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