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JEE Physics
Thermodynamics Previous Year Questions With Solutions

Thermodynamics Previous Year Questions With Solutions

1.0Introduction

Thermodynamics is a branch of physics that studies heat, energy, and work, focusing on how energy transfers and transforms, especially in systems involving temperature. It is based on four main laws. The Zeroth Law defines temperature and thermal equilibrium. The First Law states that energy is conserved—it can change forms but cannot be created or destroyed. The Second Law introduces entropy, showing that energy naturally spreads out, increasing disorder. The Third Law states that as a system nears absolute zero, its entropy approaches a minimum. Thermodynamics is essential in understanding engines, refrigerators, power generation, and biological processes.

2.0Key Concepts to Remember

Zeroth Law of Thermodynamics(ZLOT)

  • According to the Zeroth Law of Thermodynamics, two objects (or systems) are in thermal equilibrium if they are separately in thermal equilibrium with a third object (such as a thermometer) and hence have the same temperature.
  • ZLOT defines temperature only.

First Law of Thermodynamics (FLOT)

  • Suppose some heat is supplied to a system capable of doing external work.In that case,the heat absorbed by the system is equivalent to the combined effect of the increase in its internal energy and the external work it executes.      

ΔQ=ΔU+ΔW

  • FLOT introduces the concept of internal energy.
  • The first law of thermodynamics is based on the law of energy conservation.

Sign Convention used in Physics Thermodynamics

ΔQ=ΔU+ΔW

Sign Convention used in Physics Thermodynamics

Second Law of Thermodynamics

  1. Kelvin planck statement : It states that in a cyclic process total heat can not be converted into mechanical work.
  2. Claussius statement : It is impossible to have net heat flow from a low temperature body to a high temperature body.

Carnot Cycle

The enclosed area represents work done by the engine.

1.    It is a hypothetical engine because the working substance is an ideal gas.

2.    Its efficiency is maximum but not 100%.

3.    This cyclic process has two isothermal and two adiabatic processes

3.0Important Formulas

Heat and Work

W=JQwhere J=4.18calorieJoule​1cal=4.18≈4.2Joule⇒1cal>1joule

Internal Energy

dU=dUk​+dUp​

Work done by Thermodynamic System

dW=PdVW=∫Vi​Vf​​dW=∫Vi​Vf​​PdV

Work done by  gas is equal to the area under P-V graph

First Law of Thermodynamics

ΔQ=ΔU+ΔW

Isochoric Process

Volume=ConstantEquation of state⇒TP​=constantWork done w=∫Vi​Vf​​PdV=0(∵dV=0)Form of First Law: Q=ΔU=μCv​ΔTSlope of the P-V curve dVdP​=∞

Isobaric Process

Pressure=constantEquation of state⇒TV​=constant(V∝T)Work done w=∫Vi​Vf​​PdV=P(Vf​−Vi​)Form of First Law: Q=ΔU+P(Vf​−Vi​)Cp​(Tf​−Ti​)=μCv​(Tf​−Ti​)+P(Vf​−Vi​)Slope of the PV curve (dVdP​)isobaric​=0

Isothermal Process

Temperature= constantEquation of state PV=ConstantWork done w=μRTloge​[V1​V2​​]=2.303μRTlog10​[P2​P1​​]Form of First Law: ΔU=0Slope of isothermal curve [dVdP​]isothermal​=−VP​

Adiabatic Process

No heat exchange between system and surroundingsWork done W=γ−1μR​[P1​V1​−P2​V2​]=γ−1μR​(T1​−T2​)Slope of adiabatic curve [dVdP​]adiabatic​=−VγP​

Polytropic Process

PVx=Constant,C=Cv​+1−xR​For isobaric x=0,C=Cp​=γ−1R​+R=Cv​+RFor isothermal x=1,C=∞For adiabatic x=γ,C=0

Efficiency of Heat Engine

η=InputUseful output​⇒η=heat givenNet work​=Q1​w​=Q1​Q1​−Q2​​⇒η=T1​T1​−T2​​×100

Coefficient of Performance (COP) of Refrigerator

COP(β)=WorkHeat absorbed from Cold​=WQ2​​=Q1​−Q2​Q2​​COP(β)=(T1​−Tr​Tr​​)Relation between (β and η):η=1+β1​,β=η1​−1

4.0Past Year Questions with Solutions on Thermodynamics:JEE (Mains)

Q-1. 0.08 kg air is heated at constant volume through 5°C. The specific heat of air at constant volume is 0.17 kcal/kg°C and J = 4.18 joule/cal. The change in its internal energy is approximately.

(1 ) 318 J             (2) 298 J             (3) 284 J             (4) 142 J

Solution: Ans(3)

Q=ΔU as work done is zero [constant volume]ΔU=msΔTΔU=0.08×(1.70×4.18)×5≈284J


Q-2. The average kinetic energy of a monatomic molecule is 0.414 eV at temperature :

(Use kB​=1.38×10−23mol⋅KJ​)

(1) 3000 K             (2) 3200 K                 (3) 1600 K               (4) 1500 K

Solution: Ans(2)

For monoatomic molecular degree of freedom= 3

Kavg​=23​kB​TT=3×1.38×10−230.414×1.6×10−19×2​=3200K


Q-3. The total kinetic energy of 1 mole of oxygen at 27°C is :

[Use universal gas constant (R)= 8.31 J/mole K]

(1) 6845.5 J                  (2) 5942.0 J                 (3) 6232.5 J                 (4) 5670.5J

Solution: Ans(3)

Kinetic Energy = 2f​nRT

K.E=25​ ✕ 1 ✕ 8.31 ✕ 300 J=6232.5 J


Q-4.During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature.  The ratio of Cv​Cp​​ for the gas is :

(1)35​         (2)23​          (3)79​         (4)79​     

Solution: Ans(2)

P∝T3⇒PT−3=constant

PVγ=constant

P(PnRT​)γ=constantP1−γTγ=constant⇒PTγ1−γ​=constantPT1−=constant1−=−3=−3+33=2=32γ1−γ​=−3⇒1−γ=−3γ⇒1=−2γ⇒γ=23​


Q-5.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 :

A thermodynamic system is taken from an original state A to an intermediate state B by a linear process as shown in the figure.

