Thermodynamics Formula Sheet — JEE Main Physics

Every key Thermodynamics formula, definition and theorem for JEE Main Physics in one place — with common examiner traps and worked examples. Free to read; blurt from memory, then check your gaps.

Syllabus — topics coveredNTA · 7 sub-topics

Thermal equilibrium and definition of temperature
Heat, work and internal energy
First law of thermodynamics
Isothermal and adiabatic processes
Second law of thermodynamics
Reversible and irreversible processes
Heat engine and refrigerator

Fundamentals & First Law

Zeroth law: If A and B are each in thermal equilibrium with C, then A and B are in thermal equilibrium with each other — this defines

temperature

as the property shared at equilibrium.

State variables

▸P,V,T,U describe an equilibrium state; linked by $P V = μ RT$ .
▸ $Intensive$ : $P, T, ρ$ (size-independent). $Extensive$ : U,V,M.
▸U is a $state function$ (path-independent); Q and W are $not$ (path functions).

Internal energy (ideal gas)

U = \frac{f}{2} μ RT

f =

degrees of freedom,

μ =

number of moles,

R = 8.314 J mol^{- 1} K^{- 1}

T =

absolute temp (K); U depends only on T

Gas in a cylinder: heat Q added at the base, work W done by the piston, change in internal energy U — Heat in raises $U$ and/or does work.

First law of thermodynamics

Δ Q = Δ U + Δ W

Δ Q =

heat added to system,

Δ U =

change in internal energy,

Δ W =

work done by system; an energy-conservation statement

Sign convention

▸ $Δ Q > 0$ : heat $absorbed$ by the system (heat out $\Rightarrow Δ Q < 0$ ).
▸ $Δ W > 0$ : work done $by$ the system (expansion; compression $\Rightarrow Δ W < 0$ ).
▸ $Δ U > 0$ : temperature rises (ideal gas).

Δ U

for an ideal gas

Δ U = μ C_{V} Δ T

C_{V} =

molar heat capacity at constant V; holds for

every

process (not only isochoric), because

U = U (T)

only

⚡ Shortcut · Where the heat goes at constant

P

Q : Δ U : W = C_{P} : C_{V} : R

, so

\frac{Δ U}{Q} = \frac{1}{γ}

and

\frac{W}{Q} = 1 - \frac{1}{γ} = \frac{2}{f + 2}

. Monatomic:

\frac{3}{5}

to U,

\frac{2}{5}

as work; diatomic:

\frac{5}{7}

and

\frac{2}{7}

🚫 Examiner Trap · First law & sign convention

(1) Q and W are

path

functions — never write '

Δ Q

' as a state change. (2) Sign slip:

Δ W

is work done

by

the gas; many texts write

Δ Q = Δ U - W_{o n}

— pick one convention and keep it. (3)

Δ U = μ C_{V} Δ T

even for isobaric/isothermal/adiabatic,

not

only constant-V. (4) U of an ideal gas depends on T alone —

not

on P or V separately.

Work Done & P–V Diagrams

Work done by a gas

W = \int_{V_{1}}^{V_{2}} P d V = area under the P - V curve

V_{1} \to V_{2}

along the actual path; expansion (

V_{2} > V_{1}

)

\Rightarrow W > 0

, compression

\Rightarrow W < 0

Area under a P-V curve between two states equals the work done by the gas — Work is path-dependent — it is the area beneath the curve.

★ Remember

Both Q and W depend on the

path

between two states; only

Δ U

is path-independent.

🎯 Exam · Quasi-static process

An infinitely slow process keeping the system in equilibrium throughout — the only kind for which P,T are well defined at every step and

W = \int P d V

applies.

