Q 11 :

A monoatomic gas having γ=53 is stored in a thermally insulated container and the gas is suddenly compressed to (18)th of its initial volume. The ratio of final pressure and initial pressure is: (γ is the ratio of specific heats of the gas at constant pressure and at constant volume)          [2025]

  • 16

     

  • 40

     

  • 32

     

  • 28

     

(3)

PiViγ=PfVfγ

 PfPi=(ViVf)γ=(8)5/3=32



Q 12 :

An ideal gas initially at 0°C temperature, is compressed suddenly to one fourth of its volume. If the ratio of specific heat at constant pressure to that at constant volume is 3/2, the change in temperature due to the thermodynamics process is ________ K.          [2025]



(273)

        TVγ1 = constant

                    γ=32

273×V0.5=T(V4)0.5

                    T=546 K  T=273 K



Q 13 :

An ideal gas has undergone through the cyclic process as shown in the figure. Work done by the gas in the entire cycle is ________ ×101 J. (Take π = 3.14)          [2025]



(314)

Area of circle, W=π(V2)·(P2)

W=π4(500300)×103(350150)×106

W=31.4 Joule

W=314×101 Joule



Q 14 :

In an isothermal change, the change in pressure and volume of a gas can be represented for three different temperatures; T3>T2>T1 as           [2023]

  •  

  •  

  •  

  •  

(1)

For isothermal process, P-V graph is a rectangular hyperbola.

As the dotted line is an isobaric line, which implies T3>T2>T1 as the volume is increasing.



Q 15 :

Match List I with List II :              [2023]

  List I   List II
A. Isothermal Process I. Work done by the gas decreases internal energy
B. Adiabatic Process II. No change in internal energy
C. Isochoric Process III. The heat absorbed goes partly to increase internal energy and partly to do work
D. Isobaric Process IV. No work is done on or by the gas

 

Choose the correct answer from the options given below :

  • A-I, B-II, C-IV, D-III

     

  • A-II, B-I, C-IV, D-III

     

  • A-I, B-II, C-III, D-IV

     

  • A-II, B-I, C-III, D-IV

     

(2)

ΔU=nCVΔT

For isothermal process, T is constant.  

So, ΔU=0

AII

Adiabatic process

ΔQ=0

ΔQ=ΔU+ΔW

ΔU=-ΔW

Work done by gas is positive,  

so ΔU is negative.

BI

For isochoric process, ΔW=0

CIV

For isobaric process, 

ΔW=PΔV0

ΔU=nCVΔT0

Heat absorbed goes partly to increase internal energy and partly to do work.



Q 16 :

Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R.     [2023]

Assertion A: If dQ and dW represent the heat supplied to the system and the work done on the system respectively, then according to the first law of thermodynamics dQ=dUdW.

Reason R: First law of thermodynamics is based on law of conservation of energy.

In the light of the above statements, choose the correct answer from the option given below:

  • A is correct but R is not correct

     

  • A is not correct but R is correct

     

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

     

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

     

(3)

First law of thermodynamics is based on the law of conservation of energy and it can be written as

dQ=dU-dW

where dW is the work done on the system.



Q 17 :

The pressure (P) and temperature (T) relationship of an ideal gas obeys the equation PT2=constant. The volume expansion coefficient of the gas will be         [2023]

  • 3T2

     

  • 3T2

     

  • 3T3

     

  • 3T

     

(4)

PT2=constant,   Using PV = nRT

P=nRTV

PT2=nRTV×T2=constant=K

 T3=KV

So, ddT(KV)=3T2

KdVdT=3T2

 dV=3T2KdT=VγdT

dV=VγdT

 γV=3T2K

 γ=3T2KV=3T2T3=3T



Q 18 :

Heat is given to an ideal gas in an isothermal process.

A. Internal energy of the gas will decrease.
B. Internal energy of the gas will increase.
C. Internal energy of the gas will not change.
D. The gas will do positive work.
E. The gas will do negative work.  

Choose the correct answer from the options given below:              [2023]

  • A and E only

     

  • B and D only

     

  • C and E only

     

  • C and D only

     

(4)

dQ=dU+dW

 dU=nCVdT

dU=0 (For isothermal)

 U=constant

Also, dQ>0 (supplied).

