JEE Mains Physics · Thermodynamics

Adiabatic Processes

With no heat exchanged, a gas follows PV^γ = constant, and all the work it does comes out of its internal energy, so W = −ΔU.

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Why this matters

Thirty PYQs, the largest page in the chapter, with four numerical answers and five from 2026. Fourteen use PV^γ = constant or its temperature forms to find a pressure, a temperature, a volume ratio or γ itself. Ten find the work, which depends only on the change in temperature. Six test the ideas: how steep an adiabat is next to an isotherm, why compression heats the gas, and why the molar heat capacity is zero.

Concept 1 of 3: The adiabatic relations PV^γ, TV^(γ−1) and P^(1−γ)T^γ

With no heat exchanged, a gas that is compressed must heat up and one that expands must cool. That temperature change makes the pressure change faster than on an isotherm: PV^γ = constant instead of PV = constant. PV = nRT then swaps any one variable for the temperature.

Definition

$PV^{\gamma} = \text{const}$ , $TV^{\gamma - 1} = \text{const}$ , $P^{1 - \gamma}T^{\gamma} = \text{const}$ , so $P \propto T^{\gamma/(\gamma - 1)}$ .
Density form: $\rho \propto 1/V$ , so $P \propto \rho^{\gamma}$ and $T \propto \rho^{\gamma - 1}$ .
"Sudden", "insulated", "thermally non-conducting walls" and "constant entropy" all mean adiabatic.
Compressed to $V/k$ from pressure P: isothermally the pressure becomes kP, adiabatically $k^{\gamma}P$ .
γ from data: match powers in $\left(\dfrac{V_1}{V_2}\right)^{\gamma} = \dfrac{P_2}{P_1}$ , or in $\left(\dfrac{V_1}{V_2}\right)^{\gamma - 1} = \dfrac{T_2}{T_1}$ . Then $\gamma = 1 + \dfrac{2}{f}$ names the gas: 5/3 monoatomic, 7/5 rigid diatomic.
A chain of processes: work out the pressure or temperature leg by leg, using PV = const on the isothermal legs and $PV^{\gamma} = \text{const}$ on the adiabatic ones.
Use absolute temperatures in $TV^{\gamma - 1}$ .

Adiabatic relations

PV^{\gamma} = \text{const} \qquad TV^{\gamma - 1} = \text{const} \qquad P^{1 - \gamma}T^{\gamma} = \text{const}

Worked example

A gas with γ = 4/3 at 600 K and 3.2 atm expands adiabatically to 8 times its volume. Find its final temperature and pressure.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 10 April 2023 · Q11Moderate

Example 1 · Thermodynamics · Adiabatic Processes

Consider two containers

A

and

B

containing monoatomic gases at the same Pressure (P), Volume (V) and Temperature (T). The gas in A is compressed isothermally to

\frac{1}{8}

of its original volume while the gas

B

is compressed adiabatically to

\frac{1}{8}

of its original volume. The ratio of final pressure of gas in

B

to that of gas in

A

is:

A sudden change is adiabatic

There is no time for heat to flow, so use PV^γ = constant. The isothermal answer from PV = constant is always among the options.

Kelvin in TV^(γ−1)

A temperature ratio needs absolute temperatures. A gas at 27 °C is at 300 K; doubling 27 °C is not doubling the temperature.

Drill 13 more on the adiabatic relations pv^γ, tv^(γ−1) and p^(1−γ)t^γ

Concept 2 of 3: Work done in an adiabatic process

With Q = 0, the first law leaves W = −ΔU. All the work a gas does comes out of its internal energy, so for a given gas and amount the work depends only on how much the temperature changes: W = nCv(T₁ − T₂). Writing Cv = R/(γ − 1) gives the usual formula.

Definition

$W = -\Delta U = nC_V(T_1 - T_2) = \dfrac{nR(T_1 - T_2)}{\gamma - 1} = \dfrac{P_1V_1 - P_2V_2}{\gamma - 1}$ .
Expansion: T falls, W > 0, ΔU < 0. Compression: T rises, W < 0 (work is done on the gas), ΔU > 0.
"Work done on the gas" is −W.
If only volumes or pressures are given, first find the missing end value from the adiabatic relations.
$C_V = \dfrac{f}{2}R$ , counting 2 for each vibrational mode.

Adiabatic work (done by the gas)

W = \frac{nR(T_1 - T_2)}{\gamma - 1} = \frac{P_1V_1 - P_2V_2}{\gamma - 1} = -\Delta U

Worked example

2 mol of a monoatomic ideal gas expands adiabatically and cools from 500 K to 350 K. Find the work done by the gas and ΔU. (

R = 8.3\ \text{J mol}^{-1}\text{K}^{-1}

)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 2 Apr 2026 Shift 1 · Q22Moderate

Example 2 · Thermodynamics · Adiabatic Processes

A vessel contains

0.15{\text{ }m}^{3}

of a gas at pressure 8 bar and temperature

140^{\circ}C

with

c_{p} = 3R

and

c_{v} = 2R

. It is expanded adiabatically till pressure falls to 1 bar. The work done during this process is

\_\_\_\_

k J . (R is gas constant)

The sign in a compression

In an adiabatic compression the gas does negative work and warms up. "Work done on the gas" is the positive number; read which one the question asks for.

