Boost your JEE prep with Major Test Series

Home

JEE Physics

Thermodynamics Previous Year Questions With Solutions

Join ALLEN!

(Session 2026 - 27)

Name

Mobile Number

Class

Choose class

Goal

Choose your goal

Preferred Programs

Preferred Mode

State

Choose State

I agree to

I authorise ALLEN Career Institute Pvt Ltd to send me regular updates via Phone calls, Whatsapp, SMS, Robocalls (Automated Calls), Emails, or on postal address.

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.

JEE Main Previous Year Solved Questions on Thermodynamics
JEE Adv Previous Year Solved Questions on Thermodynamics I
JEE Adv Previous Year Solved Questions on Thermodynamics II

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.

$Δ Q = Δ U + Δ W$

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

$Δ Q = Δ U + Δ W$

Second Law of Thermodynamics

Kelvin planck statement : It states that in a cyclic process total heat can not be converted into mechanical work.
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

Related Video:

3.0Important Formulas

Heat and Work	$W = J Q where J = 4.18 \frac{Joule}{calorie} 1 cal = 4.18 \approx 4.2 Joule \Rightarrow 1 cal > 1 joule$
Internal Energy	$d U = d U_{k} + d U_{p}$
Work done by Thermodynamic System	$d W = P d V W = \int_{V_{i}}^{V_{f}} d W = \int_{V_{i}}^{V_{f}} P d V$ Work done by gas is equal to the area under P-V graph
First Law of Thermodynamics	$Δ Q = Δ U + Δ W$
Isochoric Process	$V o l u m e = C o n s t an t Equation of state \Rightarrow \frac{P}{T} = constant Work done w = \int_{V_{i}}^{V_{f}} P d V = 0 (∵ d V = 0) Form of First Law: Q = Δ U = μ C_{v} Δ T Slope of the P-V curve \frac{d P}{d V} = \infty$
Isobaric Process	$P ress u re = co n s t an t Equation of state \Rightarrow \frac{V}{T} = constant (V \propto T) Work done w = \int_{V_{i}}^{V_{f}} P d V = P (V_{f} - V_{i}) Form of First Law: Q = Δ U + P (V_{f} - V_{i}) C_{p} (T_{f} - T_{i}) = μ C_{v} (T_{f} - T_{i}) + P (V_{f} - V_{i}) Slope of the PV curve (\frac{d P}{d V})_{isobaric} = 0$
Isothermal Process	$Temperature= constant Equation of state PV=Constant Work done w = μ RT lo g_{e} [\frac{V _{2}}{V _{1}}] = 2.303 μ RT lo g_{10} [\frac{P _{1}}{P _{2}}] Form of First Law: Δ U = 0 Slope of isothermal curve [\frac{d P}{d V}]_{isothermal} = - \frac{P}{V}$
Adiabatic Process	$No heat exchange between system and surroundings Work done W = \frac{μ R}{γ - 1} [P_{1} V_{1} - P_{2} V_{2}] = \frac{μ R}{γ - 1} (T_{1} - T_{2}) Slope of adiabatic curve [\frac{d P}{d V}]_{adiabatic} = - \frac{γ P}{V}$
Polytropic Process	$P V^{x} = Constant, C = C_{v} + \frac{R}{1 - x} For isobaric x = 0, C = C_{p} = \frac{R}{γ - 1} + R = C_{v} + R For isothermal x = 1, C = \infty For adiabatic x = γ, C = 0$
Efficiency of Heat Engine	$η = \frac{Useful output}{Input} \Rightarrow η = \frac{Net work}{heat given} = \frac{w}{Q _{1}} = \frac{Q _{1} - Q _{2}}{Q _{1}} \Rightarrow η = \frac{T _{1} - T _{2}}{T _{1}} \times 100$
Coefficient of Performance (COP) of Refrigerator	$COP (β) = \frac{Heat absorbed from Cold}{Work} = \frac{Q _{2}}{W} = \frac{Q _{2}}{Q _{1} - Q _{2}} COP (β) = (\frac{T _{r}}{T _{1} - T _{r}}) Relation between (β and η) : η = \frac{1}{1 + β}, β = \frac{1}{η} - 1$

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)

$Q = Δ U as work done is zero [constant volume] Δ U = m s Δ T Δ U = 0.08 \times (1.70 \times 4.18) \times 5 \approx 284 J$

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

$(Use k_{B} = 1.38 \times 1 0^{- 23} \frac{J}{mol \cdot K})$

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

Solution: Ans(2)

For monoatomic molecular degree of freedom= 3

$K_{avg} = \frac{3}{2} k_{B} T T = \frac{0.414 \times 1.6 \times 1 0 ^{- 19} \times 2}{3 \times 1.38 \times 1 0 ^{- 23}} = 3200 K$

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 = $\frac{f}{2} n RT$

$K . E = \frac{5}{2} ✕ 1 ✕ 8.31 ✕ 300 J = 6232.5 J$

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 $\frac{C _{p}}{C _{v}}$ for the gas is :

$(1) \frac{5}{3} (2) \frac{3}{2} (3) \frac{9}{7} (4) \frac{9}{7}$

Solution: Ans(2)

$P \propto T^{3} \Rightarrow P T^{- 3} = constant$

$P V^{γ} = constant$

$P^{(\frac{n RT}{P})^{γ}} = constant P^{1 - γ} T^{γ} = constant \Rightarrow P T^{\frac{1 - γ}{γ}} = constant PT 1 - = co n s t an t 1 - = - 3 = - 3 + 3 3 = 2 = 32 \frac{1 - γ}{γ} = - 3 \Rightarrow 1 - γ = - 3 γ \Rightarrow 1 = - 2 γ \Rightarrow γ = \frac{3}{2}$

