JEE Mains Chemistry · Chemical Thermodynamics

Work of Expansion

The work of an expanding or compressed gas: external pressure times ΔV when that pressure is steady, −nRT ln(V₂/V₁) when the change is reversible, zero into a vacuum, and the area under the path on a p–V graph.

All Chemical Thermodynamics notes JEE Mains Chemistry strategy

Why this matters

Twenty PYQs, half of them numerical, and six from 2026, more than any other page. Seven use the reversible isothermal or adiabatic formula; nine use a constant external pressure, a free expansion or the order of single-stage and reversible work; four read work as an area on a p–V graph.

Concept 1 of 3: Reversible isothermal and adiabatic work

In a reversible expansion the outside pressure is kept only just below the gas pressure, so the gas does the most work it can. For an ideal gas at constant temperature, U and H do not change. Every joule of work the gas does is paid for by heat it takes in: q = −w.

Definition

$w_{\mathrm{rev}} = -nRT\ln\frac{V_2}{V_1} = -2.303\,nRT\log\frac{V_2}{V_1}$ .
At constant temperature $\frac{V_2}{V_1} = \frac{p_1}{p_2}$ , and $nRT = p_1V_1$ .
Isothermal ideal gas: $\Delta U = 0$ , $\Delta H = 0$ , $q = -w$ .
Reversible adiabatic: $q = 0$ , so $w = \Delta U = nC_v\Delta T$ . The gas cools as it expands.
$\ln 2 = 0.693$ , $\ln 10 = 2.303$ . 1 L bar = 100 J; 1 L atm = 101.3 J.

Reversible isothermal work

w_{\mathrm{rev}} = -2.303\,nRT\log\frac{V_2}{V_1} = -2.303\,nRT\log\frac{p_1}{p_2}

Worked example

3 mol of an ideal gas at 400 K expands reversibly and isothermally from 4 L to 16 L. Find

w

q

\Delta U

and

\Delta H

. (

R = 8.314

J K⁻¹ mol⁻¹,

\log 4 = 0.602

)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 28 Jan 2026 Shift 1 · Q26Moderate

Example 1 · Chemical Thermodynamics · Work of Expansion

20.0dm^{3}

of an ideal gas ' X ' at 600 K and 0.5 MPa undergoes isothermal reversible expansion until pressure of the gas is 0.2 MPa . Which of the following option is correct? (Given:

\log2 = 0.3010

and

\log5 = 0.6989

)

log for ln

nRT\log\frac{V_2}{V_1}

without the 2.303 is 2.303 times too small. For

nRT = 5

kJ and a volume ratio of 4 it gives

5 \times 0.602 = 3.0

kJ against the correct

5 \times 1.386 = 6.9

kJ, and the small value is always an option.

q and w with the same sign

In an isothermal change of an ideal gas,

q = -w

. An option where

q

and

w

have the same sign, or where

\Delta U

equals the work, is wrong.

Drill 6 more on reversible isothermal and adiabatic work

Concept 2 of 3: Work against a constant external pressure and free expansion

Against a steady outside pressure, the work is that pressure times the change in volume. If the outside is a vacuum, nothing resists the gas, so no work is done, however much the volume grows. Splitting a change into more, smaller steps brings it closer to reversible.

Definition

$w = -p_{\mathrm{ext}}(V_2 - V_1)$ : negative for expansion, positive for compression.
Free expansion (into a vacuum): $p_{\mathrm{ext}} = 0$ , so $w = 0$ . For an ideal gas also $q = 0$ , $\Delta U = 0$ and $\Delta T = 0$ (Joule's experiment).
Isothermal ideal gas: $\Delta U = 0$ , so $q = -w = p_{\mathrm{ext}}\Delta V$ .
Size of the work between the same two states: single-stage compression > multi-stage compression > reversible (the same for compression and expansion) > multi-stage expansion > single-stage expansion.
1 kPa dm³ = 1 J; 1 L bar = 100 J; 1 L atm = 101.3 J.

