


Sol. Answer (4) Cylinder A

Cylinder B

Free piston i .e., at

Fixed piston i .e., at

constant pressure

constant volume

Q = U nC P T = = nC V T ' C P T 0 = C V (T )' )'

T  

C P C V

T0

 T0 

5 3

T0

20. A sample sample of an an ideal gas undergoes undergoes an isotherma isothermall expansion. expansion. IfIf dQ, dU and and dW represent represent the amount of heat supplied, the change in internal energy and the work done respectively, then (1)) dQ = +ve, dU = (1 = +ve, dW = = +ve

(2)

dQ = +ve, dU = = 0, dW = = +ve

(3)) dQ = +ve, dU = (3 = +ve, dW = = 0

(4)

dQ = –ve, dU = = –ve, dW = = –ve

Sol. Answer (2) dQ = positive, dU = = zero, dW = = positive ∵

dQ = dU + dW

21.. In the 21 the dia diagr gram am sho shown wn Qiaf = 80 cal and W iaf = 50 cal. If W = = –30 cal for the curved path fi, value of Q for path fi, will be P a

f

i V

(0, 0) (1) 60 cal

(2)

30 cal

(3)

–30 cal

(4)

–60 cal

Sol. Answer (4) From process iaf Find U first, Q = W + + U

a

f

80 = 50 + U 30 cal = U

i

Use this U for for process if

Q = W + U Q = – 30 + (–30) = – 60 cal 22. A mass mass of dry dry air at N.T N.T.P .P.. is compresse compressed d to

1 32

th of its original volume suddenly. If  = = 1.4, the final pressure

would be (1) 32 atm

(2)

128 atm


(3)

1 32

attm a

(4)

150 atm


Thermodynamics 185


Sol. Answer (2) Process carried out suddenly so process is adiabatic.





 K



 P2V2

PV

P1V

1



P2

 V   P 1  1   V 2 

P 2

 V 1   (1 atm)   V 1 / 32 

7/5

P 2 = 1 atm × (25)7/5

= 128 atm 23.. Two sa 23 samp mple les s A and B of a gas initially at the same temperature and pressure, are compressed from volume V to

V

2

( A isothermally and B adiabatically). The final pressure

(1)) P A > P B (1

(2)) (2

P A = P B

(3)) (3

P A < P B

(4)) (4

P A = 2 P B

(3)

1.4  10 5 N/m 2

(4)

2.75  10 5 N/m 2

(3)) (3

Q = W

(4)) (4

Q = –U

Sol. Answer (3) P P B

B

P A

A

V V

V

2

i .e., P A < P B

24. The adiaba adiabatic tic elastici elasticity ty of a diato diatomic mic gas gas at NTP NTP is (1) Zero

(2)

1  10 5 N/m 2

Sol. Answer (3) Adiabat Adiabatic ic elasti elastici city ty = P 

7 5

 1.01 105

5 1.41 414 4  10  1.

2

N/m N/m

25.. For an iso 25 isome metri tric c proc process ess (1)) W = – U (1

(2)) (2

Q = U

Sol. Answer (2) For an isometric process, (i .e., isochoric) workdone = zero So Q = U Aakash Educational Services Pvt. Ltd.-Regd.


186

Thermodynami cs

Solutions of Assignment 1

26. A mixtu mixture re of gases at NTP NTP for for w which hich  = = 1.5 is suddenly compressed to

9

th of its original volume. The final

temperature of mixture is (1) 300°C

(2)

546°C

(3)

420°C

(4)

872°C

(4)

Adiabatic

Sol. Answer (2) –1 TV  –1 = constant –1 –1 T 1V 1  –1 = T 2V 2  –1

1

T2

 V   T 1  1   V 2 

T 2

 V   (273 K)  1  V / 9 

1.5 1.5 1

T 2 = (273 K) × 3

= 819 K = 546°C 27. In which which process process P-V diagram diagram is a straigh straightt line parallel parallel to the volume volume axis? axis? (1) Isochoric

(2)

Isobaric

Sol. Answer (2)

(3)

Isothermal

P

V

Process having a constant pressure, so isobaric process. 28. T h e P - V plots for two gases during adiabatic processes are shown in the figure. The graphs 1 and 2 should correspond respectively to P

1 2

(0, 0) (1) O2 a an nd He

(2)

He and O2

(3)

V

O2 a an nd CO

(4)

N2 and O2

Sol. Answer (1) PV  = constant [equation of graphs]

P

So for more  less less the rate of change or slope of graph and  is is less for diatomic.

1 (diatomic)

So graph 1 for O2

2 (monoatomic)

Graph 2 for He. Aakash Educational Services Pvt. Ltd.-Regd.

V


Thermodynamics 187


29. The pressu pressure re and volum volume e of a gas gas are change changed d as shown shown in the the P-V diagram diagram in this figure. The temperature of the gas will P A

B

D

C V

(1)) Incr (1 Increase ease as it goes from A to B

(2)

Incr In crea ease se as it go goes es fr from om B to C

(3) Remain constant during these changes

(4)

Decrease as it goes from D to A

Sol. Answer (1) In the process A  B

P

Pressure is constant. PV

So

A

B

T 2

 nRT

T 3

D

 T

V

T 1

C T 4 T 5

and volume is increasing so temperature also increases.

T5

>

T4

>

T 3

>

T2

>

T1

V

30.. Th 30 The e fig figur ure e sho shows ws P -V diagram diagram of a thermodynamic cycle. Which corresponding curve is correct? P B

C

A

D V

(0, 0) P

C

P

B A

(1)) (1 O

D

(2)) (2 T

P

D

C

A O

P

D

D

A

A

B

(3)) (3

B O

T

C T

(4)) (4 O

C B T

Sol. Answer (1) A  B V = = constant ∵

P B

PV = RT A P



R V

T

O

C

D T

Compare with y = = mx P -T graph is a straight line which must passes from origin A  B volume constant, P-increasing, T-increasing. Aakash Educational Services Pvt. Ltd.-Regd.


188

Thermodynami cs


B  C pressure pressure constant, volume - increasing, temperature - increasing B  C P = = constant, origin P -T graph graph is a straightline parallel to v -axis -axis C  D V = = constant then P



R V

T

P -T graph graph is straight line must passes from origin D  A P = constant P -T graph graph is a straightline parallel to T -axis. -axis.

31. During the thermody thermodynamic namic process process shown in in figure figure for an ideal ideal gas P

V

(1)) T = (1 = 0

(2)

Q = 0

(3)

W < 0

(4)

> 0 U >

Sol. Answer (4) For a straight P -V graph graph line P  V

P

If pressure increases, volume increases then T also also increases [PV  T ] So T  0 V

Volume increasing so work is positive, W > > 0 and temperature also increasing so Q > 0 ∵

Q = U + W

So U > 0 32. For P-V diagram of a thermodynamic cycle as shown in figure, process BC and and DA are isothermal. Which of the corresponding graphs is correct? P

A

B

C D V

B

A

P

P C

(1)) (1

(2)) (2 T

(0,0)


T

V

C

B

(3)) (3

D

C

V

C

D

(0,0)

D

B

A

A

(4)) (4

A

(0,0)

B

T

D

(0,0)

T


Thermodynamics 189


Sol. Answer (2) From A  B, volume increasing, increasing, pressure pressure constant P

ressure B   C , P re



1

⇒

Volume

B

A

Temperature constant

C D

A Same for D  C  D

(0,0)

T

pressure decreasing, volume constant

So P  T 33. Wor Work k done for the the proces process s shown shown in the figur figure e is V B(30

A(10

kPa, 25 cc)

kPa, 10 cc) P

(1) 1 J

(2)

1.5 J

(3)

4.5 J

(4)

0.3 J

Sol. Answer (4) v

Area under graph and V axis axis = work done

w

25 cc

1   (30  10)  1 03  (25  10) 1 06 2

10 cc

= 0.3 J

B

A

P

34. During During which of the follow following ing thermody thermodynami namic c process process represent represented ed by PV diagram diagram the heat energy absorbed by system may be equal to area under PV graph? graph? P

P

A

(1)) (1

P

A

(2)) (2

(3)) (3

B

(4)

B V

O

A

O

B

V

O

All of these

V

Sol. Answer (4) P

P A

B T O

∵

V

P

A

O

A

B T V

O

B T V

T is is constant in all cases.



