How much of JEE Main comes from Thermodynamics?

JEEGenie estimates about 6% of JEE Main marks come from Thermodynamics, based on recent papers. The share varies from year to year.

Which formulas matter most in Thermodynamics?

The Thermodynamics sheet on this page has 18 formula entries, starting with First Law of Thermodynamics, Third Law of Thermodynamics, Work Done in Thermodynamic Processes, Internal Energy of Ideal Gas, Molar Specific Heats. Each one comes with when to use it and the trap to avoid.

What topics does Thermodynamics cover for JEE?

Thermodynamics covers Zeroth and First Law, Thermodynamic Processes, Heat Engines and Efficiency, Second Law of Thermodynamics, Calorimetry, Heat Transfer.

How should I revise Thermodynamics before the exam?

Learn each formula together with its condition of use, read the trap listed beside it, then practise questions and review every mistake. The practice link on this page opens Thermodynamics questions on JEEGenie.

How many previous-year questions are there on Thermodynamics?

JEEGenie has 848 previous-year questions on Thermodynamics, and this page shows a few with worked explanations.

Is practice on Thermodynamics free on JEEGenie?

Sign-up is free and includes a 7-day Pro trial. After the trial, free accounts keep a limited daily quota.

Thermodynamics for JEE Main & Advanced — formulas, what is asked, and traps

JEEGenie estimates about 6% of JEE Main marks come from Thermodynamics. This page covers 18 formula entries, each with when to use it, what JEE asks and the traps. A common slip: Confusing thermal equilibrium with mechanical equilibrium (equal pressure) or chemical equilibrium.

Key points to know

Zeroth Law of Thermodynamics. If two systems are each in thermal equilibrium with a third system, they are in thermal equilibrium with each other.
First Law of Thermodynamics. Energy is conserved; the change in internal energy of a system equals heat added minus work done by the system.
Second Law of Thermodynamics (Kelvin-Planck & Clausius). Kelvin-Planck: No process is possible whose sole result is the absorption of heat from a reservoir and complete conversion into work. Clausius: No process is possible whose sole result is the transfer of heat from a colder to a hotter body.
Third Law of Thermodynamics. As temperature approaches absolute zero, the entropy of a perfect crystal approaches zero.
Entropy. Entropy is a measure of disorder; change in entropy for a reversible process is $Δ S = \int \frac{d Q_{rev}}{T}$ .
Thermodynamic Processes Summary. Key processes: isothermal ( $Δ T = 0$ , $Δ U = 0$ ), adiabatic ( $Q = 0$ ), isobaric ( $Δ P = 0$ ), isochoric ( $Δ V = 0$ ), cyclic ( $Δ U = 0$ ).
Degrees of Freedom and $γ$ . For monatomic gas: $f = 3$ , $C_{v} = 3 R / 2$ , $γ = 5 / 3$ ; diatomic: $f = 5$ , $C_{v} = 5 R / 2$ , $γ = 7 / 5$ ; polyatomic: $f = 6$ , $C_{v} = 3 R$ , $γ = 4 / 3$ .
Phase Change and Latent Heat. During phase change, temperature remains constant; heat absorbed/released is $Q = m L$ .
Free Expansion of Ideal Gas. In free expansion (into vacuum), $W = 0$ , $Q = 0$ , so $Δ U = 0$ and $Δ T = 0$ for ideal gas.
Mixing of Gases and Temperature. When two gases mix without heat loss, final temperature is found by equating total internal energy change to zero.
Polytropic Processes. A polytropic process follows $P V^{n} = constant$ , where $n$ is polytropic index; work done is $W = \frac{P_{2} V_{2} - P_{1} V_{1}}{1 - n}$ .
Heat Capacity in Polytropic Processes. For a polytropic process, molar heat capacity $C = C_{v} + \frac{R}{1 - n}$ .
Otto Cycle (Petrol Engine). Otto cycle consists of two adiabatic and two isochoric processes; efficiency depends on compression ratio $r = V_{1} / V_{2}$ .
Diesel Cycle. Diesel cycle has one isobaric process instead of isochoric; efficiency depends on compression ratio $r$ and cutoff ratio $ρ$ .

