Thermodynamics Formula Sheet — NEET Physics

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

Syllabus — topics coveredNCERT · 4 sub-topics

Thermal equilibrium, zeroth law, concept of temperature
Heat, work & internal energy, first law of thermodynamics
Isothermal & adiabatic processes
Second law of thermodynamics, reversible & irreversible processes

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

★ Remember

Over any process

Δ U

depends only on the end states. For an ideal gas

Δ U = μ C_{V} Δ T

— true for

every

process, not just isochoric.

🚫 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

Isothermal & Adiabatic Processes

Isothermal (

T

constant)

P V = const; Δ U = 0, Q = W = μ RT ln \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

Heat Engines & Refrigerators

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

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.

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Frequently Asked Questions

What are the most important Thermodynamics formulas for NEET?

This Thermodynamics formula sheet covers all the high-yield Physics formulas, definitions and facts you need for NEET, across Thermal equilibrium, zeroth law, concept of temperature, Heat, work & internal energy, first law of thermodynamics, Isothermal & adiabatic processes, Second law of thermodynamics, reversible & irreversible processes — each shown with the key result and, where useful, a worked example.

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Blurt the Thermodynamics key points 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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