Q 1 :

A sample of gas at temperature T is adiabatically expanded to double its volume. Adiabatic constant for the gas is γ=3/2. The work done by the gas in the process is (n = 1 mole)               [2024]

  • RT[1-22]

     

  • RT[22-1]

     

  • RT[2-2]

     

  • RT[2-2]

     

(4)   

         PV=nRTP=nRTV

          Also, PVγ= constant

         nRTVVγ= constant TVγ-1= constant

          TVγ-1=T2·(2V)γ-1

           T2=T·(12)1/2=T2

            W=nR(T1-T2)γ-1=2nR(T-T2)=RT(2-2)

          

 



Q 2 :

During an adiabatic process, if the pressure of a gas is found to be proportional to the cube of its absolute temperature, then the ratio of CPCV for the gas is                          [2024]

  • 32

     

  • 75

     

  • 53

     

  • 97

     

(1) 

         P=T3PT-3= constant

         PVT=nR= constant from ideal gas equation T=PVnR

         P(PVnR)-3= constant P-2V-3=constant

        PV3/2= constant                                   ... (1)

          Process equation for adiabatic process is 

          PVγ= constant                                           ... (2)

         Comparing equation (1) and (2)

         CPCV=γ=32

 



Q 3 :

A total of 48 J heat is given to one mole of helium kept in a cylinder. The temperature of helium increases by 2°C. The work done by the gas is (Given: R=8.31 J k-1 mol-1)                 [2024]

  • 48 J

     

  • 23.1 J

     

  • 24.9 J

     

  • 72.9 J

     

(2)   

        Given : Q=48 J

         1st law of thermodynamics, Q=U+W

          48=nCvT+W

           48=(1)(3R2)(2)+W

           W=48-3×R

           W=48-3×(8.3)       

             W=23.1 J

 



Q 4 :

A diatomic gas (γ=1.4) does 100 J of work in an isobaric expansion. The heat given to the gas is                     [2024]

  • 150 J

     

  • 250 J

     

  • 490 J

     

  • 350 J

     

(4)   

        W=100 J

        Q=nCpT=n(7R2)T=72nRT

        PV=nRTPV=nRT= work done (W)

        Q=72nRT=72W

         Q=72×100=350 J

 



Q 5 :

The volume of an ideal gas (γ=1.5) is changed adiabatically from 5 litres to 4 litres. The ratio of initial pressure to final pressure is:           [2024]

  • 45

     

  • 1625

     

  • 855

     

  • 25

     

(3)   

        For adiabatic process, PVγ=k

         P1V13/2=P2V23/2

          P1×(5)3/2=P2×(4)3/2

          P1P2=(45)3/2=(2)3(125)1/2=855

 



Q 6 :

A sample of 1 mole gas at temperature T is adiabatically expanded to double its volume. If adiabatic constant for the gas is γ=32, then the work done by the gas in the process is:                 [2024]

  • RT[2-2]

     

  • RT[2-2]

     

  • RT[2+2]

     

  • TR[2+2]

     

(1)    

           WD by gas in adiabatic process

           W=P2V2-P1V11-γ=nR[T2-T1]1-γ

           T1V1γ-1=T2V2γ-1

            T×(V)1/2=T2×(2V)1/2T2=T2

             W=R[T2-T]1-32=2R2[2T-T]

              W=RT[2-2]

 



Q 7 :

During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio of Cp/Cv for the gas is                         [2024]

  • 53

     

  • 32

     

  • 75

     

  • 97

     

(2)   

           PT3PT-3= constant

            PVγ= const

           P(nRTP)γ= const

           P1-γTγ= const PTγ1-γ= const

            γ1-γ=-3

             γ=-3+3γγ=32

 



Q 8 :

The pressure and volume of an ideal gas are related as PV3/2=K (Constant). The work done when the gas is taken from state A(P1,V1,T1) to state B(P2,V2,T2) is                 [2024]

  • 2(P1V1-P2V2)

     

  • 2(P2V2-P1V1)

     

  • 2(P1V1-P2V2)

     

  • 2(P2V2-P1V1)

     

(1)   

         For PVx= constant

         W=nRT1-x

          Here x=32

                W=P2V2-P1V1-12

          =2(P1V1-P2V2)



Q 9 :

A diatomic gas (γ=1.4) does 200 J of work when it is expanded isobarically. The heat given to the gas in the process is               [2024]

  • 850 J

     

  • 800 J

     

  • 600 J

     

  • 700 J

     

(4)   

            Q=PV+nCVT

               =PV+nR(γ-1)T=PV+PVγ-1

               =PV(1+1γ-1)

                =200(1+10.4)

                200×3.5=700 J

 



Q 10 :

The heat absorbed by a system in going through the given cyclic process is:               [2024]

  • 431.2 J

     