Hence dW>0



Q 19 :

The pressure of a gas changes linearly with volume from A to B as shown in the figure. If no heat is supplied to or extracted from the gas, then the change in the internal energy of the gas will be                [2023]

  • 6 J

     

  • Zero

     

  • −4.5 J

     

  • 4.5 J

     

(4)

As Δq=0

Δu=-W

W=PdV

Δu=-W=30×103×150×10-6

       =4500×10-3=4.5 J



Q 20 :

A sample of gas at temperature T is adiabatically expanded to double its volume. The work done by the gas in the process is (given, γ=32)               [2023]

  • W=TR[2-2]

     

  • W=TR[2-2]

     

  • W=RT[2-2]

     

  • W=TR[2-2]

     

(4)

T1V1γ-1=T2V2γ-1

TV1/2=T2(2V)1/2T2=T2

W=R(T1-T2)γ-1=R(T-T2)1/2=RT(2-2)



Q 21 :

A source supplies heat to a system at the rate of 1000 W. If the system performs work at a rate of 200 W, then the rate at which the internal energy of the system increases is       [2023]

  • 800 W

     

  • 1200 W

     

  • 600 W

     

  • 500 W

     

(1)

dQ=dU+dW

dUdt=dQdt-dWdt

dUdt=1000-200=800 W



Q 22 :

Given below are two statements:                                   [2023]

Statement I: If heat is added to a system, its temperature must increase.

Statement II: If positive work is done by a system in a thermodynamic process, its volume must increase.

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

  • Both Statement I and Statement II are true

     

  • Statement I is false but Statement II is true

     

  • Both Statement I and Statement II are false

     

  • Statement I is true but Statement II is false

     

(2)

Statement I: ΔQ>0

According to the first law of thermodynamics,

ΔQ=ΔU+W

If ΔQ>0, ΔU<0 and W>0 is also possible.  

Hence ΔT<0, so temperature decreases.

Statement I is false

Statement II: W>0

 PdV>0

Therefore, volume of the system must increase during positive work done by the system.

Statement II is true



Q 23 :

Consider two containers A and B containing monoatomic gases at the same pressure (P), volume (V) and temperature (T). The gas in A is compressed isothermally to 18 of its original volume, while the gas in B is compressed adiabatically to 18 of its original volume. The ratio of final pressure of gas in B to that of gas in A is:      [2023]

  • 8

     

  • 81/2

     

  • 18

     

  • 4

     

(4)

Isothermal process, T=constant

PV=nRT=constant

P1V1=P2V2

PV=PA(V8)

PA=8P

Adiabatic process, PVγ=constant

γ for monoatomic gas is 53

P1V1γ=P2V2γ

PBP=(V1V2)γ=(VV/8)5/3

PB=32P

PBPA=32P8P=4



Q 24 :

A gas is compressed adiabatically, which one of the following statement is NOT true?                      [2023]

  • There is no heat supplied to the system

     

  • The temperature of the gas increases

     

  • The change in the internal energy is equal to the work done on the gas

     

  • There is no change in the internal energy

     

(4)

(1) ΔQ=0

(2) ΔQ=ΔU+ΔWΔU=-ΔW

Adiabatic compression (V)

ΔW=-ve  ΔU=+ve

ΔUT

ΔU0



Q 25 :

1 kg of water at 100°C is converted into steam at 100°C by boiling at atmospheric pressure. The volume of water changes from 1.00×10-3m3 as a liquid to 1.671 m3 as steam. The change in internal energy of the system during the process will be (Given latent heat of vaporisation =2257 kJ/kg, atmospheric pressure =1×105 Pa)         [2023]

  • + 2090 kJ

     

  • - 2090 kJ

     

  • - 2426 kJ

     

  • + 2476 kJ

     

(1)

ΔQ=ΔU+ΔW

  ΔU=ΔQ-ΔW=mLv-PΔV

                =(1 kg)(2257×103J/kg)

-(1×105 Pa)(1.671 m3-1×10-3 m3)

           =2257×103 J-167×103 J=2090 kJ



Q 26 :

The thermodynamic process, in which internal energy of the system remains constant is                [2023]

  • Isochoric

     

  • Isothermal

     

  • Adiabatic

     

  • Isobaric

     

(2)

T=constantU=constant



Q 27 :

The initial pressure and volume of an ideal gas are P0 and V0. The final pressure of the gas when the gas is suddenly compressed to volume V04 will be:  (Given γ= ratio of specific heats at constant pressure and at constant volume)                [2023]

  • P0(4)1/γ

     

  • P0(4)γ

     

  • P0

     

  • 4P0

     

(2)

As gas is suddenly compressed, the process is adiabatic.

Equation of gas for adiabatic process is PVγ=constant

  P1V1γ=P2V2γ

  P0V0γ=P2(V04)γ

  P2=P0(4)γ

Option (2) is correct.