Work in simple processes

Isobaric (

P

const)

W = P Δ V = μ R Δ T

Isochoric (

V

const)

W = 0

Isothermal

W = μ RT ln \frac{V _{2}}{V _{1}}

Adiabatic

W = \frac{μ R ( T _{1} - T _{2} )}{γ - 1}

Comparative: work for the same

V_{1} \to V_{2}

Process	$W$ formula	$Δ U$
Isobaric	$P Δ V = μ R Δ T$	$μ C_{V} Δ T$
Isochoric	$0$	$μ C_{V} Δ T = Q$
Isothermal	$μ RT ln \frac{V _{2}}{V _{1}}$	$0$
Adiabatic	$\frac{μ R ( T _{1} - T _{2} )}{γ - 1}$	$- W$

⚡ Shortcut · Area = work, on sight

Read W straight off the P–V plot as the

signed area

under the path. Equal end-states but a higher path

\Rightarrow

∣ W ∣

. A vertical line (V const) encloses zero area

\Rightarrow W = 0

; a horizontal line (P const) gives

W = P Δ V

instantly.

🚫 Examiner Trap · Work & P–V diagrams

(1) W is the area under the curve

only

for a quasi-static path — free/irreversible expansion has

W = 0

against vacuum. (2) Same two end-points,

different paths \Rightarrow

different W (and Q). (3) Isochoric

W = 0

but

Q \neq = 0

(it equals

Δ U

). (4) Use

Δ T

in kelvin for

W = μ R Δ T

Other graphs & peak temperature

▸V–T graph: an isobaric process is a straight line through the origin ( $V = \frac{μ R}{P} T$ ), and a steeper line means LOWER P. P–T graph: an isochoric process is a line through the origin, and a steeper line means SMALLER V.
▸To redraw a cycle on another graph, identify which variable is constant on each leg first; only a $P$ – $V$ graph shows work as area.
▸Straight-line path $P = P_{0} - kV$ : $T = \frac{P _{0} V - k V ^{2}}{μ R}$ peaks at $V = \frac{P _{0}}{2 k}$ (the midpoint of the line), giving $T_{ma x} = \frac{P _{0}^{2}}{4 k μ R}$ .
▸Straight line through the origin ( $P \propto V$ , i.e. $n = - 1$ ): $C = C_{V} + \frac{R}{2}$ .

Isothermal & Adiabatic Processes

Isothermal (

T

constant)

P V = const; Δ U = 0, Q = W = μ RT ln \frac{V _{2}}{V _{1}}

equivalently

W = μ RT ln \frac{P _{1}}{P _{2}} = 2.303 μ RT lo g_{10} \frac{V _{2}}{V _{1}}

; slope

\frac{d P}{d V} = - \frac{P}{V}

; isothermal bulk modulus

B_{T} = P

; needs a slow process in contact with a reservoir

Adiabatic (

Q = 0

)

P V^{γ} = const; T V^{γ - 1} = const; P^{1 - γ} T^{γ} = const

γ = C_{P} / C_{V}

(adiabatic index); adiabatic bulk modulus

B_{S} = γ P

; no heat exchange (

Q = 0

)

Adiabatic work

W = \frac{P _{1} V _{1} - P _{2} V _{2}}{γ - 1} = μ C_{V} (T_{1} - T_{2})

(P_{1}, V_{1}, T_{1}) \to (P_{2}, V_{2}, T_{2})

; expansion cools the gas, compression heats it;

W = - Δ U

On a P-V diagram the adiabatic curve through a point is steeper than the isothermal curve — Adiabatic curve is steeper: $slope_{a d ia} = γ \times slope_{i so}$ .

Comparative: isothermal vs adiabatic

Feature	Isothermal	Adiabatic
Constant	T (so $Δ U = 0$ )	$Q = 0$
Law	$P V =$ const	$P V^{γ} =$ const
$P$ – $V$ slope	$- P / V$	$- γ P / V$ (steeper)
Bulk modulus	$B_{T} = P$	$B_{S} = γ P$
$Q$	$Q = W$	$0$
First law	$Q = W$	$W = - Δ U$

🎯 Exam

At the same point an adiabatic is

γ

times steeper than an isothermal. For a free (irreversible) expansion into vacuum

W = 0, Q = 0 \Rightarrow Δ U = 0

— temperature unchanged for an ideal gas.

⚡ Shortcut · Polytropic master form

Both are

P V^{n} =

const:

n = 1

isothermal,

n = γ

adiabatic,

n = 0

isobaric,

n = \infty

isochoric. Molar heat

C = C_{V} + \frac{R}{1 - n}

, work

W = \frac{P _{1} V _{1} - P _{2} V _{2}}{n - 1}

(

n \neq = 1

) — one formula covers all four straight lines/curves.