Formula sheet

Name	Formula	When to use
First Law of Thermodynamics	$Δ U = Q - W$	For any thermodynamic process to relate heat, work, and internal energy change.
Third Law of Thermodynamics	$\lim_{T \to 0} S = 0$	To define absolute entropy scale and understand behavior near absolute zero.
Work Done in Thermodynamic Processes	Isothermal: $W = n R T \ln (\frac{V_{f}}{V_{i}})$ ; Adiabatic: $W = \frac{P_{i} V_{i} - P_{f} V_{f}}{γ - 1}$ ; Isobaric: $W = P Δ V$ ; Isochoric: $W = 0$	To calculate work in any given thermodynamic process.
Internal Energy of Ideal Gas	$Δ U = n C_{v} Δ T$	To find change in internal energy when temperature changes, especially in adiabatic or isochoric processes.
Molar Specific Heats	$C_{p} - C_{v} = R$ ; $γ = \frac{C_{p}}{C_{v}}$	To relate heat capacities and compute $γ$ for adiabatic processes.
Adiabatic Process Relations	$P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$ ; $T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1}$	To relate pressure, volume, and temperature in adiabatic processes.
Heat Engine Efficiency	$η = 1 - \frac{Q_{c}}{Q_{h}}$ ; for Carnot engine: $η_{Carnot} = 1 - \frac{T_{c}}{T_{h}}$	To calculate efficiency of any heat engine or compare with Carnot efficiency.
Coefficient of Performance (Refrigerator & Heat Pump)	${COP}_{R} = \frac{Q_{c}}{Q_{h} - Q_{c}} = \frac{T_{c}}{T_{h} - T_{c}}$ (Carnot); ${COP}_{H P} = \frac{Q_{h}}{Q_{h} - Q_{c}} = \frac{T_{h}}{T_{h} - T_{c}}$ (Carnot)	To evaluate performance of refrigerators and heat pumps.
Entropy	$Δ S = \int \frac{d Q_{rev}}{T}$ ; for isothermal: $Δ S = \frac{Q}{T}$	To determine spontaneity of processes and calculate entropy changes.
Carnot Cycle Analysis	$η = 1 - \frac{T_{c}}{T_{h}}$	To find maximum possible efficiency of an engine operating between two temperatures.
Entropy Change Calculations	$Δ S = n C_{v} \ln (\frac{T_{f}}{T_{i}}) + n R \ln (\frac{V_{f}}{V_{i}})$ for ideal gas	To compute entropy change for ideal gas processes, phase changes, or mixing.
Degrees of Freedom and $γ$	$C_{v} = \frac{f}{2} R$ ; $γ = 1 + \frac{2}{f}$	To compute specific heats and $γ$ for different gases.
Phase Change and Latent Heat	$Q = m L$ (where $L$ is latent heat)	To calculate heat transfer during melting, boiling, or condensation.
Mixing of Gases and Temperature	$n_{1} C_{v 1} (T_{f} - T_{1}) + n_{2} C_{v 2} (T_{f} - T_{2}) = 0$	To find final temperature when gases are mixed in an insulated container.
Polytropic Processes	$W = \frac{P_{2} V_{2} - P_{1} V_{1}}{1 - n}$ (for $n \neq 1$ ); for $n = 1$ , isothermal	For processes that are not strictly isothermal, adiabatic, etc., but follow $P V^{n} = constant$ .
Heat Capacity in Polytropic Processes	$C = C_{v} + \frac{R}{1 - n}$	To find heat capacity for a given polytropic process.
Otto Cycle (Petrol Engine)	$η = 1 - \frac{1}{r^{γ - 1}}$	To analyze spark-ignition internal combustion engines.
Diesel Cycle	$η = 1 - \frac{1}{r^{γ - 1}} (\frac{ρ^{γ} - 1}{γ (ρ - 1)})$	To analyze compression-ignition engines.