190

Thermodynami cs


35. The specif specific ic heat heat of a gas in a polytrop polytropic ic process process is given given by (1)) (1

R

 – 1



R

R

(2)) (2

N – 1

1 – 



R

(3)) (3

1 – N

R

 – 1

–

R N – 1

(4)) (4

R

1 – 

–

R

1 – N

Sol. Answer (3)

∵

 C V 

C

R



1  N

R

  1



R N  1

36. For a certain certain process, process, pressure pressure of diatomic gas varies varies according according to the relation P = = aV 2, where a is constant. What is the molar heat capacity of the gas for this process? (1)) (1

17R

(2)) (2

6

6R

(3)) (3

17

13R 6

(4)) (4

16R 7

Sol. Answer (1) P = aV 2 PV –2 = a

Compare with PV N = constant then N = = – 2

Polytropic process ∵

C

 C V 







R

  1

R 1  N



1  N

R

7    1 5  5R 2



 

R



of diatomic diatomic



7 5



R

1  ( 2)

R

1 2



17R 6

37. In a thermody thermodynami namic c process process two moles moles of a monatom monatomic ic ideal gas gas obeys P  V 2 . If temperature of the gas increases from 300 K to 400 K, then find work done by the gas (where R = = universal gas constant). (1) 200 R

(2)

–200 R

(3)

–100 R

(4)

–400 R

Sol. Answer (2) P  V –2 PV 2 = constant

Compare with PV N = constant then N = = 2

 R   T  1  N 

W

 

W





R (T2  T 1 ) 1  N 2  R (400  300) (1  2)

= –200 R Aakash Educational Services Pvt. Ltd.-Regd.


Thermodynamics 191


38.. En 38 Entro tropy py of of a syst system em decr decreas eases es (1)) When heat is supplied to (1 to a system system at constant constant temperature temperature (2)) When heat is taken out from the system (2 system at constant constant temperature temperature (3) At equ equili ilibr brium ium (4)) In any (4 any spont spontaneo aneous us proces process s Sol. Answer (2) Entropy of a system decreases when heat is taken out of the system at constant temperature. 39. If during an an adiabatic process process the pressure of mixture of of gases is found found to be proportional to square of its absolute temperature. The ratio of C P /C V for mixture of gases is (1) 2

(2)

1.5 1.

(3)

1.67 1.

(4)

2.1 2.

Sol. Answer (1) P  T 2 PT –2 = constant C P

compare with

     constant  1   

PT 

 2

C V

40. If the the effic efficiency iency of a carn carnot ot engine engine is , then the coefficient of performance of a heat pump working between the same temperatures will be (1) 1 – 

(2)) (2

1



(3)) (3



1



(4)

1+

1



Sol. Answer (3) Coefficient of performance of heat pump =

1 1  efficiency of Carnot engine 

41. In a Carnot Carnot engine, engine, when heat heat is absorbed absorbed from the source, source, temperatur temperature e of source source (1) Increases

(2)

Decreases

(3) Remains constant

(4)

Cannot say

Sol. Answer (3) Even when heat is taken out temperature stays the same. i .e., heat capacity of surface is infinite. 42. A Carnot Carnot engine working working between between 300 300 K and 600 K has a work work output output of 800 J per cycle. cycle. The amount amount of heat heat energy supplied to engine from the source in each cycle is (1) 800 J

(2)

1600 J

(3)

3200 J

(4)

6400 J

Sol. Answer (2) W = = 800 J W Q

 1

800

Q

T 2 T 1

 1

300 600

1600 J = Q Aakash Educational Services Pvt. Ltd.-Regd.


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43. An ideal heat heat engine operate operates s on Carnot Carnot cycle between between 227°C 227°C and 127°C. 127°C. It absorbs absorbs 6 × 104 cal at the higher temperature. The amount of heat converted into work equals to (1) 4.8 × 1 10 04 c ca al

(2)

3.5 × 104 c ca al

(3)

1.6 × 104 c ca al

1.2 × 104 cal

(4)

Sol. Answer (4) W

6  10

4

 1

400 500

W = = 1.2 × 10 4 cal

44. The maxim maximum um possibl possible e efficien efficiency cy of a heat heat engine engine is (1) 100% (2)) (2

(3)) (3

T 1 T 2 T 1 T 2

1



T



(4)) Depende (4 Dependent nt upon upon the temper temperature ature of source source (T 1) and sink (T 2) and is equal to  1  2  T



1



Sol. Answer (4)

  1

T 2 T 1

So it depends on source and sink temperature. 45. A frictionless frictionless heat engine can be 100% efficient only if its exhaust temperatu temperature re is (1) Equal to its input temperature

(2)

Less than its input temperature

(3) 0 K

(4)

0°C

Sol. Answer (3) ∵

  1

T 2 T 1

If exhaust temperature zero kelvin then  = 100%. 46. A reversible reversible engine and an irreversi irreversible ble engine are working working between between the same temperat temperatures. ures. The efficienc efficiency y of the (1) Two engines are same

(2)

Reversible engine is greater

(3) Irreversible engine is greater

(4)

Two engines cannot be compared

Sol. Answer (2) Efficiency of reversible engine is greater, because there is no loss of heat. 47. Which of the the following following can be be coefficient coefficient of perfo performanc rmance e of refrigerat refrigerator? or? (1) 1

(2)

0.5

(3)

9

(4)

All of these

Sol. Answer (4)

 

1 

 1



1



Thermodynamics 193


 is less than 1 so 

1



1



1

1 0

 >0 48. The temperature temperature inside and outside a refrigerator refrigerator are are 273 K and 300 K respectively respectively.. Assuming Assuming that the the refrigerator cycle is reversible, for every joule of work done, the heat delivered to the surrounding will be nearly (1) 11 J

(2)

22 J

(3)

33 J

(4)

50 J

Sol. Answer (1)

  1  

T 2 T 1

1 



;   1



100 9

273 300

 1



91 9

9 100

∼ 11 J

Q W

For W = = 1 J Q= Q = 11 J

49. By opening opening the the door of a refrige refrigerator rator placed inside inside a room you you (1)) Can cool (1 cool the the room room to certai certain n degree degree (2)) Can cool it to the temper (2 temperature ature inside inside the refrigera refrigerator tor (3)) Ulti (3 Ultimat mately ely warm warm the room room slightly slightly (4)) Can neither (4 neither cool nor nor warm warm the the room room Sol. Answer (3) Ultimately warm the room because work is being done by the refrigerator. 50. A Carnot engine whose whose sink is at 300 K has has an efficiency efficiency of 40%. 40%. By how how much should should the temperatur temperature e of source be increased to as to increase its efficiency by 50% of original efficiency? (1) 150 K

(2)

250 K

(3)

300 K

(4)

450 K

Sol. Answer (2) 40 300  1 T 1 100 T 1

 500 K

50% increase in efficiency 150  0.4  0.6 100 new efficiency efficiency = 0.6 =

60 100

60 300  1 T 1 100 T 1

750 K  750

Difference between 2 Temperatures = 250 K Aakash Educational Services Pvt. Ltd.-Regd.


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Thermodynami cs

Solutions of Assignment SECTION - B

Objective Type Questions 1.

A cont containe ainerr is filled filled with with 20 moles moles of an an ideal diatom diatomic ic gas at at absolute absolute tempe temperatu rature re T . When heat is supplied to gas temperature remains constant but 8 moles dissociate into atoms. Heat energy given to gas is (1) 4RT

(2)

6RT

(3)

3RT

(4)

5RT

Sol. Answer (1) Heat supplied = U = = U final – U initial Uinitial

U 



 2

5 2

 20  RT , U final 

5 2

  20  8  RT 

3 2

  2  8  RT

 8  RT

= 4RT

 Heat energy given is 4RT . 2.

Liquid oxygen oxygen at at 50 K is heate heated d to 300 K at constan constantt pressure pressure of of 1 atm. atm. The rate rate of heatin heating g is constant constant.. Which Which one of the following graphs represents the variation of temperature with time?

(1)) (1

e r u t a r e p m e T


(2)) (2

(3)) (3

Time


Time

(4)) (4


Time

Time

Sol. Answer (3) Liquid oxygen when heated will observe a rise in temperature as well as change in state one time, which can be represented as

3.


Forr an Fo an iso isobar baric ic pr proc ocess ess,, the the ra ratio tio of Q (amount of heat supplied) to the W (work done by the gas) is

    

C P 



 C V 

(1))  (1

– 1  –

(2)) (2

(3)

   1

(4)) (4

   1

Sol. Answer (4) For isobaric process Q = nC P T and and W = = nR T So,

CP Q CP    R CP  CV W

1 1

C V



  1

 C P    ∵  CV 

C P



Thermodynamics 195


4.