🚫 Examiner Trap · Isothermal & adiabatic

(1)

Δ U = 0

holds for isothermal

only

Q = 0

for adiabatic

only

— do not swap them. (2) The adiabatic is

steeper

(factor

γ

), never the isothermal. (3) Adiabatic

free

expansion (into vacuum) is irreversible:

Δ U = 0

and T unchanged, so

P V^{γ} =

const does

not

apply. (4) Adiabatic expansion lowers T; do not assume T constant just because

Q = 0

Internal Energy, Cp, Cv & Equipartition

Molar specific heats

C_{V} = \frac{1}{μ} \frac{d U}{d T} = \frac{f}{2} R, C_{P} = C_{V} + R

C_{V}, C_{P} =

molar heat capacities at constant V, P;

f =

degrees of freedom; Mayer's relation

C_{P} - C_{V} = R

Adiabatic exponent

γ = \frac{C _{P}}{C _{V}} = 1 + \frac{2}{f}

f =

degrees of freedom;

γ > 1

always, and decreases as f grows

★ Remember · Equipartition of energy

Each degree of freedom carries average energy

\frac{1}{2} k_{B} T

per molecule; internal energy

U = \frac{f}{2} μ RT

⚡ Shortcut · Get

γ

from f instantly

γ = 1 + 2/ f

: monatomic

f = 3 \Rightarrow \frac{5}{3}

, diatomic

f = 5 \Rightarrow \frac{7}{5}

, polyatomic

f = 6 \Rightarrow \frac{4}{3}

. For a mixture use

\frac{μ _{1} + μ _{2}}{γ _{mi x} - 1} = \frac{μ _{1}}{γ _{1} - 1} + \frac{μ _{2}}{γ _{2} - 1}

Gas	$f$	$C_{V}$	$C_{P}$	$γ$
Monatomic (He, Ar)	3	$\frac{3}{2} R$	$\frac{5}{2} R$	$\frac{5}{3} \approx 1.67$
Diatomic (H₂, O₂)	5	$\frac{5}{2} R$	$\frac{7}{2} R$	$\frac{7}{5} = 1.40$
Polyatomic (non-linear)	6	$3 R$	$4 R$	$\frac{4}{3} \approx 1.33$

Refinements

▸Diatomic: add $2$ to f ( $\to 7$ ) if vibration is active at high T.
▸Mixture: $C_{V} = \frac{μ _{1} C _{V 1} + μ _{2} C _{V 2}}{μ _{1} + μ _{2}}$ .
▸For a solid, $C \approx 3 R$ (Dulong–Petit).

Degrees of freedom

f

▸Monatomic: $3$ translational. Diatomic: $3$ trans $+ 2$ rot $= 5$ .
▸Linear $\Rightarrow 2$ rotational; non-linear $\Rightarrow 3$ rotational.
▸Each active vibrational mode adds $2$ ( $\frac{1}{2}$ KE $+ \frac{1}{2}$ PE).

🚫 Examiner Trap · Heat capacities & equipartition

(1)

C_{P} > C_{V}

always (extra R for expansion work) — never

C_{V} > C_{P}

. (2)

C_{P} - C_{V} = R

is per

mole

; for specific heats divide by molar mass. (3) Use the f that is

active

at that T — vibration is usually frozen at room T, so diatomic

f = 5

not

7

. (4) Process-dependent heat

C = C_{V} + \frac{R}{1 - n}

can be

0

(adiabatic) or even negative — 'C' is not always

C_{P}

C_{V}

Cyclic Processes & the Second Law

Cyclic process

Δ U = 0 \Rightarrow Q_{n e t} = W_{n e t} = area enclosed

Δ U = 0

because the system returns to its initial state; clockwise loop

\Rightarrow W_{n e t} > 0

, anticlockwise

\Rightarrow W_{n e t} < 0

Closed loop on a P-V diagram; the enclosed area equals the net work done in one cycle — Net work in a cycle = enclosed P–V area.