Open the Thermodynamics formula page

What JEE asks

Zeroth Law of Thermodynamics. JEE often tests the conceptual understanding of thermal equilibrium and its relation to temperature measurement.
First Law of Thermodynamics. JEE frequently asks to apply the first law to cyclic processes ( $Δ U = 0$ ), adiabatic processes ( $Q = 0$ ), and isothermal processes ( $Δ U = 0$ for ideal gas).
Second Law of Thermodynamics (Kelvin-Planck & Clausius). JEE tests conceptual understanding of why perpetual motion machines of the second kind are impossible.
Third Law of Thermodynamics. JEE rarely asks directly but may link to entropy change calculations.
Work Done in Thermodynamic Processes. JEE often gives a $P - V$ diagram and asks to compute work as area under curve, or uses these formulas in numerical problems.
Internal Energy of Ideal Gas. JEE uses this to link $Δ U$ with $Q$ and $W$ via first law.
Molar Specific Heats. JEE often asks to compute $C_{p}$ or $C_{v}$ for polyatomic gases or mixtures.
Adiabatic Process Relations. JEE frequently asks to find final state variables after adiabatic compression/expansion.
Heat Engine Efficiency. JEE often asks to compute efficiency from given $Q_{h}$ and $Q_{c}$ , or find Carnot efficiency from temperatures.
Coefficient of Performance (Refrigerator & Heat Pump). JEE may ask to compute COP given temperatures or heat transfers.
Entropy. JEE tests entropy change calculations for isothermal expansion, mixing, and phase changes.
Solving Problems Using First Law and $P - V$ Diagrams. JEE often presents a cycle (e.g., Carnot, Otto) and asks to compute efficiency or work.
Carnot Cycle Analysis. JEE often asks to compute Carnot efficiency or compare with actual engine efficiency.
Entropy Change Calculations. JEE may ask to find entropy change of universe for irreversible processes.
Thermodynamic Processes Summary. JEE often describes a process and asks to find work, heat, or internal energy change.
Degrees of Freedom and $γ$ . JEE often asks to find $C_{v}$ , $C_{p}$ , or $γ$ for a given gas.
Phase Change and Latent Heat. JEE may combine phase change with first law or entropy change.
Free Expansion of Ideal Gas. JEE often tests that temperature remains constant for ideal gas in free expansion.
Mixing of Gases and Temperature. JEE may ask to compute final temperature or entropy change in mixing.
Polytropic Processes. JEE may ask to find work or heat in a polytropic process.
Heat Capacity in Polytropic Processes. JEE may ask to compute heat transfer using $Q = n C Δ T$ for polytropic processes.
Otto Cycle (Petrol Engine). JEE may ask to compute efficiency given compression ratio.
Diesel Cycle. JEE may ask to compare efficiencies of Otto and Diesel cycles.

Common traps

Confusing thermal equilibrium with mechanical equilibrium (equal pressure) or chemical equilibrium. Zeroth Law of Thermodynamics
Using wrong sign convention: $W$ is work done BY the system; if work is done ON the system, $W$ is negative. First Law of Thermodynamics
Thinking the second law forbids any conversion of heat to work; it forbids 100% efficiency. Second Law of Thermodynamics (Kelvin-Planck & Clausius)
Thinking entropy becomes zero for all substances at 0 K; only perfect crystals. Third Law of Thermodynamics
Using isothermal formula for adiabatic processes; they have different $P - V$ relations. Work Done in Thermodynamic Processes
Using $C_{p}$ instead of $C_{v}$ for internal energy change. Internal Energy of Ideal Gas
Using $C_{p} - C_{v} = R$ for real gases; it's approximate. Molar Specific Heats
Using isothermal relation $P V = constant$ for adiabatic processes. Adiabatic Process Relations
Using $T_{c} / T_{h}$ directly without converting to Kelvin. Heat Engine Efficiency
Confusing COP of refrigerator with heat pump; they have different numerators. Coefficient of Performance (Refrigerator & Heat Pump)
Using $d Q$ instead of $d Q_{rev}$ ; entropy change is defined for reversible paths. Entropy
Assuming $Δ U = 0$ for a non-cyclic process; it is zero only for cyclic processes (or isothermal ideal-gas processes). Solving Problems Using First Law and $P - V$ Diagrams
Thinking Carnot cycle is the only reversible cycle; it's the most efficient between two temperatures. Carnot Cycle Analysis
Using irreversible path directly; must use reversible path. Entropy Change Calculations

Topics in Thermodynamics

Zeroth and First Law
Thermodynamic Processes
Heat Engines and Efficiency
Second Law of Thermodynamics
Calorimetry
Heat Transfer

Previous-year questions on Thermodynamics

JEEGenie has 848 previous-year questions on Thermodynamics. A few worked examples follow.

Year	Questions
2026	37
2025	54
2024	45
2023	58
2022	58
2021	73
2020	48
2019	98
2018	23
2017	24
2016	20
2015	16
2014	19
2013	22
2012	15
2011	9
2010	8
2009	11
2008	4
2007	12
2006	8
2005	11
2004	18
2003	14
2002	16
2001	10
2000	8
1999	5
1998	9
1997	21
1996	7
1995	4
1994	5
1993	5
1992	6
1991	3
1990	2
1989	8
1988	5
1987	5
1986	3
1985	4
1984	2
1983	2
1982	4
1981	5
1979	2
1978	2

Question 1 easy

Heat is supplied to a diatomic gas at constant pressure. Then the ratio of

Δ Q : Δ U : Δ W

is ______.