Divide by γ − 1, not by γ

W = nRΔT/(γ − 1). Dividing by γ gives a much smaller wrong value, and it is often one of the options.

Drill 9 more on work done in an adiabatic process

Concept 3 of 3: True and false statements about adiabatic processes

Assertion and statement questions on adiabatic processes test a handful of ideas. Q = 0 is the definition; the internal energy does change; the curve is steeper than an isotherm; and the molar heat capacity is zero, because the temperature changes with no heat at all.

Definition

Slopes at the same point: isothermal $\dfrac{dP}{dV} = -\dfrac{P}{V}$ ; adiabatic $\dfrac{dP}{dV} = -\gamma\dfrac{P}{V}$ .
On an adiabat $\dfrac{dP}{P} = -\gamma\dfrac{dV}{V}$ ; on an isotherm $\dfrac{dP}{P} = -\dfrac{dV}{V}$ .
So for the same rise in pressure, the volume falls more on an isotherm than on an adiabat.
A free expansion into a vacuum has Q = 0 and W = 0, so ΔU = 0: it is adiabatic but irreversible, and PV^γ = const does not describe it.

Claim about an adiabatic process	True or false	Why
No heat crosses the boundary	True	That is the definition: Q = 0
The internal energy stays constant	False	ΔU = −W, which is not zero unless no work is done Q = 0 does not stop U changing: the work comes out of U.
An adiabatic compression raises the temperature	True	The work done on the gas goes into U, and U rises with T
Its P–V curve is steeper than the isotherm through the same point	True	Its slope is γ times the isothermal slope
The molar heat capacity is zero	True	C = Q/(nΔT) with Q = 0 while ΔT is not zero
The product TV stays constant	False	It is TV^(γ−1) that stays constant
A free expansion into a vacuum follows PV^γ = constant	False	It is irreversible; an ideal gas keeps its temperature

Every row follows from Q = 0 and Q = ΔU + W.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 29 Jan 2025 · Q77Moderate

Example 3 · Thermodynamics · Adiabatic Processes

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : With the increase in the pressure of an ideal gas, the volume falls off more rapidly in an isothermal process in comparison to the adiabatic process. Reason (R) : In isothermal process,

PV =

constant, while in adiabatic process

PV^{\gamma}=

constant. Here

\gamma

is the ratio of specific heats, P is the pressure and V is the volume of the ideal gas. In the light of the above statements, choose the correct answer from the options given below :

Adiabatic is not isothermal

Q = 0 does not make ΔT = 0. With no heat to make up for the work, the temperature must change in an adiabatic process.

The steeper curve is the adiabat

Through any point the adiabat is steeper than the isotherm by the factor γ. On a P–V diagram showing both, the steeper one is the adiabat.

Drill 5 more on true and false statements about adiabatic processes

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

The adiabatic relations PV^γ, TV^(γ−1) and P^(1−γ)T^γ
Adiabatic relations
$PV^{\gamma} = \text{const} \qquad TV^{\gamma - 1} = \text{const} \qquad P^{1 - \gamma}T^{\gamma} = \text{const}$
Work done in an adiabatic process
Adiabatic work (done by the gas)
$W = \frac{nR(T_1 - T_2)}{\gamma - 1} = \frac{P_1V_1 - P_2V_2}{\gamma - 1} = -\Delta U$

Reference tables (1)

True and false statements about adiabatic processes7 rows

Claim about an adiabatic process	True or false	Why
No heat crosses the boundary	True	That is the definition: Q = 0
The internal energy stays constant	False	ΔU = −W, which is not zero unless no work is done Q = 0 does not stop U changing: the work comes out of U.
An adiabatic compression raises the temperature	True	The work done on the gas goes into U, and U rises with T
Its P–V curve is steeper than the isotherm through the same point	True	Its slope is γ times the isothermal slope
The molar heat capacity is zero	True	C = Q/(nΔT) with Q = 0 while ΔT is not zero
The product TV stays constant	False	It is TV^(γ−1) that stays constant
A free expansion into a vacuum follows PV^γ = constant	False	It is irreversible; an ideal gas keeps its temperature

Every row follows from Q = 0 and Q = ΔU + W.

Watch out for (6)

A sudden change is adiabatic
→ The adiabatic relations PV^γ, TV^(γ−1) and P^(1−γ)T^γ
Kelvin in TV^(γ−1)
→ The adiabatic relations PV^γ, TV^(γ−1) and P^(1−γ)T^γ
The sign in a compression
→ Work done in an adiabatic process
Divide by γ − 1, not by γ
→ Work done in an adiabatic process
Adiabatic is not isothermal
→ True and false statements about adiabatic processes
The steeper curve is the adiabat
→ True and false statements about adiabatic processes

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