  • 616 J

     

  • 61.6 J

     

  • 19.6 J

     

(3)

ΔU=0 (Cyclic process)

ΔQ=W=area of P-V curve

ΔQ=π×(140×103 Pa)×(140×10-6 m3)=61.6 J

 



Q 11 :

Two different adiabatic paths for the same gas intersect two isothermal curves as shown in P - V diagram. The relation between the ratio VaVd and the ratio VbVc is      [2024]

  • VaVd=VbVc

     

  • VaVdVbVc

     

  • VaVd=(VbVc)2

     

  • VaVd=(VbVc)-1

     

(1)

For adiabatic process

TVγ-1=constant

Ta·Vaγ-1=Td·Vγ-1

(VaVd)γ-1=TdTa                                 ...(i)

Tb·Vbγ-1=Tc·Vcγ-1

(VbVc)γ-1=TcTb                           ...(ii)

As, Td=Tc and Ta=TbVaVd=VbVc



Q 12 :

A real gas within a closed chamber at 27°C undergoes the cyclic process as shown in figure. The gas obeys PV3=RT equation for the path A to B. The net work done in the complete cycle is (assuming R = 8 J/mol K)             [2024]

  • 225 J

     

  • 205 J

     

  • 20 J

     

  • - 20 J

     

(2)

For A to B:

Given PV3=RTP=RTV3

Work done WAB=PdV

WAB=RTV3dV=RT[V-2-2]VA=2VB=4

=-RT2[1V2]=-RT2[116-14]

=-RT2(-316)=3RT32

WAB=332×8×300=225 J

WBC (Isobaric process)

WBC=PΔV=10(2-4)=-20 J

WAC (Isochoric process) = 0

Wnet=WAB+WBC+WCA=225-20+0=205 J



Q 13 :

A thermodynamic system is taken from an original state A to an intermediate state B by a linear process as shown in the figure. Its volume is then reduced to the original value from B to C by an isobaric process. The total work done by the gas from A to B and B to C would be          [2024]

  • 33800 J

     

  • 2200 J

     

  • 800 J

     

  • 1200 J

     

(3)

Work done AB=12(8000+4000) Dyne/cm2×4 m3

                          =(6000 Dyne/cm2)×4 m3

Work done BC=(4000 Dyne/cm2)×4 m3

Total work done =2000 Dyne/cm2×4 m3

         =2×103×1105Ncm2×4 m3

         =2×10-2×N10-4 m2×4 m3

          =2×102×4 Nm=800 J



Q 14 :

Choose the correct statement for processes A and B shown in figure.               [2024]

  • PVγ=k for process B and PV=k for process A.

     

  • PV=k for process B and A.

     

  • Pγ1/Tγ=k for process B and 𝑇=𝑘 for process A.

     

  • Tγ/Pγ1=k for process A and PV=k for process B.

     

(1)

Slope of isothermal < Slope of adiabatic

(Slope)A<(Slope)B

Process B  adiabatic  PVγ=K

Process A isothermalPV=K



Q 15 :

Given are statements for certain thermodynamic variables,

(A) Internal energy, volume (V) and mass (M) are extensive variables.

(B) Pressure (P), temperature (T) and density (ρ) are intensive variables.

(C) Volume (V), temperature (T) and density (ρ) are intensive variables.

(D) Mass (M), temperature(T) and internal energy are extensive variables.

Choose the correct answer from the options given below:          [2025]

  • (C) and (D) only

     

  • (D) and (A) only

     

  • (A) and (B) only

     

  • (B) and (C) only

     

(3)

Extensive variables depend on size and amount of system.

Extensive : Volume, mass, internal energy

Intensive : Pressure, temperature, density 



Q 16 :

Match the List-I with List-II

  List-I   List-II
A. Pressure varies inversely with volume of an ideal gas. I. Adiabatic process
B. Heat absorbed goes partly to increase internal energy and partly to do work. II. Isochoric process
C. Heat is neither absorbed nor released by a system. III. Isothermal process
D. No work is done on or by a gas. IV. Isobaric process

 

Choose the correct answer from the options given below:          [2025]

  • A-I, B-IV, C-II, D-III

     

  • A-III, B-I, C-IV, D-II

     

  • A-I, B-III, C-II, D-IV

     

  • A-III, B-IV, C-I, D-II

     

(4)

PV=μRT

(A) P1V  isothermal

(B) Q=U+W  isobaric

(C) Q=0  adiabatic

(D) W=0  isochoric



Q 17 :

Using the given P-V diagram, the work done by an ideal gas along the path ABCD is          [2025]

  • 4P0V0

     

  • 3P0V0

     

  • 4P0V0

     

  • 3P0V0

     

(4)

Area under graph will be magnitude of graph and being counter clock wise graph it would be negative

Area=2P0×V0+P0V0=3P0V0

W=3P0V0



Q 18 :

An ideal gas goes from an initial state to final state. During the process, the pressure of gas increases linearly with temperature.