3 moles of an ideal ideal gas are are contained contained withi within n a cylinder cylinder by a frictionle frictionless ss piston piston and are initia initially lly at tempera temperatur ture e T . The pressure of the gas remains constant while it is heated and its volume doubles. If R is molar gas constant, the work done by the gas in increasing its volume is 3

(1)) (1

2

RT ln 2

(2)

3RT ln ln 2

(3)

3 2

RT

(4)

3RT

Sol. Answer (4) W = P V

= PV = nRT = 3RT 5.

Two mo mole les s of of a ga gas s at at tem tempe pera ratu ture re T and and volume V are are heated to twice its volume at constant pressure. If C p C v

(1)) (1

  then increase in internal energy of the gas is RT

(2)) (2

  1

2RT

(3)) (3

  1

2RT 3(   1)

(4)) (4

2T

  1

Sol. Answer (2)

Q = U + W Q – W = U  W  W  U   1

 1  (P.V ) nRT 2RT U  W      1   1    1    1  6.

A triatomic, triatomic, diatomic and monatomic monatomic gas is supplie supplied d same same amou amount nt of of heat heat at at const constant ant pre pressu ssure, re, then then (1)) Fractiona (1 Fractionall energy used to to change internal internal energy is maximum maximum in monatomic monatomic gas (2)) Fractional energy used to change internal (2 internal energy energy is maximum in diatomic gas (3)) Fractional energy used to change internal (3 internal energy energy is maximum in triatomic triatomic gases (4)) Fractiona (4 Fractionall energy used to change internal internal energy is same same in all the three three gases

Sol. Answer (3)

U nCV T   Q nCP T

 U   Q   

mono



C V C P

1

 mono

1  U   Q      dia dia

5

1  U    Q    tria  tria

3





1



3 5

7

4

Fractional energy used to change internal energy is maximum in Triatomic gas. Aakash Educational Services Pvt. Ltd.-Regd.


196

7.

Thermodynami cs


105 calories calories of heat is required to raise raise the temperat temperature ure of 3 moles moles of an ideal ideal gas at constant constant pressure pressure from 30°°C to 35° 30 35°C. The amount of heat required in calories to raise the temperature of the gas through the range



C p



C v

(60°°C to 65° (60 65°C) at constant volume is     (1) 50 cal

(2)



 1.4 

75 cal



(3)

70 cal

(4)

90 cal

Sol. Answer (2) At constant pressure heat absorbed = Q = nC P T …(1) At constant volume heat absorbed = U = = nC V T

…(2)

Dividing (1) by (2), 105 Q C P     1.4 ⇒  1.4 U CV U

 U V = 75 cal 8.

To an ideal triatom triatomic ic gas 800 cal hea heatt energy energy is given at consta constant nt pressure pressure.. If vibration vibrational al mode is neglect neglected, ed, then energy used by gas in work done against surroundings is (1) 20 200 cal

(2)

300 cal

(3)

400 cal

(4)

60 cal

Sol. Answer (1) Heat at constant pressure

Q = nC P T Heat for doing work

W = nR T Then

nR T W  nCP T Q

W 800

W 800

W 800

   1    



 1

 1

1

 3 4

= 200 cal W = 9.

A clo close sed d cyli cylind ndri rica call vess vessel el con conta tain ins s N moles moles of an ideal diatomic gas at a temperature T . On supplying heat, temperature remains same, but n moles get dissociated into atoms. The heat supplied is (1)) (1

5 2

(N  n )RT

5

(2)) (2

2

nRT

(3)) (3

1 2

nRT

(4)) (4

3 2

nRT

Sol. Answer (3) Heat supplied = U = U f iinn al al  U i ni ni ti ti al al

Total internal energy initially =

5 2

NRT

[Only diatomic gas is present]

Total internal energy when Aakash Educational Services Pvt. Ltd.-Regd.


Thermodynamics 197


'n' moles get dissociated =

5 2

 N  n  RT



3 5  5  N  n  RT   2n  RT   NRT 2 2  2

U  

3 2

 2n  RT

[diatomic and monoatomic both are present]

Solving this we get U 

1

nRT

2

 Heat supplied is

1 2

nRT .

10. Figure shows shows the isotherms isotherms of of a fixed mass of an ideal gas at three temperat temperatures ures T A, T B and T C , then 1 V

A B C

P

O

(1)) T A > T B > T C (1

(2)) (2

T A < T B < T C

(3)) (3

T B < T A < T C

(4)) (4

T A = T B = T C

Sol. Answer (2) ∵

RT V

∵

∵

A

1 V

PV = RT

B

 P

C

For constant

1

So, P  T

V

O

P C > P B > P A then

P A P B

P C

P

T C > T B > T A

11.

An ideal monatomic monatomic gas at 300 300 K expands adiabatically adiabatically to 8 times its volume. volume. What is the the final temperature? temperature? (1) 75 K

(2)

300 K

(3)

560 K

(4)

340 K

Sol. Answer (1) Adiabatic expansion for monoatomic gas   for 1

T1V1

5 3

1  T2V2 

1

T2

 V   T 1  1   V 2 

 V   300  1   8V 1 

5/ 3



300 4

 75 K



198

Thermodynami cs


12.. Slo 12 Slope pe of isot isother herm m for for a gas (hav (having ing  

5 3

) is 3 × 105 N/m2. If the same gas is undergoing adiabatic change

then adiabatic elasticity at that instant is (1) 3 × 105 N/m2

5 × 105 N/m2

(2)

(3)

6 × 105 N/m2

(4)

10 × 105 N/m2

Sol. Answer (2) P Adiabatic elasticity =  P



5 3

5 5  3  10 10  5  10 10

2

N/ N/m

13.. A gas may 13 may expand either either adiabaticall adiabatically y or isothermally isothermally.. A number of P –V curves are drawn for the two processes over different range of pressure and vol ume. It will be found that (1)) An adiabatic (1 adiabatic curve curve and an isothermal isothermal curve curve may intersect (2)) Two adiabatic (2 adiabatic curves curves do not interse intersect ct (3)) Two isothermal (3 isothermal curves do not intersect (4) Al Alll of of the these se Sol. Answer (4) Slope for isothermal and adiabatic are not same so they will intersect. 14.. The var 14 variat iation ion of pres pressur sure e P with with volume V for for an ideal monatomic gas during an adiabatic process is shown in figure. At point A the magnitude of rate of change of pressure with volume is P

A

3 P 0

O

(1)) (1

3 P 0

(2)) (2

5 V 0

V

2 V 0

5 P 0

(3)) (3

3 V 0

3 P 0

(4)) (4

2 V 0

5 P 0 2 V 0

Sol. Answer (4) PV  = constant P  V –  dP P dP dV

P dV

    

P

3 P 0

A

V





5 3



3 P 0

O

2 V 0

V

2 V 0

5 P 0 2 V 0

 dP    dV 

Then 

V

5 P 0 2V 0



Thermodynamics 199


15. Fig Figure ure show shows, s, the adia adiabat batic ic curve curve on on a log T and and log V scale scale performed on ideal gas. The gas is log T 5 4 3 2 1 O

A B

1 2 3 4 5

log V

(1) Monatomic

(2)

Diatomic

(3) Polyatomic

(4)

Mixture of monatomic and diatomic

Sol. Answer (1) 1

TV

 K

logT     1 logV  0 logT      1 logV y = – (  – – 1) x y x

     1  slope =

– 1) =   – (  –  

24 4 1

2 3

5 3

 Monoatomic. 16. A cyclic process on on an ideal ideal monatomic monatomic gas is shown in figure. figure. The correct statement is P

B

C

A V

(1)) Work done by gas (1 gas in proc process ess AB is more than that in the process BC (2) Net heat heat energy energy has has been supplie supplied d to the the system system (3)) Temperature of the gas (3 gas is maximum maximum at state state B (4) In pr proc oces ess s CA, hea heatt energ energy y is abs absorb orbed ed by sys system tem Sol. Answer (2) It is a cyclic system   U = 0 and work done is (+)ive, so heat is supplied to system. 17. A diatomi diatomic c gas under undergoes goes a proce process ss represe represented nted by PV 1.3 = const constant. ant. Choos Choose e the incor incorrect rect stat statemen ementt (1)) The gas expands by absorbing heat from (1 from the surrounding surroundings s (2)) The gas (2 gas cools cools down down during during expansi expansion on (3)) The work (3 work done by surroundings surroundings during expansion expansion of the the gas is negative negative (4) No None ne of of the these se Aakash Educational Services Pvt. Ltd.-Regd.


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Thermodynami cs


Sol. Answer (4) PV 1.3 = K W



P2V2

 P1V1

∵

1  N

N > > 1, so W is is negative.