Reversible vs irreversible

▸Reversible: quasi-static $+$ no dissipation (friction/viscosity) — an idealisation.
▸All real processes (free expansion, finite $Δ T$ heat flow, friction) are irreversible.
▸Irreversible $\Rightarrow$ total entropy of system $+$ surroundings increases.

	Reversible	Irreversible
Speed	infinitely slow	finite/real
Dissipation	none	friction/viscosity
Equilibrium	every step	only end-states
$Δ S_{u ni v}$	$= 0$	$> 0$

Second law of thermodynamics

🎯 Exam · Kelvin–Planck statement

No process is possible whose sole result is the complete conversion of heat from a reservoir into work (

Q_{2} \neq = 0

, so

η < 1

🎯 Exam · Clausius statement

No process is possible whose sole result is the transfer of heat from a colder to a hotter body (a refrigerator needs work,

W \neq = 0

★ Remember

The two statements are equivalent. The first law allows

η = 1

; the second law forbids it.

🚫 Examiner Trap · Cyclic & second law

(1) In one full cycle

Δ U = 0

, so

Q_{n e t} = W_{n e t}

— but Q and W on each leg are

not

zero. (2) Clockwise loop on P–V does net

positive

work (engine); anticlockwise is a refrigerator. (3) The second law

forbids

η = 1

, even though the first law allows it. (4) Kelvin–Planck

\neq =

η < 1

in one step' — heat can fully convert to work in a

single

expansion (isothermal), but not in a

cycle

Entropy change

Δ S = \int \frac{d Q _{re v}}{T}; Δ S_{g a s} = μ C_{V} ln \frac{T _{2}}{T _{1}} + μ R ln \frac{V _{2}}{V _{1}}

S is a state function, so evaluate along ANY reversible path between the same end states. Reservoir at T gaining Q:

Q / T

. Body of heat capacity C:

C ln \frac{T _{f}}{T _{i}}

. Melting/boiling:

m L / T

. Reversible adiabatic:

0

. Not named in the syllabus, but questions reach it through the second law

🎯 Exam · Entropy of the universe

Heat Q conducted from

T_{1}

T_{2}

(

T_{1} > T_{2}

Δ S = \frac{Q}{T _{2}} - \frac{Q}{T _{1}} > 0

. Free expansion

V_{1} \to V_{2}

W = Q = 0

yet

Δ S = μ R ln \frac{V _{2}}{V _{1}} > 0

— use the reversible isothermal path to compute it, never

\int d Q / T

along the actual path.

Heat Engines & Refrigerators

★ Remember · Heat engines & refrigerators — not in JEE Main 2026 syllabus

NTA dropped heat engines and refrigerators from the 2026 syllabus. This page is kept because older papers and practice questions still use it, and the second law (which IS in the syllabus) is stated in terms of engines.

Heat engine efficiency

η = \frac{W}{Q _{1}} = 1 - \frac{Q _{2}}{Q _{1}}

Q_{1} =

heat absorbed from source,

Q_{2} =

heat rejected to sink,

W = Q_{1} - Q_{2} =

net work output;

0 \leq η < 1

Heat engine: Q1 from hot reservoir, work W out, Q2 rejected to cold reservoir — Engine: heat $\to$ work, $η < 1$ .

Refrigerator / heat pump (COP)

β_{re f} = \frac{Q _{2}}{W} = \frac{Q _{2}}{Q _{1} - Q _{2}}; β_{p u m p} = \frac{Q _{1}}{W}

β =

coefficient of performance;

Q_{2} =

heat drawn from cold space,

W =

work input;

β_{p u m p} = β_{re f} + 1

and COP can exceed 1

Refrigerator: Q2 extracted from cold reservoir, work W input, Q1 dumped to hot reservoir — Reverse of an engine: work pumps heat uphill.

Comparative: engine vs fridge vs heat pump

Device	Useful out	Performance
Heat engine	work $W$	$η = \frac{W}{Q _{1}} = 1 - \frac{Q _{2}}{Q _{1}}$
Refrigerator	$Q_{2}$ (cooling)	$β_{re f} = \frac{Q _{2}}{W}$
Heat pump	$Q_{1}$ (heating)	$β_{p u m p} = \frac{Q _{1}}{W}$

★ Remember

A refrigerator is a heat engine run in reverse:

Q_{1} = Q_{2} + W

delivered to the hot reservoir.