Q 28 :

A thermodynamic system is taken through a cyclic process. The total work done in the process is                 [2023]

  • 100 J

     

  • 300 J

     

  • Zero

     

  • 200 J

     

(2)

On P-V scale, area of loop = work done

  W=+12(2)×300W=300 J



Q 29 :

A thermodynamic system is taken through the cyclic process ABC as shown in the figure. The total work done by the system during the cycle ABC is ________ J.   [2026]



(300)

Work done=Area bounded by cycle

=12×3×200=300 J



Q 30 :

10 mole of an ideal gas is undergoing the process shown in the figure. The heat involved in the process from P1 to P2 is α Joule (P1=21.7 Pa and P2=30 PaCv=21 J/K.mol, R=8.3 J/mol.K). The value of α is _______.             [2026]

  • 28

     

  • 24

     

  • 15

     

  • 21

     

(4)

ΔQ=nCvΔT  (isochoric)

=CvRnRΔT=CvR(P2-P1)V

=218.3×(30-21.7)×1=21 J



Q 31 :

When 300 J of heat is given to an ideal gas with Cp=72R its temperature raises from 20°C to 50°C keeping its volume constant. The mass of the gas is (approximately) _______ g.    (R=8.314  J/mol.K)                   [2026]



(481)

For a constant volume process, the heat supplied (Q) is given by the formula:

Q=nCvΔT

Where:

1. n = number of moles of the gas.
2. Cv = molar specific heat capacity at constant volume.
3. ΔT = change in temperature =50°C-20°C=30°C=30K

We are given the molar specific heat at constant pressure, Cp=72R. Using Mayer's relation (Cp-Cv=R):

Cv=Cp-R

Cv=72R-R=52R

Using the formula of heat,

300=n×(52×8.314)×30

n=300×25×8.314×300.481 moles

So, the amount of the gas is 0.481 moles. As the molar mass of the gas is not given in the question, we can't determine the mass of the gas directly.



Q 32 :

One mole of an ideal diatomic gas expands from volume V to 2V isothermally at a temperature 27°C and does W joule of work. If the gas undergoes same magnitude of expansion adiabatically from 27°C doing the same amount of work W, then its final temperature will be (close to) ______ °C.

(log2=0.693)   [2026]

  • −189

     

  • −117

     

  • −30

     

  • −56

     

(4)

For isothermal process

Wisothermal=nRTln(V2V1)

     =1·R·300·ln(2)

      =300R(0.693)    (1)

Now for adiabatic process,

It is given work done in isothermal = work done in adiabatic

Wadiabatic=nR(T1-T2)γ-1    (2)

(1)=(2)

nR(300-Tfinal)1.4-1=300R(0.693)

Tfinal=216.84 K

=-56.3°C



Q 33 :

The internal energy of a monoatomic gas is 3nRT. One mole of helium is kept in a cylinder having internal cross sectional area of 17 cm² and fitted with a light movable frictionless piston. The gas is heated slowly by supplying 126 J heat. If the temperature rises by 4 °C, then the piston will move ____ cm.
(atmospheric pressure = 105 Pa)   [2026]

  • 15.5

     

  • 1.45

     

  • 14.5

     

  • 1.55

     

(1)

ΔU=3nRΔT

ΔU=3×1×253×4=100 Joule

ΔQ=126

W=26=PΔV

26=105×17×10-4Δx

Δx=26170=15.3



Q 34 :

A diatomic gas (γ=1.4) does 100 J of work when it is expanded isobarically. Then the heat given to the gas ______ J.      [2026]



(350)

w=100 J=nRΔT  for isobaric process.

Q=nCpΔT=(f2+1)nRΔT

=72·(100)=350 Joule.



Q 35 :

An insulated cylinder of volume 60 cm3 is filled with a gas at 27°C and 2 atmospheric pressure. Then the gas is compressed making the final volume as 20 cm3 while allowing the temperature to rise to 77°C. The final pressure is __________ atmospheric pressure.      [2026]



(7)

PV=nRT

P1V1RT1=P2V2RT2

2×105×60300=P2×20350

P2=2×105×7×36

P2=7×105=7 atm
 



Q 36 :

In the following p-V diagram, the equation of state along the curved path is given by (V-2)2=4ap, where a is a constant. The total work done in the closed path is   [2026]

  • -1a

     

  • +13a

     

  • 12a

     

  • -13a

     

(4)

w=Area of parabola

=23(Area of rectangle AC31A)

=23P0(3-1)=4P03

When V=1

(1-2)2=4aP0

P0=14a

w=43P0=43·14a=13a

wgas=-13a