Heat supplied by surrounding heat goes to do work.

 Down when expands. 18.. If a gas 18 gas is is tak taken en fro from m A to C through through B then heat absorbed by the gas is 8 J. Heat absorbed by the gas in A C taking it from to directly directly is P (kPa) (kPa)

20

C B

10

A

200 (1) 8 J

(2)

400

9J

(3)

V (cc) (cc)

11 J

(4)

12 J

Sol. Answer (2) When taken through ABC [ [ U + + work = heat absorbed] Heat absorbed = area under graph + U = = 8 U  8 

10  200 1000

6

when taken directly to C W + U = Q

 10  200 1 2000   1000  2  10 6Q  Q = 9 J 1 000   19.. The 19 The pr proc oces ess s CD is shown in the diagram. As system is taken from C to D, what happens to the temperature of the system? P 3 p0

C

p0

D v 0

3v 0

V

(1) Tem empe pera ratu ture re fi firs rstt de decr crea ease ses s and and th then en in incr crea ease ses s

(2)

Tem empe pera ratu ture re fi firs rstt in incr crea ease ses s and and th then en de decr crea ease ses s

(3) Temperature de decreases co continuously

(4)

Temperature in increases co continuously

Sol. Answer (2)

P T 1T 2 T 3

T 3 > T 2 > T 1

So from C  D Temperature first increases then decreases. V Aakash Educational Services Pvt. Ltd.-Regd.


Thermodynamics 201


20. A P -T graph graph is shown for a cyclic process. Select correct statement regarding this P C

B A

D T

O

(1) Du Duri ring ng pro proce cess ss CD, work done by gas is negative (2) Du Durin ring g pro proces cess s AB, work done by the gas is positive (3) Du Durin ring g proc process ess BC internal internal energy of system increases (4) Du Durin ring g proc process ess BC internal internal energy of the system decreases Sol. Answer (3) In process BC (isochoric (isochoric process) where T is is (+)ive. So U = = nC V T increases T is positive  U increases

∵

21. A hydrogen cylinder is designed to withstand an an internal pressure pressure of 100 atm. At 27°C, 27°C, hydrogen is pumped into the cylinder which exerts a pressure of 20 atm. At what temperature does the danger of explosion first sets in? (1) 500 K

(2)

1500 K

(3)

1000 K

(4)

2000 K

Sol. Answer (2) Constant volume process PV = nRT P1 T1



20 300

P 2 T 2



100 T 2

T 2 = 1500 K

22.. An ide 22 ideal al gas gas of of volu volume me V and pressure P expands isothermally to volume 16 V and then compressed adiabatically to volume V . The final pressure of gas is [  = 1.5]  = (1)) P (1

(2)

3P

(3)

4P

(4)

6P

Sol. Answer (3) Isothermal expansion P 1V 1 = P 2V 2 PV = = 16 V × P ' P

 P '

16



P1V

1

 P2V2

[adiabatic compression]

P '(16 '(16 V )1.5 = P " ( V )1.5 P

16

 161.5  P "  4P



202

Thermodynami cs


23.. The 23 The pre press ssur ure e P of of an ideal diatomic gas varies with its absolute temperature T as as shown in figure. The molar heat capacity of gas during this process is [R is is gas constant] P

T

(1) 1.7 R

(2)

3.25 R

(3)

2.5 R

(4)

4.2 R

Sol. Answer (3) CV of diatomic =

5

R

2

24. An ideal ideal gas expa expands nds accor according ding to the law law P 2V = = constant. The internal energy of the gas (1) Increases continuously

(2)

Decreases continuously

(3) Remain constant

(4)

First increases and then decreases

Sol. Answer (1) P 2V = K

or PV –2 = K N = = –2 C

 C V 

R

1  N

 positive quantity

C > > 0 W > > 0 ∵

[gas is expanding]

PV –2 = K so TV –3 = constant

will increases if V increases.  T will

 T > 0 So U = = C T > > 0 It will increase continuously. 25. The The varia variati tion on of of pres pressu sure re P with with volume V for for an ideal diatomic gas is parabolic as shown in the figure. The molar specific heat of the gas during this process is P

V

O

(1)) (1

9R

(2)) (2

5

17R

(3)) (3

6

3R 4

(4)) (4

8R 5

Sol. Answer (2) P = aV –2 C 

 C 

R

  1

So, PV 2 = constant then N = 2



R 1  N

∵

= 1.4  =

17R 6



Thermodynamics 203


26. Neon gas of a given given mass expands expands isotherma isothermally lly to double double volume. What What should should be the further further fractional fractional decrease in pressure, so that the gas when adiabatically compressed from that state, reaches the original state? (1) 1 – 2–2/3

(2)

1 – 31/3

21/3

(3)

(4)

32/3

Sol. Answer (1) P1V1

 P2V2

PV

 P ' 2V

P

2



1

2

P

 P '

P1V P

[for isothermal]

 P2V2 5/3

  2V 

 P 2 .  2 

[for adiabatic] 5/3

 P2 V 

[  for neon = 5/3 ]  for

2/3

Fractional decrease



P2

 P

P2



P2

 P 2 .  2  P 2

2/3

 1  2 2/3

27. When 1 kg of ice ice at 0°C melts melts to water water at 0° C, C, the resulting resulting change in its its entropy, entropy, taking latent latent heat of ice to be 80 cal/°C is (1) 293 cal/K

(2)

273 cal/K

8 × 104 ca cal/K

(3)

(4)

80 cal/K

Sol. Answer (1)

∵

 Q   ml f     T    T 

Entropy  

y 

1 00 00 0  8 0 273

 293 cal/K

28.. Car 28 Carnot not cyc cycle le is is plot plotted ted in P -V graph. graph. Which portion represents an isothermal expansion? P A B

C

D

V

(1)) AB (1

(2)) (2

BC

(3)) (3

CD

(4)) (4

DA

Sol. Answer (1) AB is isothermal expansion. BC is is adiabatic expansion CD is isothermal compression

 A = adiabatic compression. Aakash Educational Services Pvt. Ltd.-Regd.


204

Thermodynami cs


29. Efficiency Efficiency of a heat engine working working between between a given source and and sink is 0.5. Coefficient Coefficient of of performance performance of the refrigerator working between the same source and the sink will be (1) 1

(2)

0.5

(3)

1.5

(4)

2

Sol. Answer (1)

 0.5

1 1 



1

=1

1 

30. A heat heat engine engine rejects rejects 600 cal to the sink sink at 27°C. 27°C. Amount Amount of work work done by the the engine will will be (Temperature of source is 227°C & J = = 4.2 J/cal) (1) 16 1680 J

(2)

840 J

(3)

2520 J

(4)

None of these

Sol. Answer (1) 1

1

W Q1

T 2 T1

300

Q1



500

W Q1

2



Q1 

W



5 5W 2

W = Q1 – Q2 Then Q2 = Q1 – W ∵

Q2 

5W 2

Then W 

 W 

2Q2 3



3W 2

2  600 3

= 400 cal = 400 × 4.2 J = 1680 J SECTION - C

Previous Years Questions 1.

A Ca Carno rnott engi engine ne hav having ing an ef effic ficien iency cy of

1 10

as heat engine, is used as a refrigerator. If the work done on the

system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is [NEET-2017] (1) 1 J

(2)

90 J 90

(3)

99 J 99

(4)

1 00 J 10

Sol. Answer (2)



=

1 



1



1

10 1

10



=9



=

9



10 1 10

Q2 W

Q2 = 9 × 10 = 90 J Aakash Educational Services Pvt. Ltd.-Regd.


Thermodynamics 205


2.

Thermo The rmodyna dynamic mic proc processe esses s are indi indicat cated ed in the the follow following ing diagr diagram. am. P IV

i

f

I

III

f

II

f 700 K 500 K 300 K

f

V

Match the following Column-I P.

Column-II

Process I

a.

Adiabatic

Q. Process II

b.

Isobaric

R. Process III

c.

Isochoric

S. Process IV

d.

Isothermal [NEET-2017]

(1) P  a, Q  c, R  d, S  b

(2)

P  c, Q  a, R  d, S  b

(3) P  c, Q  d, R  b, S  a

(4)

P  d, Q  b, R  a, S  c

Sol. Answer (2) Process Proc ess I = Isochori Isochoric c II = Adiab Adiabati atic c III = Isothermal IV = Isobaric Isobaric 3.

