Isothermal vs Adiabatic Expansion of an Ideal Gas

By Aniket Bhardwaj · 15 October 2026 · Thermodynamics

An ideal gas that expands pushes against its surroundings and does work. How much work it does depends on how the expansion is carried out. Two limiting cases are asked again and again in physical chemistry: the isothermal reversible expansion (temperature fixed, heat flows in) and the adiabatic reversible expansion (no heat flows, so the gas cools). Comparing them is a standard route to understanding the first law.

Sign convention

This article uses ΔU = q + w, so work done by the gas is negative. Some textbooks use ΔU = q − w, where work done by the gas is positive. The magnitudes below are the same in both conventions; only the sign differs. Use the convention in your own syllabus.

The formulas for n moles of an ideal gas

Isothermal, reversible: ΔU = 0, q = −w, w = −nRT ln(V₂/V₁)
Adiabatic, reversible: q = 0, w = ΔU = nC_v(T₂ − T₁)
T₁V₁^γ−1 = T₂V₂^γ−1, with γ = C_p/C_v
Free expansion into vacuum: w = 0, q = 0, ΔU = 0 (ideal gas)

For an ideal monatomic gas C_v = (3/2)R and γ = 5/3. For a diatomic gas at ordinary temperature C_v = (5/2)R and γ = 7/5. The adiabatic equation holds only for a reversible process on an ideal gas with constant heat capacity.

Worked example 1: isothermal expansion

Question. One mole of an ideal monatomic gas at 300 K expands reversibly and isothermally from 10.0 L to 20.0 L. Find w, q and ΔU. (R = 8.314 J mol⁻¹ K⁻¹.)

Step 1. V₂/V₁ = 2, ln 2 = 0.6931.

Step 2. w = −nRT ln 2 = −1 × 8.314 × 300 × 0.6931 = −1729 J.

Step 3. For an ideal gas ΔU depends only on T, so ΔU = 0. Then q = −w = +1729 J.

The gas takes 1729 J from the surroundings as heat and returns it all as work.

Worked example 2: adiabatic expansion between the same volumes

Question. The same gas (1 mol, monatomic, 300 K, 10.0 L) now expands reversibly and adiabatically to 20.0 L. Find the final temperature, w, q and ΔU.

Step 1. γ − 1 = 5/3 − 1 = 2/3. T₂ = T₁(V₁/V₂)^2/3 = 300 × (0.5)^0.6667 = 300 × 0.6300 = 189.0 K.

Step 2. ΔT = 189.0 − 300 = −111.0 K.

Step 3. w = ΔU = nC_vΔT = 1 × 1.5 × 8.314 × (−111.0) = −1384 J. And q = 0.

Compare. The gas does 1384 J of work in the adiabatic case but 1729 J in the isothermal case, for the same volume change. In the adiabatic case the energy comes from the gas itself, so it cools. In the isothermal case, heat from the surroundings makes up the difference. On a P–V graph, the adiabatic curve falls more steeply than the isothermal curve, and the area under it is smaller.

Worked example 3: pressure check

Question. For the gas in example 2, find the starting and final pressures, and confirm that the adiabatic final pressure is lower than the isothermal one.

Step 1. P₁ = nRT₁/V₁ = (1 × 8.314 × 300) ÷ (0.0100 m³) = 2.494 × 10⁵ Pa, which is about 2.46 atm (1 atm = 101325 Pa).

Step 2 (isothermal). P₂ = nRT/V₂ = 8.314 × 300 ÷ 0.0200 = 1.247 × 10⁵ Pa, about 1.23 atm. This is simply half of P₁.

Step 3 (adiabatic). P₂ = nRT₂/V₂ = 8.314 × 189.0 ÷ 0.0200 = 7.857 × 10⁴ Pa, about 0.775 atm.

Conclusion. The adiabatic gas ends at a lower pressure (0.775 atm) than the isothermal gas (1.23 atm) because it is also colder.

Side by side

Quantity (1 mol, 300 K, 10 L → 20 L)	Isothermal	Adiabatic
q	+1729 J	0
w (by-gas negative)	−1729 J	−1384 J
ΔU	0	−1384 J
Final temperature	300 K	189.0 K
Final pressure	1.23 atm	0.775 atm

Common mistakes

Setting q = 0 for an isothermal process. Isothermal means ΔT = 0 and ΔU = 0, not q = 0. Heat flows.
Using PV = constant for an adiabatic change. That is Boyle's law for isothermal changes. Adiabatic reversible changes follow PV^γ = constant.
Using the adiabatic formula for free expansion. Free expansion is not reversible, so the formula does not apply; for an ideal gas T does not change.
Using log₁₀ where ln is needed. The isothermal work uses natural log, or multiply by 2.303 with log₁₀.
Using the wrong γ. Monatomic: 5/3. Diatomic: 7/5.
Mixing sign conventions. Decide ΔU = q + w or q − w first and keep to it.
Using litres and Pa together. Convert V to m³ when P is in Pa.

Exam relevance

Question type	What to do
Work in a reversible isothermal process	−nRT ln(V₂/V₁) or −nRT ln(P₁/P₂)
Final T after reversible adiabatic change	Use T V^γ−1 = constant
Work in adiabatic change	nC_vΔT
Compare two paths	Same initial state and volume change, different w and q (path functions)