A. The work done by gas during the process is zero.

B. The heat added to gas is different from change in its internal energy.

C. The volume of the gas is increased.

D. The internal energy of the gas is increase.

E. The process is isochoric (constant volume process)

Choose the correct answer from the options given below:          [2025]

  • A, B, C, D Only

     

  • A, D, E Only

     

  • E Only

     

  • A, C Only

     

(2)

Given for ideal gas.

P = constant T

 PT1 = constant

 PPV = constant  V is constant

So, isochoric process

Work done by gas will be zero.



Q 19 :

Given below are two statemets. One is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A) : In an insulated container, a gas is adiabatically shrunk to half of its initial volume. The temperature of the gas decreases.

Reason (R) : Free expansion of an ideal gas is an irreversible and an adiabatic process.

In the light of the above statements, choose the correct answer from the options given below:          [2025]

  • Both (A) and (R) are true and (R) is the correct explanation of (A)

     

  • (A) is true but (R) is false

     

  • (A) is false but (R) is true

     

  • Both (A) and (R) are true but (R) is NOT the correct explanation of (A)

     

(3)

(A) : T1V1γ1=T2V2γ1

Temperature increases when volume decreases.

(R) : Free expansion is assumed fast, so the pressure is adiabatic.



Q 20 :

The magnitude of heat exchanged by a system for the given cyclic process ABCA (as shown in figure) is (in SI unit)           [2025]

  • 10 π

     

  • π

     

  • zero

     

  • 40 π

     

(2)

W=12πR2

    =12×π×(2002×103)×2002×106=10π2=5πJ



Q 21 :

The work done in an adiabatic change in an ideal gas depends upon only:          [2025]

  • change in its pressure

     

  • change in its specific heat

     

  • change in its volume

     

  • change in its temperature

     

(4)

Work done in adiabatic process,

W=nRT1γ



Q 22 :

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A) : With the increase in the pressure of an ideal gas, the volume falls off more rapidly in an isothermal process in comparison to the adiabatic process.

Reason (R) : In isothermal process, PV = constant, while in adiabatic process PVγ = constant. Here γ is the ratio of specific heats, P is the pressure and V is the volume of the ideal gas.

In the light of the above statements, choose the correct answer from the options given below:          [2025]

  • Both (A) and (R) are true but (R) is NOT the correct explanation of (A)

     

  • (A) is true but (R) is false

     

  • Both (A) and (R) are true and (R) is the correct explanation of (A)

     

  • (A) is false but (R) is true

     

(4)

For isothermal process

PV = constant

 PdV+VdP=0

 dVV=dPP

 dV=(VP)dP

For adiabatic PVγ = constant

 PγVγ1dV+VγdP=0

 dV=VγP(dP)

Magnitude of |VP| is greater than |VγP|

So, volume falls more rapidly in adiabatic.

2nd statement is the process description of the isothermal and adiabatic.



Q 23 :

A poly-atomic molecule (CV = 3RCP = 4R, where R is gas constant) goes from phase space point A (PA=105Pa,VA=4×106m3) to point B (PB=5×104Pa,VB=6×106m3) to point C (PC=104Pa,VC=8×106m3). A to B is an adiabatic path and B to C is an isothermal path.

The net heat absorbed per unit mole by the system is:.          [2025]

  • 500 R (ln 3 + ln 4)

     

  • 450 R (ln 4 – ln 3)

     

  • 500 R ln 2

     

  • 400 R ln 4

     

(2)

For process A  B

(Q)AB=0 (adiabatic)

For process B  C

(Q)BC=(W)BC=nRT ln(VCVB)

                     = 450 R [ln (4) – ln 3]       

(Q)net=(Q)AB+(Q)BC

                     = 450 R [ln (4) – ln (3)]



Q 24 :

Identify the characteristics of an adiabatic process in a monoatomic gas.

(A) Internal energy is constant.

(B) Work done in the process is equal to the change in internal energy.

(C) The product of temperature and volume is a constant.

(D) The product of pressure and volume is a constant.

(E) The work done to change the temperature from T1 to T2 is proportional to (T2T1)

Choose the correct answer from the options given below:          [2025]

  • (A), (C), (D) only

     

  • (A), (C), (E) only

     

  • (B), (E) only

     

  • (B), (D) only

     

(3)

For adiabatic process, Q = 0

Q=U+W=0  W=U

W=nCvT  |W|=nCvT T2T1

B, E are correct



Q 25 :

During the melting of a slab of ice at 273 K at atmospheric pressure:          [2025]

  • Internal energy of ice-water system remains unchanged.

     

  • Positive work is done by the ice-water system on the atmosphere.