One mole mole of of an ideal mona monatomi tomic c gas unde undergoe rgoes s a process process desc described ribed by the the equation equationPV 3 = constant. The [NEET (Phase-2)-2016] heat capacity of the gas during this process is (1)) (1

3 2

R

5

(2)) (2

2

R

(3)

2R

(4)) (4

R

Sol, Answer (4) PV 3 = constant polytropic process with n = 3 C =

=



C v

R

1

r

R 1 n



R

1 n

=

R

5 3

4.

1



R

1 3



R

The tem temper peratu ature re ins inside ide a ref refrig rigera erator tor is t 2°C and the room temperature is t 1°C. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be [NEET (Phase-2) - 2016] (1)) (1

(3)) (3

t 1 t1



t 2

t 2



273

t1

 t 2


(2)) (2

(4)) (4

t 1



t1

273



t 2

t1  t 2 t 1  273


206

Thermodynami cs


Sol. Answer (2) K 

Q2 W

1



t 1 t 2

Q2 

1

t 2W t1  t 2

Q1  Q2  W 

= 5.

t2W t1  t 2 t1W t1

 t2

W 

t 1



t1

273

 t2

A refrigera refrigerator tor works works between between 4°C 4°C and 30°C. 30°C. It is required required to remove remove 600 600 calories calories of heat heat every second second in in order to keep the temperature of the refrigerated space constant. The power required is [Take 1 cal = 4.2 J) [NEET-2016] (1) 23 2365 W

(2)

2.365 W

(3)

23.65 W

(4)

236.5 W

Sol. Answer (4) T 2 = 4°C = 277 K T 1 = 303 K Q2 = 600 cal Q1 Q2





T 1 T 2

Q2  W Q2



T 1 T 2

W = = 236.5 W

6.

A gas is compre compressed ssed isothe isothermal rmally ly to half half its initial initial volume volume.. The same same gas is compr compressed essed s separ eparatel ately y through through an adiabatic process until its volume is again reduced to half. Then [NEET-2016] (1)) Which of the case (whether (1 (whether compression compression through through isothermal isothermal or through adiabatic adiabatic process) requires requires more work will depend upon the atomicity of the gas (2)) Compres (2 Compressing sing the gas gas isothermally isothermally will require require more more work to to be done (3)) Compre (3 Compressing ssing the gas through through adiabatic adiabatic process process will require require more work work to be done (4)) Compress (4 Compressing ing the gas isothermally isothermally or adiabatically adiabatically will require require the same amount amount of work

Sol. Answer (3) P

Adiabat Adi abatic ic Isothermal

v

2

v

V



Thermodynamics 207


7.

4.0 g of a gas gas occupies occupies 22.4 22.4 litres litres at NTP NTP.. The speci specific fic heat heat capacity capacity of of the gas gas at constan constantt volume volume is 5.0 –1 –1 –1 JK mol . If the speed of sound in this gas at NTP is 952 ms , then the heat capacity at constant pressure is (Take gas constant R = = 8.3 JK–1 mol –1) [Re-AIPMT-2015] (1) 8.5 J JK K–1 mol–1

(2)

8.0 JK–1 mol–1

(3) 7.5 J JK K–1 mol–1

(4)

7.0 JK–1 mol–1

Sol. Answer (2) 8.

The coeffici coefficient ent of perfor performance mance of of a refrigerat refrigerator or is 5. If the the temperat temperature ure inside inside freezer freezer is –20°C, –20°C, the the temperatu temperature re of the surroundings to which it rejects heat is [Re-AIPMT-2015] (1) 21°C

(2)

31°C

(3)

41°C

(4)

11°C

Sol. Answer (2) 9.

An ideal ideal gas is compre compressed ssed to to half its its initial initial volume volume by means means of severa severall processes processes.. Which of of the proces process s results in the maximum work done on the gas? [Re-AIPMT-2015] (1) Isothermal

(2)

Adiabatic

(3)

Isobaric

(4)

Isochoric

Sol. Answer (2) 10. One mole mole of an an ideal diatomic gas undergoes undergoes a transition transition from A to B along a path AB as shown in the figure P (in (in

kPa) 5

A

B

2

4

V

6 3

(in (i nm)

The change in internal energy of the gas during the transition is

[AIPMT-2015]

(1) –12 kJ (2) 20 kJ kJ (3) –20 kJ (4) 20 J Sol. Answer (3)

U = nC V T



R



 n  (T2  T 1 )    1  



n[RT2



5

 RT1 ]

  1 n(P2V2

A

B

2

 P1V1 )

  1

0 3



(kPa) P (kPa)

4

6

3

(m V (m

)

3

1(2  6  10 1 0  5  4  10 10 ) 7 / 5 1

 8  103 2/5

 20  103

= –20 kJ Aakash Educational Services Pvt. Ltd.-Regd.


208

Thermodynami cs


11. A Car Carnot not engine engine,, having having an eff efficien iciency cy of  =

10

as heat engine, is used as a refrigerator. If the work done

on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is [AIPMT-2015] (1) 1 J

(2)

100 J

(3) 99 J

(4)

90 J

Sol. Answer (4) ∵



1



 1

1 (1 / 10)

Q2 W

1

(10  1) 

Q2

10

Q2

10

Q2 = 90 J

12. Figure below shows two two paths that that may be taken by by a gas to to go from from a state state A to a state C . In process AB, 400 J of heat is added to the system and in process BC , 100 J of heat is added to the system. The heat [AIPMT-2015] absorbed by the system in the process AC will will be P B

4

6 × 10 10 Pa

C

A

4

2 × 10 10 Pa

V 2 × 10

–3

m

3

4 × 10

–3

m

(1) 300 J

(2)

380 J

(3) 500 J

(4)

460 J

3

Sol. Answer (4) In process ABC ∵

Q = U + W

So, U = Q – W = (400 + 100) – (6 × 10 4 × 2 × 10–3) U = = 500 – 120 U =

U = 380 J In process AC

Q = U + W

1   380    (2  10 4  6 10 4 )  2 10 3  2  = 380 + 80 = 460 J Aakash Educational Services Pvt. Ltd.-Regd.


Thermodynamics 209


13. A mon monoat oatom omic ic gas gas at a pre pressu ssure re P , having a volume V expands expands isothermally to a volume 2V and and then adiabatically [AIPMT-2014] to a volume 16V . the final pressure of the gas is: (take  = = 5/3) (1) 64P

(2)

32P

(3)) (3

P /64

(4)

16 1 6P

Sol. Answer (3) In isothermal process P 1V 1 = P 2V 2 PV = = P 2(2V ) P 2



P

...(1)

2

In adiabatic process 



2

3

P 2V = P 3V

 P      (2V )  P3 (16V ) 2 P 3



 2v     2  16V  P



P   1 

5/ 3



  2 8

P

64

14. A thermod thermodynam ynamic ic system system under undergoes goes cyclic cyclic proce process ss ABCDA as shown in figure. The work done by the system [AIPMT-2014] in the cycle is P C

B

3 P 0 2 P 0 P 0

D

A V 0

(1)) P 0 V 0 (1

(2)

2P 0 V 0

2V 0 V P0V 0

(3)) (3

(4)

2

Zero

Sol. Answer (4) P

W = = area enclosed by AODA + by area enclosed OBCO

C

1   1     (2V0  V0 )  P0     (2V0 V0 )P0  2   2  =0

B

3 P 0 O

2 P 0

clockwise  AODA is a clockwise   while DBCO is anticlockwise

P 0

D

A V 0

2V 0 V

15.. A gas 15 gas is take taken n thro through ugh the cyc cycle le A  B  C   A, as shown. What is the net work done by the gas? P (10 ( 10

5

Pa)) Pa 7

B

6 5 4 3

A

2

C

1 0 1

2

4

6

8

V (10

–3

m3) [NEET-2013]

(1) 1000 J

(2)

Zero


(3)

– 2000 J

(4)

2000 J


210

Thermodynami cs


Sol. Answer (1) ∵

Cyclic curve is clockwise i .e., W = = +ve

W = area enclosed



1 2

 5  10 3  4  105

= 1000 J 16. The molar molar specific specific heats of an ideal ideal gas at constant constant pressur pressure e and volume volume are denoted denoted byC p and C v respectively. If  

(1)) (1

C p

and R is is the universal gas constant, then C v is equal to

C v R

(2)) (2

(   1)

(   1)

(3)) (3

R

 R

[NEET-2013] 1

 1  

(4)) (4

Sol. Answer (1)

 

C p C v

We know C p – C v = R So

C v



R

  1

17. During an adiabatic adiabatic process, the pressure pressure of a gas is found found to be proportional proportional to the cube of its temperature. temperature. The ratio of

C p C v

for the gas is:

(1) 2

(2)

