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A gas fills one half of an insulated box and the other half is a vacuum. You pull out the partition. What happens to the temperature? Many students say the gas does work, or that it must cool because the process is adiabatic. For an ideal gas, both are wrong. This post shows why Q, W and ΔU are all zero and works two examples.
Where is free expansion in the syllabus?
The NCERT Class 11 chapter Thermodynamics uses free expansion as an example of a system that is not in equilibrium. Fig. 11.6 shows a partition "suddenly removed leading to free expansion of the gas", and the text says the gas then passes through states that are not equilibrium states and "cannot be described by state variables". Later the chapter states plainly: "The free expansion of a gas is irreversible."
The Syllabus for JEE (Main) 2026 lists the first law, isothermal and adiabatic processes, and reversible and irreversible processes under Unit 8. The NEET (UG) 2026 syllabus lists the first law and isothermal and adiabatic processes. Check the syllabus for your own year.
Why is the work zero?
A gas does work when it pushes something back: a piston, the atmosphere. In a free expansion it moves into empty space inside a rigid container. Nothing resists it, so the external pressure is zero and W = 0, however much the volume grows.
Why are Q and ΔU zero?
The container is insulated, so Q = 0. NCERT writes the first law as ΔQ = ΔU + ΔW, with ΔW the work done by the system. With ΔQ = 0 and ΔW = 0, ΔU = 0.
NCERT also states that "for an ideal gas, internal energy depends only on temperature". If U does not change, T does not change. The pressure falls only because the same molecules now spread over a larger volume.
Worked example 1: one mole into a vacuum
One mole of a monatomic ideal gas at 400 K occupies 5.0 L. It is connected to an evacuated 10.0 L vessel, everything is insulated, and the valve is opened. Take R = 8.314 J/(mol K).
W = 0, Q = 0, so ΔU = 0 and the final temperature is 400 K.
Initial pressure: P₁ = nRT/V = 8.314 × 400 ÷ 0.0050 ≈ 6.65 × 10⁵ Pa.
Final pressure: P₂ = 8.314 × 400 ÷ 0.015 ≈ 2.22 × 10⁵ Pa, one third of P₁, as Boyle's law predicts for equal temperatures.
The trap: applying TV^(γ−1) = constant gives 400 × (1/3)^(2/3) ≈ 192 K. That answer is wrong because the process is not quasi-static.
Now reach the same final state another way: a slow, reversible isothermal expansion at 400 K from 5.0 L to 15.0 L. Our post on isothermal and adiabatic processes gives W = nRT ln(V₂/V₁) = 8.314 × 400 × ln 3 ≈ 3.65 kJ, and Q = 3.65 kJ. Same start, same end, ΔU = 0 both times, but Q and W differ completely. Heat and work depend on the path; internal energy does not.
Chemistry adds one more number. Entropy is a state function, listed in the JEE (Main) chemistry unit on thermodynamics, so its change is the same for both paths: ΔS = nR ln(V₂/V₁) ≈ 9.13 J/K. In the free expansion that rise happens with no heat at all, which marks the process as irreversible.
Worked example 2: two samples mixed in an insulated box
A rigid, insulated container has two parts. Part A holds 2 mol of a monatomic ideal gas at 300 K in 10 L. Part B holds 1 mol of the same gas at 600 K in 20 L. The partition is removed.
No work is done on the surroundings and no heat flows, so the total internal energy stays the same.
For one gas, U = nC_vT, so 2C_v × 300 + 1C_v × 600 = 3C_v × T_f, giving T_f = 1200 ÷ 3 = 400 K.
Before: P_A = 2 × 8.314 × 300 ÷ 0.010 ≈ 4.99 × 10⁵ Pa and P_B = 8.314 × 600 ÷ 0.020 ≈ 2.49 × 10⁵ Pa.
After: P_f = 3 × 8.314 × 400 ÷ 0.030 ≈ 3.33 × 10⁵ Pa.
The simple average works here only because both parts hold the same gas. For different gases, weight each temperature by nC_v.
Why does this question catch so many students?
Meltzer (2004) studied written answers from 653 students in three introductory university physics courses and interviewed 32 more. He found a widespread tendency to treat work and heat as if they were state functions, and found that only 20 percent or fewer were able to make effective use of the first law even after instruction. Free expansion tests exactly this: the same change of state with zero work in one path and kilojoules in another. Our post on sign conventions in thermodynamics helps if the signs of Q and W are what trip you up.
Which mistakes cost marks?
W = PΔV with the gas's own pressure. Work depends on the external pressure, which is zero here.
Using PV^γ or TV^(γ−1). These need a quasi-static process.
Assuming the gas cools because Q = 0. An ideal gas cools in an adiabatic expansion only if it does work.
Drawing the process as a curve on a PV diagram. NCERT says the intermediate states are not equilibrium states, so only the end points can be plotted.
Forgetting the word ideal. ΔT = 0 relies on U depending only on T, which NCERT states for an ideal gas.
When a problem mixes a free expansion with a later slow step, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning, so you can check which stage each equation belongs to.
A practice routine
Day 1: For example 1, write Q, W, ΔU and ΔT for both paths side by side from memory.
Day 2: Two mixing problems, one with different gases weighted by nC_v.
Day 3: One problem where a gas expands freely and is then compressed slowly back. Find the net heat and work.
Before your next session, redo example 1 with a 20 L empty vessel and check that the final pressure is one fifth of the start.
