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Isothermal and adiabatic questions look alike: a gas expands or is compressed, and you must find the work, the heat or the final temperature. The difference is in what stays fixed, and using the wrong formula is the commonest error. This post sets out both processes side by side, works through three examples with numbers, and shows how to read them on a PV diagram.
Where are these in the syllabus?
The JEE (Main) 2026 syllabus lists the first law of thermodynamics and isothermal and adiabatic processes. The JEE (Advanced) 2026 brochure lists isothermal and adiabatic processes, the bulk modulus of gases and the first law applied to ideal gases. NCERT Class 11 Chapter 11, section 11.8 covers both processes and uses the convention ΔQ = ΔU + ΔW, with ΔW the work done by the system. Check your year's bulletin.
Why do students get adiabatic questions wrong?
Loverude, Kautz and Heron (2002) asked university students about the adiabatic compression of an ideal gas. Few recognised that the first law was relevant, and fewer still used work to explain the temperature change. Most reasoned from a misreading of the ideal gas law. Meltzer (2004), studying 653 students in written questions and 32 in interviews, found that most interviewed students thought the net work in a cyclic process must be zero, and only 20% or fewer could use the first law effectively after instruction.
The cure is a habit: for every process, write the first law line before you reach for PV = nRT.
Isothermal versus adiabatic: what stays fixed?
Isothermal (T constant). For an ideal gas, ΔU = 0, so Q = W. PV = constant. Work done by the gas: W = nRT ln(V2/V1).
Adiabatic (Q = 0). ΔU = −W, so the gas cools when it expands and warms when compressed. PV^γ = constant and TV^(γ−1) = constant. Work done by the gas: W = nR(T1 − T2)/(γ − 1).
Use R = 8.314 J/(mol K). For a monatomic gas γ = 5/3; for a diatomic gas near room temperature γ = 7/5.
Worked example 1: isothermal expansion
Two moles of an ideal gas at 300 K expand slowly from 10 L to 20 L.
W = nRT ln(V2/V1) = 2 × 8.314 × 300 × ln 2 = 4988 × 0.693 ≈ 3.46 kJ.
ΔU = 0, so the gas absorbs 3.46 kJ of heat.
The pressure halves, since PV is constant.
Worked example 2: adiabatic compression
One mole of a monatomic gas at 300 K is compressed adiabatically from 8 L to 1 L.
T2 = T1 (V1/V2)^(γ−1) = 300 × 8^(2/3) = 300 × 4 = 1200 K.
W by gas = nR(T1 − T2)/(γ − 1) = 8.314 × (300 − 1200)/(2/3) = 8.314 × (−900) × 1.5 ≈ −11.2 kJ. So about 11.2 kJ of work is done on the gas.
Check with ΔU = nC_vΔT = 1 × (3/2)(8.314)(900) ≈ 11.2 kJ. With Q = 0, ΔU = −W. It matches.
The pressure rises by 8^(5/3) = 32 times, far more than the 8 times an isothermal compression would give.
How do you read the two curves on a PV diagram?
From PV = constant, the isothermal slope is dP/dV = −P/V. From PV^γ = constant, the adiabatic slope is −γP/V. Through the same point, the adiabatic curve is steeper by a factor γ. NCERT's Figure 11.8 shows two adiabatic curves connecting two isotherms.
Worked example 3. One mole of a monatomic gas at 300 K doubles its volume, once isothermally and once adiabatically.
Isothermal: W = 8.314 × 300 × ln 2 ≈ 1.73 kJ.
Adiabatic: T2 = 300 × 2^(−2/3) ≈ 189 K, so W = 8.314 × (300 − 189)/(2/3) ≈ 1.38 kJ.
The adiabatic curve falls faster, so it encloses less area, and the gas does less work. That matches the physics: with no heat coming in, the gas pays for the work from its own internal energy and cools.
Common mistakes
Using ln for an adiabatic process. The ln formula needs constant temperature.
Using P1V1 = P2V2 for an adiabatic change. It needs P1V1^γ = P2V2^γ.
Saying "no heat, so no temperature change". Adiabatic means no heat flow; the temperature changes because work is done.
Mixing conventions. Physics uses ΔQ = ΔU + ΔW with W done by the gas; chemistry often uses work done on the system. Our post on thermodynamics sign conventions compares them.
A practice routine
For six processes, write only the first law line and which quantity is zero (ΔU or Q).
Solve three isothermal and three adiabatic problems, checking each with ΔU = nC_vΔT.
Sketch both curves from the same starting point and shade the work for each.
Finish with a cycle problem; our guide to heat engines and the Carnot cycle uses all four steps.
If the curves still feel abstract, Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
Tonight, redo the adiabatic compression example and check that your ΔU matches your work.