5

(3)) (3

3

3 2

[NEET-2013] 4

(4)) (4

3

Sol. Answer (3) P  T 3 PT –3 = constant

Compare with Then,

    1   PT 

 constant

  3 1  

 

3 2

18. In the the gi give ven n (V – – T ) diagram, what is the relation between pressures P 1 and P 2?

V

[NEET-2013]

P 2 P 1 2 1


T


Thermodynamics 211


(1)) P 2 > P 1 (1

(2)) (2

P 2 < P 1

(3) Ca Cannot be predicted

(4)

P 2 = P 1

Sol. Answer (2) PV = RT

∵

V T



i .e.,

R P

Slope of V -T grap raph  ∵ Slo   V   m  tan     T

 tan 

P  

1 tan 

2 > 1 so tan2 > tan 1  P 2 < P 1 , then P 2 < P 1 19. One mole mole of an ideal ideal gas goes from from an an initial initial state state A to final state B via two processes: It first undergoes isothermal expansion from volume V to to 3 V and and then its volume is reduced from 3V to to V at at constant pressure. The correct P-V diagram diagram representing the two processes is [AIPMT (Prelims)-2012] A

(1)) (1

A

(2)) (2

P B

(3)) (3

P

3V

(4)

P

B

V

A

B

V

V

V

V

3V

V

3V

V

3V

V

P B

A

Sol. Answer (2) A

P B

3V

V V

20. A thermody thermodynami namic c system system is taken taken throu through gh the cycle cycle ABCD as shown in figure. Heat rejected by the gas during the cycle is [AIPMT (Prelims)-2012] D

C

A

B

V

3V

2 P e r u s s e r P

P

Volume

(1)) (1

1 2

PV

(2)) (2

PV

(3)

2 PV

(4)

4 PV

Sol. Answer (3) Heat rejected = workdone by surrounding = area of PV graph graph = P × × 2 V = = 2PV



212

Thermodynami cs


21. An ide ideal al gas goe goes s from from sta state te A to state B via three different processes as indicated in the P -V diagram diagram A

1 2

P

3

B

V

If Q1, Q2, Q3 indicate the heat absorbed by the gas along the three processes and U 1, U 2, U 3 indicate the change in internal energy along the three processes respectively, then [AIPMT (Mains)-2012] (1)) Q1 > Q2 > Q3 and U 1 = U 2 = U 3 (1

(2)) (2

Q3 > Q2 > Q1 and U 1 = U 2 = U 3

(3)) Q1 = Q2 = Q3 and U 1 > U 2 > U 3 (3

(4)) (4

Q3 > Q2 > Q1 and U 1 > U 2 > U 3

Sol. Answer (1) Q1 > Q2 > Q3 and U 1 = U 2 = U 3

22. During an isotherm isothermal al expansion, expansion, a confined confined ideal gas does does –150 J of work work against its its surroundings. surroundings. This This implies implies that [AIPMT (Prelims)-201 (Prelims)-2011] 1] (1)) 150 J of heat (1 heat has been added to the the gas gas (2)) 150 J of heat (2 heat has been been remov removed ed from from the gas (3)) 300 J of heat has (3 has been been added to the gas gas (4)) No heat is transferred (4 transferred because the the process process is isothermal Sol. Answer (2) It implies 150 J heat has been removed from the gas. 23.. A mas 23 mass s of di diat atom omic ic gas gas  ( = 1.4) at a pressure of 2 atmospheres is compressed adiabatically so that its temperature rises from 27°C to 927°C. The pressure of the gas in the final state is [AIPMT (Mains)-2011] (1) 256 atm

(2)

8 atm

(3)

28 atm

(4)

68.7 atm

Sol. Answer (1)

P

1

P2 P1

P 2

2

P 2 2 P 2

T



 C then,

 T    1   T 2 

 constant

 1

1.4

 300     1200   1   4

    1   PT 

7

11.4

2

 26  256 atm

24. If U and W represent represent the increase in internal energy and work done by the system respectively in a thermodynamical thermodyna mical process, which of the following is true? [AIPMT (Prelims)-2010] (1)) U = (1 = – W , in a isothermal process

(2)

U = – W , in a adiabatic process

(3)) U = (3 = W , in a isothermal process

(4)

= W , in a adiabatic process U =



Thermodynamics 213


Sol. Answer (2) In adiabatic process Q = 0 So U = – W

[∵ Q = W + U ]

25. If C p and C v denote the specific heats (per unit mass) of an ideal gas of molecular weight M , where R is is the [AIPMT (Mains)-2010] molar gas constant (1)) C p – C v = R /M 2 (1

(2)) (2

C p – C v = R

(3)) C p – C v = R /M (3

(4)) (4

C p – C v = MR

Sol. Answer (3) C p

 C v 

R M

Because C p & C v are given per unit mass And C p – C v = R is is for 1 mole So here we use R /M where where M is is molecular mass.

26.. A monoa 26 monoato tomic mic gas at pres pressur sure e P 1 and V 1 is compressed adiabatically to

1 8

th its original volume. What is the

final pressure of the gas?

[AIPMT (Mains)-2010]

(1) 64 P 1

(2)) (2

P 1

(3) 16 P 1

(4)

32 P 1

Sol. Answer (4) ∵

PV  = constant 

P2

5  V   P1  1   P  8  3  32P1  V 2 

27. In thermody thermodynami namic c processes processes which which of the following following stateme statements nts is not true? true?

[AIPMT (Prelims)-2009]

(1)) In an isochoric (1 isochoric process process pressure pressure remain remains s constant constant (2)) In an isothermal process the (2 the temperature temperature remains constant constant (3)) In an adiab (3 adiabatic atic proc process ess PV  = constant (4)) In an adiabatic (4 adiabatic process the system is insulated from the surroundings Sol. Answer (1) In isochoric processes volume remains constant. 28. The internal internal energy change in a system that that has absorbed absorbed 2 Kcals of heat and done 500 J of work work is [AIPMT (Prelims)-2009] (1) 6400 J

(2)

5400 J

(3)

7900 J

(4)

8900 J

Sol. Answer (3) 2 × 4.2 × 1000 = dU + + 500  dU = = 7900 J 29. If Q, E and W denote denote respectively the heat added, change in internal energy and the work done in a closed cycle process, then [AIPMT (Prelims)-2008] (1)) Q = 0 (1

(2)

W = = 0


(3)

Q = W = = 0

(4)

E =0


214

Thermodynami cs


Sol. Answer (4) E = = change in U and in cyclic cyclic process process U = 0

 E = 0 30. At 10°C the value of the density of a fixed mass of an ideal gas divided divided by its pressure pressure is x . At 110°C this ratio is [AIPMT (Prelims)-2008] 283

(1)) (1

383

(2)) (2

x

383

(3)) (3

x

283

10 x

(4)) (4

110

x

Sol. Answer (1)

 P

 x

M PV

at 10°C

 x

Molecular mass × number of moles R  T

1



T

383 283

x

 

 x

 x



x



x

283 383

x

31. An engine has an efficiency of of 1/6. When When the temperature temperature of sink sink is reduced by 62°C, its efficiency is doubled. doubled. Temperature of the source is [AIPMT (Prelims)-2007] (1) 99°C

(2)

124°C

(3)

37°C

(4)

62°C

Sol. Answer (1) 1 3 1 3

T L  62

 1

T H 5

 1

6



 T L   T H

62 T H

5



6

T H = 372° K = 99°C T L = 37°C

32. A Carnot engine whose whose sink is at 300 K has has an efficiency efficiency of 40%. 40%. By how how much should should the temperatur temperature e of source be increased so as to increase its efficiency by 50% of original efficiency? [AIPMT (Prelims)-2006] (1) 275 K

(2)

325 K

(3) 250 K

(4)

380 K

Sol. Answer (3)

  1

T 2

W here

T 1

T 2

 Sink Temperature

T 1

 Source Temperature

Temperature of sink is given to be 300 K.

 = 0.4 Aakash Educational Services Pvt. Ltd.-Regd.


Thermodynamics 215


So

 1

0.4

300 T 1

 T 1 = 500 K Now,  is increased by 50%.



 

150 100

 

15 10

 0.4  0.6

To maintain same sink temperature new source temperature is 0.6

 1

300 T 1

T 1 = 750 K

 Increase in temperature = 750 – 500 = 250 K

33. The molar molar specific specific heat heat at const constant ant pressu pressure re of an ideal ideal gas gas is

7 2

R .