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 :

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

Solution: Ans(Bonus)

$Work Done AB: = \frac{1}{2} (800 + 600) dyne/cm^{2} \times 4 m^{3} = (600 dyne/cm^{2}) \times 4 m^{3} Work Done BC: = (4000 dyne/cm^{2}) \times 4 m^{3} = 2 \times 1 0^{3} \times \frac{1}{1 0 ^{5}} N/cm^{2} \times 4 m^{3} = 2✕102✕ 4 N m = 800 J = 2 \times 1 0^{2} \times 4 Nm = 800 J$

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. P_A and P_B are the pressures of the gases in A and B respectively, then $\frac{P _{A}}{P _{B}}$ is

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

Solution: Ans(1)

$\frac{P _{A} V _{A}}{P _{B} V _{B}} = \frac{n _{A} R T _{A}}{n _{B} R T _{B}} Given V_{A} = V_{B}, and T_{A} = T_{B} \frac{P _{A}}{P _{B}} = \frac{n _{A}}{n _{B}}, \frac{P _{A}}{P _{B}} = \frac{1/2}{1/32} = 16$

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)

$K E = \frac{1}{2} k_{B} T$

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

$(1) P_{2} < P_{1} (2) P_{2} > P_{1} (3) P_{1} = P_{2} (4) P_{1} > P_{2}$

Solution: Ans(4)

$P V = n RT \Rightarrow V = (\frac{n R}{P}) T Slope = \frac{n R}{P}, slope \propto \frac{1}{P} \Rightarrow (slope)_{2} > (slope)_{1} P_{1} > P_{2}$

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)

$U = n C V_{T} U = n_{1} C_{V 1} T + n_{2} C_{V 2} T U = 8 \times \frac{3 R}{2} \times T + 6 \times \frac{5 R}{2} \times T = 27 RT$

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 :

$(1) \frac{9}{4} R (2) \frac{7}{4} R (3) \frac{3}{2} R (4) \frac{5}{2} R$

Solution: Ans(1)

$C_{v} = \frac{n _{1} C _{v 1} + n _{2} C _{v 2}}{n _{1} + n _{2}} = \frac{2 \times \frac{3 R}{2} + 6 \times \frac{5 R}{2}}{2 + 6} = \frac{9}{4} R$

Q-11.The pressure and volume of an ideal gas are related as PV^3/2 = K (Constant). The work done when the gas is taken from state A (P₁, V₁, T₁) to state
B (P₂, V₂, T₂) is :

$(1) 2 (P_{1} V_{1} - P_{2} V_{2}) (2) 2 (P_{2} V_{2} - P_{1} V_{1}) (3) 2 (P_{1} V_{1} - P_{2} V_{2}) (4) 2 (P_{2} V_{2} - P_{1} V_{1})$

Solution: Ans(1)

$P V^{x} = constant I f w or k d o n e b y t h e g a s i s a s k e d w = \frac{n R Δ T}{1 - x} Here x = 32 Here x = \frac{3}{2}, W = \frac{P _{1} V _{1} - P _{2} V _{2}}{1 - \frac{3}{2}} = 2 (P_{1} V_{1} - P_{2} V_{2})$

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)

$γ = 1 + \frac{f}{2} = 1.4 \Rightarrow \frac{2}{f} = 0.4 \Rightarrow f = 5 W = n R Δ T = 200 J Q = (\frac{f + 2}{2}) n R Δ T \Rightarrow Q = \frac{7}{2} \times 200 = 700 J$

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)

$v_{rms} = \frac{3 RT}{M} \frac{v _{1}}{v _{2}} = \frac{M _{2}}{M _{1}} \Rightarrow \frac{2}{v _{2}} = \frac{32}{2} v_{2} = 0.5 km/s$

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

$(1) C_{B} = \infty, C_{A} = 0 (2) C_{A} = 0 and C_{B} = \infty (3) C_{P} > C_{V} > C_{A} = C_{B} (4) C_{A} > C_{P} > C_{V}$

Solution: Ans(Bonus)

For Process A

$lo g P = γ lo g V \Rightarrow P = V^{γ}, (γ > 1)$

$P V^{- γ} = constant$

$C_{A} = C_{V} + \frac{R}{1 + γ} ........ (1)$

Likewise for process B

$\to P V^{- 1} = constant C_{A} = C_{V} + \frac{R}{1 + 1} C_{A} = C_{V} + \frac{R}{2} ........ (2) C_{P} = C_{V} + R ......... (3) B y (1), (2) an d (3) C_{P} > C_{B} > C_{A} > C_{V} (No Answer Matching)$

Q-15.If three moles of monoatomic gas $γ = \frac{5}{3}$ is mixed with two moles of a diatomic gas =53, $γ = \frac{5}{3}$ 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)

$f_{1} = 3, f_{2} = 3, n_{1} = 3, n_{2} = 2 f_{mixture} = \frac{n _{1} f _{1} + n _{2} f _{2}}{n _{1} + n _{2}} = \frac{9 + 10}{5} = \frac{19}{5} γ_{mixture} = 1 + \frac{2 \times 5}{19} = \frac{29}{19} = 1.52$

Join ALLEN!

(Session 2026 - 27)

Name

Mobile Number

Class

Choose class

Goal

Choose your goal

Preferred Programs

Preferred Mode

State

Choose State

I agree to

I authorise ALLEN Career Institute Pvt Ltd to send me regular updates via Phone calls, Whatsapp, SMS, Robocalls (Automated Calls), Emails, or on postal address.

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.

JEE Main Previous Year Solved Questions on Thermodynamics
JEE Adv Previous Year Solved Questions on Thermodynamics I
JEE Adv Previous Year Solved Questions on Thermodynamics II