     

  • Internal energy of the ice-water system decreases.

     

  • Positive work is done on the ice-water system by the atmosphere.

     

(4)

Melting of ice requires heat therefore internal energy increases. But because of decrease in volume the work done on atmosphere is negative or atmosphere does positive work on ice-water system.



Q 26 :

A piston of mass M is hung from a massless spring whose restoring force law goes as F=kx3, where k is the spring constant of appropriate dimension. The piston separates the vertical chamber into two parts, where the bottom part is filled with 'n' moles of an ideal gas. An external work is done on the gas isothermally (at a constant temperature T) with the help of a heating filament (with negligible volume) mounted in lower part of the chamber, so that the piston goes up from a height L0 to L1, the total energy delivered by the filament is (Assume spring to be in its natural length before heating)          [2025]

  • 3nRT ln(L1L0)+2Mg(L1L0)+k3(L13L03)

     

  • nRT ln(L12L02)+Mg2(L1L0)+k4(L14L04)

     

  • nRT ln(L1L0)+Mg(L1L0)+k4(L14L04)

     

  • nRT ln(L1L0)+Mg(L1L0)+3k4(L14L04)

     

(3)

Total energy supplied = gravitational potential energy of piston + spring potential energy + work done by gas.

Q =Mg(h2h1)+L0L1kx3dx+μRT lnV2V1

   =Mg(L1L0)+k4(L14L04)+nRT ln[L1AL0A]

   =Mg(L1L0)+k4(L14L04)+nRT ln[L1L0]



Q 27 :

A gas is kept in a container having walls which are thermally non-conducting. Initially the gas has a volume of 800 cm3 and temperature 27°C. The change in temperature when the gas is adiabatically compressed to 200 cm3 is:

(Take γ = 1.5 : γ is the ratio of specific heats at constant pressure and at constant volume)                    [2025]

  • 327 K

     

  • 600 K

     

  • 522 K

     

  • 300 K

     

(4)

V1=800 cm3, V2=200 cm3, T1=300 K

for adiabatic process, TVγ1 = const.

(300)(800)1.51=T2(200)1.51

T2=300[800200]0.5=300×(22)1/2

T2=600 K

T=600300=300 K



Q 28 :

An ideal gas exists in a state with pressure P0, volume V0. It is isothermally expanded to 4 times of its initial volume (V0), then isobarically compressed to its original volume. Finally the system is heated isochorically to bring it to its initial state. The amount of heat exchanged in this process is:          [2025]

  • P0V0(2 ln 20.75)

     

  • P0V0(ln 20.75)

     

  • P0V0(ln 20.25)

     

  • P0V0(2 ln 20.25)

     

(1)

Q=U+W

For cyclic process Q = W

W=W1+W2+W3

      =P0V0 ln 4V0V0+P04(V04V0)+0

      =P0V0 ln 434P0V0

      =P0V0(2 ln 20.75)



Q 29 :

Match List I with List II

  List I   List II
(A) Isobaric (I) Q=W
(B) Isochoric (II) Q=U
(C) Adiabatic (III) Q=zero
(D) Isothermal (IV) Q=U+PV

 

Q = Heat supplied

W = Work done by the system

U = Change in internal energy

P = Pressure of the system

V = Change in volume of the system

Choose the correct answer from the options given below:         [2025]

  • (A)-(IV), (B)-(III), (C)-(II), (D)-(I)

     

  • (A)-(IV), (B)-(I), (C)-(III), (D)-(II)

     

  • (A)-(IV), (B)-(II), (C)-(III), (D)-(I)

     

  • (A)-(II), (B)-(IV), (C)-(III), (D)-(I)

     

(3)

(A) Isobaric process, Q=U+PV

(B) Isochoric process, Q=U

(C) Adiabatic process, Q=0

(D) Isothermal process, (U=0), Q=W



Q 30 :

Match List-I with List-II

  List-I   List-II
(A) Isothermal (I) W (work done) = 0
(B) Adiabatic (II) Q (supplied heat) = 0
(C) Isobaric (III) U (change in internal energy)  0
(D) Isochoric (IV) U=0

 

Choose the correct answer from the options given below:          [2025]

  • (A)-(III), (B)-(II), (C)-(I), (D)-(IV)

     

  • (A)-(IV), (B)-(I), (C)-(III), (D)-(II)

     

  • (A)-(IV), (B)-(II), (C)-(III), (D)-(I)

     

  • (A)-(II), (B)-(IV), (C)-(I), (D)-(III)

     

(3)

For Isothermal,

T = constant, T=0, U=nCvT=0          ... (IV)

For Adiabatic,

Q=0          ... (II)

For Isobaric,

p = constant, U=nCvT0,

Q0, W0          ... (III)

For Isochoric,

V = constant, W=0          ... (I)