Q 21 :

A thermodynamic system is taken through cyclic process. The total work done in the process is:

  • 100J

     

  • 300J

     

  • 200J

     

  • zero

     

(2)

work done=Area of graph

W=12(400-100)(4-2)

W=300J



Q 22 :

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:

  • 45

     

  • 855

     

  • 25

     

  • 1625

     

(2)

Adiabatic process



Q 23 :

An ideal gas undergoes a four step cycle as shown in the P-V diagram below. During this cycle, positive work is done by the gas in

  • step 1

     

  • step 2

     

  • step 3

     

  • step 4

     

(1)

Heat and work depends on the path taken to reach a specific value. Hence, heat and work are path functions.



Q 24 :

Heat is supplied to a diatomic gas at constant pressure. Then the ratio of ΔQ : ΔU : ΔW is ______.            [2026]

  • 2 : 3 : 5

     

  • 5 : 3 : 2

     

  • 2 : 5 : 7

     

  • 7 : 5 : 2

     

(4)

Diatomic gas (p=constant)

ΔQ=ΔQp=nCpΔT=72nRΔT

ΔU=nCvΔT=52nRΔT

ΔW=nRΔT

ΔQ:ΔU:ΔW::7:5:2



Q 25 :

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.

Statement I: Change in internal energy of a system containing n mole of ideal gas can be written as

ΔU=nCv(Tf-Ti)=nRγ-1(Tf-Ti), where γ=CpCv,  Ti=initial temperature,  Tf=final temperature.

Statement II: Relation between degree of freedom f and γ(=CpCv) is (γ=1+2f).             

Choose the correct answer from the options given below                    [2026]

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

     

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

     

  • A is true but R is false

     

  • A is false but R is true

     

(2)

ΔU=nCv(Tf-Ti)=nRγ-1(Tf-Ti)

γ=CpCv=(f2+1)Rf2R=1+2f



Q 26 :

Consider the following statements:

A. Zeroth law of thermodynamics gives concept of temperature

B. First law of thermodynamics gives concept of internal energy

C. In isothermal expansion of ideal gas, ΔQ≠ΔW

D. Product of intensive and extensive variables is extensive

E. The ratio of any extensive variable to mass will be an extensive variable

Choose the correct combination of statements from the options given below:                 [2026]

  • C, D and E Only

     

  • A, B and C Only

     

  • A, B and D Only

     

  • B, C and D Only

     

(3)

A. True

B. True

C. False

D. True

E. False



Q 27 :

An ideal gas at pressure P and temperature T is expanding such that PT3=constant. The coefficient of volume expansion of the gas is ______.           [2026]

  • 2T

     

  • 1T

     

  • 4T

     

  • 3T

     

(3)

PT3=k (constant)

(nRTV)T3=k⇒V=(nRk)T4

Volume expansion coefficient, γ=1V(dVdT)

γ=knRT4·(nRk)·4T3=4T



Q 28 :

A cylinder with adiabatic walls is closed at both ends and is divided into two compartments by a frictionless adiabatic piston. Ideal gas is filled in both (left and right) the compartments at same P, V, T. Heating is started from left side until pressure changes to 27P8. If initial volume of each compartment was 9 litres then the final volume in right-hand side compartment is ______ litres. (for this ideal gas CpCv=1.5)               [2026]

  • 3

     

  • 4

     

  • 14

     

  • 9

     

(2)

PLeft=PRight=27P8

For right side process in Adiabatic

PiViγ=PfVfγ

P(9)3/2=27P8×Vf3/2

27=278Vf3/2⇒Vf=4 litres



Q 29 :

One mole of diatomic gas having rotational modes only is kept in a cylinder with a piston system. The cross-section area of the cylinder is 4 cm2. The gas is heated slowly to raise the temperature by 1.2°C during which the piston moves by 25 mm. The amount of heat supplied to the gas is ______ J. (Atmospheric pressure = 100 kPa, R = 8.3 J/mol·K) (Neglect mass of the piston)                 [2026]

  • 24.8

     

  • 25

     

  • 15.04

     

  • 29.98

     

(2)

Q=ΔU+W

Q=f2nRΔT+PΔV

Diatomic gas f=5

Q=52×1×8.3×1.2+100×103(4×10-4×0.025)

Q=24.9+1=25.9 J, nearest answer is 25

Here, please note that instead of the above method, if we use Q=nCpΔT, then also no option matches.



Q 30 :

Initial pressure and volume of a monoatomic ideal gas are P and V. The change in internal energy of this gas in adiabatic expansion to volume Vfinal=27V is ______ J.        [2026]

  • -2PV(33-1)

     

  • 43PV

     

  • 34PV

     

  • -43PV

     

(4)

P1V1γ=P2V2γ

Here, γ=53

PV5/3=P2×(27V)5/3

⇒P2=P(27)5/3=P35

ΔU=nCvΔT=nRΔT(γ-1)

ΔU=P2V2-P1V1(γ-1)

⇒ΔU=P35×33·V-PV(53-1)=-43PV



Q 31 :

A vessel contains 0.15 m3 of a gas at pressure 8 bar and temperature 140°C with Cp=3R and Cv=2R. It is expanded adiabatically till pressure falls to 1 bar. The work done during this process is ________ kJ. (R is gas constant)                [2026]



(120)

γ=CpCv=32

∵  P1V1γ=P2V2γ⇒V2=(P1P2)1/γV1

⇒V2=(81)2/3×0.15=0.6 m3

Work done in adiabatic process,   W=P1V1-P2V2γ-1

W=(8×0.15-1×0.61.5-1)×105=1.2×105 J

⇒W=120 kJ



Q 32 :

5 moles of unknown gas is heated at constant volume from 10°C to 20°C. The molar specific heat of this gas at constant pressure Cp=8 cal/mol·°C and R=8.36 J/mol·°C. The change in the internal energy of the gas is __________ calorie.                 [2026]



(300)

Cp-Cv=R⇒Cv=(Cp-R)

Cv=6 cal/mol·C∘

ΔU=nCvΔT=5×6×10=300 cal

 



Q 33 :

A certain gas is isothermally compressed to (13)rd of its initial volume (V0=3 litre) by applying required pressure. If the bulk modulus of the gas is 3×105 N/m2, the magnitude of work done on the gas is _______ J.                  [2026]



(989)

Initial volume (V1)=3 litre

Final volume (V2)=V13=1 litre=10-3 m3

For isothermal process, Bulk modulus = Pressure (B=P)

Bulk modulus of gas for isothermal process is not constant.

In the given solution we have considered it as initial bulk modulus.

For isothermal process work done,   W=P1V1ln(V2V1)

W=(3×105)×(3×10-3)ln(10-33×10-3)

W=-988.75 J=989 J