Work against a constant external pressure

w = -p_{\mathrm{ext}}\,(V_2 - V_1)

Worked example

2 mol of an ideal gas at 350 K expands isothermally from 10 L to 25 L against a constant external pressure of 1.2 bar. Find

w

and

q

in joules.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 27 Jan 2024 · Q54Moderate

Example 2 · Chemical Thermodynamics · Work of Expansion

If three moles of an ideal gas at

300\text{ }K

expand isothermally from

30dm^{3}

45dm^{3}

against a constant opposing pressure of

80kPa

, then the amount of heat transferred is _______

J

A change in volume does not mean work

In a free expansion the volume does change. The work is still zero because the external pressure is zero, and

w

-p_{\mathrm{ext}}\Delta V

, not

-p_{\mathrm{gas}}\Delta V

Reversible is the extreme in two different senses

A reversible path gives the MOST work out of an expansion but needs the LEAST work for a compression. So single-stage compression needs the most work of all, and single-stage expansion gives the least.

Drill 8 more on work against a constant external pressure and free expansion

Concept 3 of 3: Work as an area on a p–V graph

On a graph of p against V, the work in any step is the area under that step. Around a closed cycle the areas partly cancel, and the net work is the area enclosed. With p up and V across, a clockwise cycle means the gas does net work; an anticlockwise one means work is done on it.

Definition

Vertical step (constant $V$ ): $w = 0$ .
Horizontal step (constant $p$ ): $w = -p\Delta V$ .
Straight sloping step: the area of a trapezium, average $p$ times $\Delta V$ .
Step along an isotherm: $w = -nRT\ln\frac{V_2}{V_1}$ , with $nRT = pV$ read from any point on it.
Whole cycle: $\Delta U = 0$ , $q = -w$ , and $|w|$ is the enclosed area.
Check which axis carries $V$ before reading any area; some graphs plot $V$ upwards.

Work around a cycle

w_{\mathrm{net}} = -\oint p\,dV,\qquad |w_{\mathrm{net}}| = \text{area enclosed}

Worked example

An ideal gas goes round the cycle A(1 bar, 2 L) → B(1 bar, 6 L) → C(3 bar, 6 L) → D(3 bar, 2 L) → A. Find the net work done on the gas in joules.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 24 Jan 2023 · Q150Moderate

Example 3 · Chemical Thermodynamics · Work of Expansion

One mole of an ideal monoatomic gas is subjected to changes as shown in the graph. The magnitude of the work done (by the system or on the system) is____

\times 10^{2}\text{ }J

(nearest integer)

V on the vertical axis

When a graph plots V upwards, read each point as (p, V) and work the steps out one by one. The rule 'clockwise means work done by the gas' reverses when the axes are swapped.

A cycle is not always the biggest area

A cycle's net work is only the area between its two paths. One long expansion sweeps the whole area under its curve down to the V axis, which is often bigger than any enclosed loop.

Drill 3 more on work as an area on a p–v graph

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 (3)

Reversible isothermal and adiabatic work
Reversible isothermal work
$w_{\mathrm{rev}} = -2.303\,nRT\log\frac{V_2}{V_1} = -2.303\,nRT\log\frac{p_1}{p_2}$
Work against a constant external pressure and free expansion
Work against a constant external pressure
$w = -p_{\mathrm{ext}}\,(V_2 - V_1)$
Work as an area on a p–V graph
Work around a cycle
$w_{\mathrm{net}} = -\oint p\,dV,\qquad |w_{\mathrm{net}}| = \text{area enclosed}$

Watch out for (6)

log for ln
→ Reversible isothermal and adiabatic work
q and w with the same sign
→ Reversible isothermal and adiabatic work
A change in volume does not mean work
→ Work against a constant external pressure and free expansion
Reversible is the extreme in two different senses
→ Work against a constant external pressure and free expansion
V on the vertical axis
→ Work as an area on a p–V graph
A cycle is not always the biggest area
→ Work as an area on a p–V graph

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