2 : 3 : 5
5 : 3 : 2
2 : 5 : 7
7 : 5 : 2

Answer: D

Explanation. For a diatomic gas, the molar specific heats are:

C_{v} = \frac{5}{2} R, C_{p} = \frac{7}{2} R

At constant pressure, the heat supplied is:

Δ Q = n C_{p} Δ T = n \cdot \frac{7}{2} R Δ T

The change in internal energy depends only on temperature change:

Δ U = n C_{v} Δ T = n \cdot \frac{5}{2} R Δ T

The work done at constant pressure is:

Δ W = P Δ V = n R Δ T

Thus, the ratio is:

Δ Q : Δ U : Δ W = \frac{7}{2} : \frac{5}{2} : 1 = 7 : 5 : 2

D

From NTA/IIT papers: JEE Main 2026 · 2 Apr Shift 1

Question 2 medium

The equation for real gas is given by

(P + \frac{a}{V^{2}}) (V - b) = R T

, where

P, V, T

and R are the pressure, volume, temperature and gas constant, respectively. The dimension of

{a b}^{- 2}

is equivalent to that of :

Compressibility
Planck's constant
Energy density
Strain

Answer: C

Explanation. The van der Waals equation is

(P + \frac{a}{V^{2}}) (V - b) = R T

. From

\frac{a}{V^{2}}

, the dimension of

a

[P] [V^{2}] = [M L^{- 1} T^{- 2}] [L^{6}] = [M L^{5} T^{- 2}]

. From

(V - b)

b

has dimension

[V] = [L^{3}]

. Thus

a b^{- 2}

has dimension

[M L^{5} T^{- 2}] [L^{- 6}] = [M L^{- 1} T^{- 2}]

, which is the dimension of pressure, i.e., energy per unit volume (energy density). Option C is correct.

From NTA/IIT papers: JEE Main 2025 · 2 Apr Shift 1

Question 3 hard

The specific heat capacity of a substance is temperature dependent and is given by the formula

C = k T

, where

k

is a constant of suitable dimensions in SI units, and

T

is the absolute temperature. If the heat required to raise the temperature of

1

kg of the substance from

- 73^{\circ}

C to

27^{\circ}

C is

n k

, the value of

n

is _____. [Given:

0

= - 273^{\circ}

C.]

Answer: 25000

Explanation. We convert the temperatures to kelvin:

- 73^{\circ} C = 200 K, 27^{\circ} C = 300 K .

Heat required for mass

m = 1 kg

Q = m \int_{T_{i}}^{T_{f}} C (T) d T = 1 \cdot \int_{200}^{300} k T d T .

Evaluate the integral:

Q = k {[\frac{T^{2}}{2}]}_{200}^{300} = \frac{k}{2} (300^{2} - 200^{2}) .

Compute:

300^{2} - 200^{2} = 90000 - 40000 = 50000,

Q = \frac{k}{2} \cdot 50000 = 25000 k .

Given

Q = n k

, we have

n k = 25000 k ⟹ n = 25000.

25000

From NTA/IIT papers: JEE Advanced 2024

Question 4 easy

A source supplies heat to a system at the rate of

1000 W

. If the system performs work at a rate of

200 W

. The rate at which internal energy of the system increases is

600 W
1200 W
500 W
800 W

Answer: D

Explanation. Using the first law of thermodynamics,

Δ

U = Q - W, where

Δ

U is the change in internal energy, Q is the heat added, and W is the work done by the system.

Δ

U = 1000 W - 200 W = 800 W.

From NTA/IIT papers: JEE Main 2023 · 6 Apr Shift 1

Frequently asked questions

How much of JEE Main comes from Thermodynamics?: JEEGenie estimates about 6% of JEE Main marks come from Thermodynamics, based on recent papers. The share varies from year to year.
Which formulas matter most in Thermodynamics?: The Thermodynamics sheet on this page has 18 formula entries, starting with First Law of Thermodynamics, Third Law of Thermodynamics, Work Done in Thermodynamic Processes, Internal Energy of Ideal Gas, Molar Specific Heats. Each one comes with when to use it and the trap to avoid.
What topics does Thermodynamics cover for JEE?: Thermodynamics covers Zeroth and First Law, Thermodynamic Processes, Heat Engines and Efficiency, Second Law of Thermodynamics, Calorimetry, Heat Transfer.
How should I revise Thermodynamics before the exam?: Learn each formula together with its condition of use, read the trap listed beside it, then practise questions and review every mistake. The practice link on this page opens Thermodynamics questions on JEEGenie.
How many previous-year questions are there on Thermodynamics?: JEEGenie has 848 previous-year questions on Thermodynamics, and this page shows a few with worked explanations.
Is practice on Thermodynamics free on JEEGenie?: Sign-up is free and includes a 7-day Pro trial. After the trial, free accounts keep a limited daily quota.