⚡ Shortcut · Carnot COP from temperatures

For a Carnot (reversible) device:

β_{re f} = \frac{T _{2}}{T _{1} - T _{2}}

β_{p u m p} = \frac{T _{1}}{T _{1} - T _{2}} = β_{re f} + 1

, and

η = 1 - \frac{T _{2}}{T _{1}}

. Always

β_{p u m p} = \frac{1}{η _{C a r n o t}}

between the same reservoirs.

🚫 Examiner Trap · Engines & refrigerators

(1) Efficiency uses

input

Q_{1}

in the denominator; COP uses

work

W — do not mix. (2)

β

can be

> 1

;

η

can

never

\geq 1

. (3)

β_{p u m p} = β_{re f} + 1

(the pump also delivers the work as heat). (4)

W = Q_{1} - Q_{2}

— count rejected heat

Q_{2}

, the most-forgotten term.

Carnot Engine

★ Remember · Carnot engine — not in JEE Main 2026 syllabus

Part of the dropped heat-engine topic; kept for older papers and practice questions. The isothermal and adiabatic legs themselves are in the syllabus (page 3).

Carnot cycle (reversible)

▸ $A \to B$ : isothermal expansion at $T_{1}$ (absorbs $Q_{1}$ ).
▸ $B \to C$ : adiabatic expansion ( $T_{1} \to T_{2}$ ).
▸ $C \to D$ : isothermal compression at $T_{2}$ (rejects $Q_{2}$ ).
▸ $D \to A$ : adiabatic compression ( $T_{2} \to T_{1}$ ).

Carnot efficiency

η = 1 - \frac{Q _{2}}{Q _{1}} = 1 - \frac{T _{2}}{T _{1}}

T_{1} =

source temp,

T_{2} =

sink temp, both in

kelvin

; depends only on the two reservoir temperatures, not the working substance

Heat–temperature ratio (Carnot)

\frac{Q _{1}}{Q _{2}} = \frac{T _{1}}{T _{2}}

T

in kelvin; defines the absolute (thermodynamic) temperature scale

Carnot cycle on a P-V diagram: two isotherms (T1, T2) joined by two adiabats forming a closed loop — Two isotherms + two adiabats.

Step	Process	Heat
$A \to B$	isothermal exp. $T_{1}$	$+ Q_{1}$
$B \to C$	adiabatic exp.	$0$
$C \to D$	isothermal comp. $T_{2}$	$- Q_{2}$
$D \to A$	adiabatic comp.	$0$

🎯 Exam · Carnot's theorem

No engine working between two temperatures can exceed the Carnot efficiency, and it is

independent

of the working substance. Hence

\frac{Q _{1}}{Q _{2}} = \frac{T _{1}}{T _{2}}

defines the absolute temperature scale.

★ Remember

η = 1

only if

T_{2} = 0

K (unattainable) or

T_{1} \to \infty

— so a perfectly efficient engine is impossible.

🚫 Examiner Trap · Carnot engine

(1) Temperatures must be in

kelvin

— using

^{\circ}

C is the classic blunder. (2)

η

depends

only

T_{1}, T_{2}

, never on the gas or on

Q_{1}

. (3)

η_{C a r n o t}

is the

maximum

; a real engine is always less. (4) Raising

T_{1}

Δ T

helps less than lowering

T_{2}

by the same

Δ T

. (5) The cycle is

reversible

— running it backward gives a Carnot refrigerator.

More JEE Main Physics formula sheets

Frequently Asked Questions

What are the most important Thermodynamics formulas for JEE Main?

This Thermodynamics formula sheet covers all the high-yield Physics formulas, definitions and theorems you need for JEE Main, across Thermal equilibrium and definition of temperature, Heat, work and internal energy, First law of thermodynamics, Isothermal and adiabatic processes, Second law of thermodynamics — each shown with the key result and, where useful, a worked example.

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How should I revise Thermodynamics formulas?

Blurt the Thermodynamics formulas from memory, then check against this sheet to find your gaps — and practise a few previous-year questions on the chapter to make sure you can apply them under time pressure.

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