The ratio of specific heat at constant

pressure to that at constant volume is : 7

(1)) (1

8

(2)) (2

5

[AIPMT (Prelims)-2006] (3)) (3

7

5 7

(4)) (4

9 7

Sol. Answer (1) CP



7 2

R 

 R  



 

7 5

[AIPMT (Prelims)-2005]

34. Which of the the followin following g processe processes s is revers reversible ible ? (1) Transfer of heat by radiation

(2)

Electrical heating of a nichrome wire

(3) Transfer of heat by conduction

(4)

Isothermal compression

Sol. Answer (4) Isothermal compression takes place slowly at constant pressure, also U is is zero so it is a reversible process. 35. An ideal gas heat engine operates in Carnot cycle cycle between 227°C and 127°C. It It absorbs 6 × 104 cal of heat at [AIPMT (Prelims)-2005] higher temperature. Amount of heat converted to work is (1) 2.4 × 1 10 04 cal

(2)

6 × 104 cal

(3)

1.2 × 104 cal

(4)

4.8 × 104 cal

Sol. Answer (3) 36. A syst system em is take taken n from from sta state te a to state c by by two paths adc and and abc as as shown in the figure. The internal energy at a is U a = 10 J. Along the path adc the the amount of heat absorbed Q1 = 50 J and the work obtained W 1 = 20 J whereas along the path abc the the heat absorbed Q2 = 36 J. The amount of work along the path abc is is d

c

a

b

P

V

(1) 6 J

(2)

10 J 10


(3)

12 J 12

(4)

36 J 36


216

Thermodynami cs


Sol. Answer (1) dQ1 adc = 50 = d U adc + 20

 dU adc = 30 = dU abc dQabc = 36 = 30 + d W abc

 dW abc = 6 J 37. Consid Consider er two two insul insulate ated d chambe chambers rs A ( , B) of same volume connected by a closed knob, S . 1 mole of perfect gas is confined in chamber A. What is the change in entropy of gas when knob S is is opened? R = = 8.31 J mol–1K–1. S A

(1) 1. 1 .46 J/K

(2)

3.46 J/K

B

(3)

5.46 J/K

(4)

7.46 J/K

Sol. Answer (3)

S  2.303 nR loge

V 2 V 1

If initially volume is taken as V , then final volume = ZV , as volume of both chambers is given to be same.

S  2.30 2.303 3  1 8.31 8.31 log loge



2V V

S  5.46 J/K 38. A Carnot engine engine has efficiency efficiency 25%. It operates operates between between reservoirs reservoirs of constant constant temperatur temperatures es with temperatur temperature e difference of 80°C. What is the temperature of the low-temperature reservoir? (1) –25°C

(2)

25°C

(3)

–33°C

(4)

33°C

Sol. Answer (3)

  1 1 4

T L T H

 1

TH



4 3

T L T H T L

Also, T H – T L = 80

 T L = 240 K = –33°C 39. In an adiabatic adiabatic change, the pressure and temperature temperature of a monatomic gas gas are related related as c P  T , where c equals equals 3

(1)) (1

5

(2)) (2

5

3

2

(3)) (3

5

(4)) (4

5 2

Sol. Answer (4) P

 TC

⇒

PT

C

K

And compare with

    1   PT 

 constant


[condition from adiabatic process] Office: Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

Thermodynamics 217


Then,  C 

C 



 1   5/3

1 5 / 3



5/3

2 / 3



5 2

40. An ideal Carnot Carnot engine, whose whose efficiency is 40%, receives heat at 500 K. K. If its efficiency efficiency is 50%, then the intake temperature for the same exhaust temperature is (1) 800 K

(2)

900 K

(3) 600 K

(4)

700 K

Sol. Answer (3)

  1

40

T 2 T 1

 1

100

T 2

500

T 2 = 300 K

If  = 50% 50

 1

100

300 T 1

 T 1 = 600 K 41. A monatomic monatomic gas initially at 18°C is compressed adiabatically to one eighth of its original volume. The temperature after compression will be (1) 1164 K

(2)

144 K

(3)

18 K

(4)

887.4 K

Sol. Answer (1) –1 TV  –1 = constant

∵

1

 V 1  T1    V 2 

 T 2

291 × (8)2/3 = T 2 T 2 = 291 × 4 = 1164 K

42. An ideal gas, undergoing undergoing adiabatic adiabatic change, change, has which of the following following pressure pressure temperatu temperature re relationship? relationship? (1) P  T1–  = constant

(2)

P1–  T  = constant

–1  (3) P  –1 T = constant

(4)

–1 P  T  –1 = constant

Sol. Answer (2) 

PV

 constant 

 T    constant  P 

P 

P

1

. T



∵    

   RT   V     P  

PV

 RT

 constant



218

Thermodynami cs


43.. A sample 43 sample of of gas expa expands nds fro from m volume volume V 1 to V 2. The amount of work done by the gas is greatest, when the expansion is (1) Adiabatic

(2)

Equal in all cases

(3)

Isothermal

(4)

Isobaric

Sol. Answer (4) Work done is maximum in isobaric process W = = P .V = P (V 2 – V 1) = nR (T 2 – T 1)

44. The efficiency efficiency of a Carnot Carnot engine engine operating operating with reservoi reservoirr temperature temperature of 100°C 100°C and – 23°C will will be 373  250

(1)) (1

373  250

(2)) (2

373

100  23

(3)) (3

373

100

(4)) (4

100  23 100

Sol. Answer (2)

  1

 Where    T 2  sink temperature  T 1  reservoir temperature 

T 2 T 1

3 73 73  2 50 50



373

45. We conside considerr a ther thermody modynami namic c system. system. If U represents the increase in its internal energy and W the work done by the system, which of the following statements is true? (1)) U = –W in an isothermal process (1

(2)

U = W in an isothermal process

(3)) U = – W in an adiabatic process (3

(4)

U = W in an adiabatic process

Sol. Answer (3) As Q = zero for adiabatic process So U = = –W for for adiabatic process 46. If the ratio ratio of specific specific heat of a gas gas at constant constant pressure pressure to that at at constant constant volume is  , the change in internal energy of a mass of gas, when the volume changes from V to to 2 V at at constant pressure P , is PV

(1)) (1

(2)) (2

(   1)

(3)) (3

PV

R

(   1)

(4)) (4

PV  PV (   1)

Sol. Answer (1)

U  nCV T 



  T     1 

n

R

nR(T2

 T1 )

 1



nRT2

 nRT1

 1



 V ) PV    1  1

P (2V

47. An ideal gas at at 27°C is compressed compressed adiabatically adiabatically to 8/27 of its original volume. volume. The rise in temperature temperature is (Take (Take  = 5/3) (1) 275 K

(2)

375 K

(3)

475 K

(4)

175 K

Sol. Answer (2) –1 TV  –1 = constant

∵

 1

T1V

1

1  T2V2

1

 V 1  T1    V 2 

 27  300    8

 T 2 5 3

1

 T 2



Thermodynamics 219


3  300   2   T 2   300 

9 4

 T 2

675 K = T 2 = 675 – 300 = 375 K  T = 48.. Two Car 48 Carno nott engi engine nes s A and B are operated in series. The engine A receives heat from the source at temperature T 1 and rejects the heat to the sink at temperature T . The second engine B receives the heat at temperature T and and rejects to its sink at temperature T 2. For what value of T the the efficiencies of the two engines are equal? T1

(1)) (1

 T 2

(2)) (2

2

T1

 T 2

(3)) (3

2

T 1T 2

(4)) (4

T1T 2

Sol. Answer (4)

 A = B

 TL   T L       TH  A  T H  B T



T1

T

2

T 2 T

 T1T2

⇒

T



T1T2

49. The (W/Q) (W/Q) of a Carnot engine engine is 1/6. Now Now the temperatu temperature re of sink is reduced reduced by 62°C, then then this ratio ratio becomes becomes twice, therefore the initial temperature of the sink and source are respectively (1) 33°C, 67°C

(2)

37°C, 99°C

(3)

67°C, 33°C

(4)

97K, 37K

Sol. Answer (2) 1 6

 1

T L

...(1)

T H

 T L  62   1   1   6  T H 

2

1 3 1 3

 1

 1

T 2  62 T H 5 6



62 T H

T H = 372°K = 99°C

TL = 37°C 50. A scientist says that the efficiency of his heat engine which works works at source source temperature temperature 127°C and sink temperature 27°C is 26%, then (1) It is impossible

(2)

It is possible but less probable

(3) It is quite probable

(4)

Data are incomplete



220

Thermodynami cs


Sol. Answer (1) T L

  1

T H

 1

300 400



25%

51. The efficiency efficiency of Carnot engine is 50% and temperatur temperature e of sink is 500 K. K. If temperature temperature of source source is kept constant and its efficiency raised to 60%, then the required temperature of sink will be (1) 100 K

(2)

600 K

(3)

400 K

(4)

500 K

Sol. Answer (3) 1 2

 1

500

6

T H

10

 103

TH

T L

T L

 1

10

3

 4  102  400 K

52. An ideal gas heat engine operates in a Carnot cycle between 227°C and 127°C. It absorbs 6 kcal kcal at the higher temperature. The amount of heat (in kcal) converted into work is equal to (1) 4.8

(2)

3.5 3.