(1) 33800 J            (2) 2200 J             (3) 600 J              (4) 1200 J

Solution: Ans(Bonus)

Work Done AB: =21​(800+600)dyne/cm2×4m3 =(600dyne/cm2)×4m3Work Done BC: =(4000dyne/cm2)×4m3=2×103×1051​N/cm2×4m3=2✕102✕ 4Nm=800J=2×102×4Nm=800J


Q-6.Two vessels A and B are of the same size and are at same temperature. A contains 1g of hydrogen and B contains 1g of oxygen. PA and PB are the pressures of the gases in A and B respectively, then PB​PA​​ is 

(1) 16          (2) 8             (3) 4             (4) 32

Solution: Ans(1)

PB​VB​PA​VA​​=nB​RTB​nA​RTA​​Given VA​=VB​, and TA​=TB​PB​PA​​=nB​nA​​,PB​PA​​=1/321/2​=16


Q-7.The parameter that remains the same for molecules of all gases at a given temperature is :

1) kinetic energy         2) momentum       3) mass           4) speed

Solution: Ans(1)

KE=21​kB​T


Q-8.The given figure represents two isobaric processes for the same mass of an ideal gas, then

The given figure represents two isobaric processes for the same mass of an ideal gas

(1)P2​<P1​(2)P2​>P1​(3)P1​=P2​(4)P1​>P2​

Solution: Ans(4)

PV=nRT⇒V=(PnR​)TSlope=PnR​,slope∝P1​⇒(slope)2​>(slope)1​P1​>P2​


Q-9.A gas mixture consists of 8 moles of argon and 6 moles of oxygen at temperature T. Neglecting all vibrational modes, the total internal energy of the system is                                

1) 29RT          2) 20 RT           3) 27 RT           4) 21RT

Solution: Ans(3)

U=nCVT​U=n1​CV1​T+n2​CV2​TU=8×23R​×T+6×25R​×T=27RT


Q-10.Two moles of a monoatomic gas are mixed with six moles of a diatomic gas. The molar specific heat of the mixture at constant volume is :

(1)49​R(2)47​R(3)23​R(4)25​R

Solution: Ans(1)

Cv​=n1​+n2​n1​Cv1​+n2​Cv2​​=2+62×23R​+6×25R​​=49​R


Q-11.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 :

(1)2(P1​V1​−P2​V2​)   (2)2(P2​V2​−P1​V1​)    (3)2(P1​V1​​−P2​V2​​)    (4)2(P2​V2​​−P1​V1​​)

Solution: Ans(1)

PVx=constantIfworkdonebythegasisaskedw=1−xnRΔT​Herex=32Here x=23​,W=1−23​P1​V1​−P2​V2​​=2(P1​V1​−P2​V2​)


Q-12.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 :

1) 850 J               2) 800 J            3) 600J            4) 700J

Solution: Ans(4)

γ=1+2f​=1.4⇒f2​=0.4⇒f=5W=nRΔT=200JQ=(2f+2​)nRΔT⇒Q=27​×200=700J

Q-13. If the root mean square velocity of hydrogen molecule at a given temperature and pressure is 2 km/s, the root mean square velocity of oxygen at the same condition in km/s is :

1) 2.0                   2) 0.5                 3) 1.5            4) 1.0

Solution: Ans. (2)

vrms​=M3RT​​v2​v1​​=M1​M2​​​⇒v2​2​=232​​v2​=0.5 km/s


Q-14.Two thermodynamical processes are shown in the figure. The molar heat capacity for process A and B are CA and CB. The molar heat capacity at constant pressure and constant volume are represented by CP and CV, respectively. Choose the correct statement. 

Two thermodynamical processes are shown in the figure. The molar heat capacity for process A and B are CA and CB.

(1)CB​=∞,CA​=0  (2)CA​=0 and CB​=∞ (3)CP​>CV​>CA​=CB​(4)CA​>CP​>CV​

Solution: Ans(Bonus)

For Process A 

logP=γlogV⇒P=Vγ,(γ>1)

PV−γ=constant

CA​=CV​+1+γR​........(1)

Likewise for process B

→PV−1=constantCA​=CV​+1+1R​CA​=CV​+2R​........(2)CP​=CV​+R.........(3)By(1),(2)and(3)CP​>CB​>CA​>CV​ (No Answer Matching)


Q-15.If three moles of monoatomic gas γ=35​  is mixed with two moles of a diatomic gas  =53, γ=35​ the value of adiabatic exponent g for the mixture is:  

1) 1.75                   2) 1.40                 3) 1.52            4) 1.35

Solution: Ans(3)

f1​=3,f2​=3,n1​=3,n2​=2fmixture​=n1​+n2​n1​f1​+n2​f2​​=59+10​=519​γmixture​=1+192×5​=1929​=1.52


Table of Contents


  • 1.0Introduction
  • 2.0Key Concepts to Remember
  • 2.1Zeroth Law of Thermodynamics(ZLOT)
  • 2.2First Law of Thermodynamics (FLOT)
  • 2.3Second Law of Thermodynamics
  • 2.4Carnot Cycle
  • 3.0Important Formulas
  • 4.0Past Year Questions with Solutions on Thermodynamics:JEE (Mains)

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