(3)

1.6 1.

(4)

1.2 1.

Sol. Answer (4) W Q1 W

6

 1  1

T L T H 400 500

 1.2 J

53. One mole mole of an ideal gas gas at an an initial initial temper temperatur ature e of T K does 6R joules joules of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is 5/3, the final temperature of gas will be (1) (T + 2.4) K

(2)

(T – 2.4) K

(3)

(T + 4) K

(4)

(T – – 4) K

Sol. Answer (4) W



nR T2  T1    1

Tfinal





    

dU

5



3

 T  4  K

54. The amount of heat energy energy required required to raise the the temperature temperature of 1 g of Helium at at NTP, from T 1 K to T 2 K is 3 Na kB (T2  T1 ) 2

(1)) (1

(2)) (2

3 Na kB (T2  T1 ) 4

(3)) (3

3 4

Na k B

 T 2     T 1 

(4)) (4

3 8

Na k B (T2

 T1 )

Sol. Answer (4) Q .



3 8

f 2

R  dT 

Na K B  T2  T1   Q

55. Which of the the following relations does does not give the equation of an adiabatic adiabatic process, process, where terms have their usual meaning? (1)) P  .T 1–  = constant (1

(2)

P 1–  T  = constant

(3)

PV  = constant

(4)

–1  –1 TV = constant

Sol. Answer (1) It is P 1–  T  = K Aakash Educational Services Pvt. Ltd.-Regd.


Thermodynamics 221


R 56. According to to C.E. van der Waal, the interatomic interatomic potential potential varies varies with the average average interatomic interatomic distance distance R ( ) as

(1)) R –1 (1

R –2

(2)) (2

(3)) (3

R –4

(4)) (4

R –6

Sol. Answer (4) According to van der Waal's formulae, interatomic potential is inversely proportion to R 6. So, U  R –6 57.. In a vesse 57 vessel, l, the the gas gas is at a pressu pressure re P . If the mass of all the molecules is halved and their speed is doubled, then the resultant pressure will be (1) 4P

(2)

2P

(3)) (3

(4)) (4

P

P /2 /2

Sol. Answer (2) P

1



P'

2

MnV

3



1 3



M

2



n 2V



2

 2

1 3

MnV

2

 2P

d) of the molecules as 58. The mean mean free path of collision of gas molecules varies with its its diameter diameter (d

(1)) d –1 (1

(2)) (2

d –2

(3)) (3

d –3

(4)) (4

d –4

(4)

Potential energy

Sol. Answer (2) 1



2

d

59. At 0 K, which which of the the followi following ng properti properties es of a gas will will be zero? zero? (1) Volume

(2)

Density

(3)

Kinetic energy

Sol. Answer (3) at 0 K V rms = 0 so K.E. = 0 60. The value value of critical temperat temperature ure in terms of van der Waals’ Waals’ constants constants a and b is given by (1)) (1

T C



8a 27Rb

(2)) (2

T C



27a 8Rb

(3)) (3

T C

(3)

2



a 2Rb

(4)) (4

T C

(4)

8



a 27Rb

Sol. Answer (1) T C



8a 27Rb

61. The degre degrees es of fre freedo edom m of a triat triatomi omic c gas is is (Consider moderate temperature) (1) 6

(2)

4

Sol. Answer (1) Degree of freedom = 3 rotational + 3 translational + 0 vibrational [T is is moderate] = 6 62. To find find out degree of freedom, freedom, the expressi expression on is (1)) (1

f



2

  1

(2)) (2

f



  1 2


(3)) (3

f



2

  1

(4)) (4

f



1

  1


222

Thermodynami cs


Sol. Answer (1) ∵

C V

Then,

f



fR



2 2CV

R



2C V

CP



 C V

2

C P C V

 1

2

 1

63. The equati equation on of state state for for 5 g of oxyge oxygen n at a pressu pressure re P and and temperature T , when occupying a volume V , will be (where R is is the gas constant) 5

(1)) PV = (1

(3)) PV = (3

RT

32

5 2

(2)) (2

PV = = 5RT

(4)) (4

PV =

5

RT

16

RT

Sol. Answer (1) ∵

m  RT  M 

PV

 nRT  

PV



5 32

RT

SECTION - D

Assertion - Reason Type Questions 1.

A : Work Work done by a gas gas in isothe isotherma rmall expansi expansion on is more more than than the the work work done done by the the gas gas in the the same same expansion adiabatically. R : Temperatu emperature re remains constant constant in isothermal expansion expansion and not in adiabatic expansion. expansion.

Sol. Answer (2) A : is true R : is true, but not correct explanation correct explanation is, in isothermal expansion. ∵

= 0 so U = = 0 T =

 Q = W all the heat goes in doing work. Whereas in adiabatic process Heat goes to work as well as in increasing internal energy.

 W isothermal > W adiabatic 2.

A : Ef Effi fici cien ency cy of he heat at en engi gine ne ca can n nev never er be 10 100% 0%.. R : Second law of thermodynamics thermodynamics puts puts a limitation on the efficiency efficiency of a heat engine.

Sol. Answer (1) A : is true R : is true, and correct explanation Aakash Educational Services Pvt. Ltd.-Regd.


Thermodynamics 223


3.

A : He Heat at ab abso sorb rbed ed in a cyc cycli lic c pro proce cess ss is ze zero ro.. R : Work done done in a cyclic proces process s is zero.

Sol. Answer (4) A : is false, in cyclic process only U = 0, Q = W . R : is false, work done is not zero only change in internal energy is zero. 4.

A : Coe Coeffi fficien cientt of per perfor forman mance ce of of a refr refrige igerat rator or is is alway always s great greater er than than 1. R : Efficien Efficiency cy of heat heat engine is greater than 1.

Sol. Answer (4) A : is false. R : is false Because efficiency of heat engine can never be equal to greater to 1.

 ∵

1

all the heat cannot be converted to work.

and coefficient of performane of refrigerator

 ∵

5.

1 





1



1

  1 so  may be less than 1.

A : Ad Adia iaba bati tic c exp expan ansi sion on ca caus uses es co cool olin ing. g. R : In adiabatic adiabatic expansion, internal energy energy is used up in doing work. work.

Sol. Answer (1) A : is true R : is true, and correct explanation 6.

A : The spe specif cific ic heat heat of an an ideal ideal gas is zero zero in an an adiab adiabati atic c proce process. ss. R : Spec Specific ific heat of a gas is process independent.

Sol. Answer (3) A : is true R : is false Because specific heat depends on the process. 7.

A : Th The e chang change e in int inter ernal nal ene energ rgy y does does not not depe depend nd on on the the path path of of proce process. ss. R : The internal internal energy of an ideal gas is independent independent of the configuration configuration of its molecules. molecules.

Sol. Answer (2) A : is true R : is true, but not the correct explanation, because internal energy depends on the temperature of the gas. Aakash Educational Services Pvt. Ltd.-Regd.


224

8.

Thermodynami cs


A : Heat supp supplied lied to to a gaseous gaseous syst system em in an isother isothermal mal proce process ss is used to do wor work k against against surr surroundi oundings. ngs. R : During isothermal isothermal process process there is no change change in internal internal energy of the the system.

Sol. Answer (1) A : true R : true and correct explanation 9.

A : In nat natur ure e all all ther thermo mody dyna namic mic pr proce ocess sses es are are irr irreve evers rsib ible. le. R : During a thermodynamic thermodynamic process process it is not possible to eliminate eliminate dissipative dissipative effects.

Sol. Answer (1) A : is true R : is true and correct explanation 10.. A : Du 10 Durin ring g a cyclic cyclic proc process ess work work done done by by the syst system em is zero zero.. R : Heat supplied to a system system in the cyclic process converts converts into internal internal energy of the system. system. Sol. Answer (4) A : is false, in cyclic process, work done is not zero, internal energy change is zero. R : is false, heat supplied converts to work as initial state is equal to final state.

 No change in internal energy.









CLS Aipmt 17 18 XIII Phy Study Package 3 SET 1 Chapter 12

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