PhysicsThermodynamicsJEE · NEET · NSEP · INPhO · IPhO

Thermodynamics · Part 7 of 10

Free Expansion of a Gas

Let a gas rush into a vacuum inside an insulated vessel. It pushes against nothing and exchanges no heat, so $Q = W = Δ U = 0$ and an ideal gas keeps its temperature. The end states obey Boyle's law, but the process itself is irreversible and has no path on a P–V diagram.

Builds on: Part 5 · Isochoric, Isobaric and Isothermal Processes, Part 6 · Adiabatic Process.

Video coming soon

Before

Gas | valve | vacuum

An insulated vessel holds gas on one side and vacuum on the other, separated by a valve.

A box with gas on the left half and vacuum on the right. — Insulated walls: no heat can flow.

The box with gas spread through both halves. — The gas fills the whole vessel at the same temperature.

After

$Q = W = Δ U = 0$

The gas expands into the vacuum with nothing to push, so $W = 0$ ; it is insulated and fast, so $Q = 0$ . Then $Δ U = 0$ and, for an ideal gas, T is unchanged.

End states only

$P_{1} V_{1} = P_{2} V_{2}$

Same T at start and end, so the end states obey Boyle's law. In between, the gas has no single P or T: there is no path, and the process is irreversible. It is not an isothermal process, and $P V^{γ}$ does not apply.

Two states on the same isotherm with no path between them. — Only the two end states can be drawn.

Summary

Key formulas

Q = 0, W = 0, Δ U = 0

Free Expansion of a Gas

P_{1} V_{1} = P_{2} V_{2}

Free Expansion of a Gas

Worked examples

One for every idea

Free Expansion of a Gas

1. An ideal gas occupies 5 L at 300 K and $1 \times 1 0^{5}$ Pa in a thermally insulated cylinder. A valve opens to an evacuated chamber and the gas freely expands to a total volume of 8 L. Find the final temperature and pressure.

Free expansion: $Q = 0$ , $W = 0$ , so $Δ U = 0$ and T stays 300 K.
Boyle for the end states: $P_{2} = \frac{P _{1} V _{1}}{V _{2}} = 1 0^{5} \times \frac{5}{8}$ .

$T = 300$ K, $P_{2} = 6.25 \times 1 0^{4}$ Pa.

JEE-style question

Your turn

An ideal gas at 27 °C, in a thermally insulated vessel, expands freely into a vacuum until its volume doubles. Its final temperature is:

(a)27 °C

(b)13.5 °C

(c)−123 °C

(d)−46 °C

Show the answer and the traps

Q = 0 and W = 0, so ΔU = 0 and T stays 27 °C: option a.

Halving the temperature, in °C or in kelvin, has no basis. −46 °C uses the adiabatic law, which is the wrong process.

Watch out

Common mistakes

Assuming the gas does work because its volume growsWork needs something to push against; a vacuum offers nothing.

Using PV^γ = constant because Q = 0Free expansion isn't a slow, reversible adiabatic process; only

P_{1} V_{1} = P_{2} V_{2}

holds for the end states.

Practice

Try these

1. An ideal gas at 2 atm in an insulated vessel expands freely into an equal evacuated vessel. Find the final pressure and the change in temperature.

1 atm; no change in temperature.

2. In the free expansion of an ideal gas, which of Q, W, ΔU, P, V and T change?

Only P (falls) and V (rises). Q = W = ΔU = 0 and T is unchanged.

Free Expansion of a Gas

Gas | valve | vacuum

Q=W=ΔU=0

P1​V1​=P2​V2​

Key formulas

One for every idea

Your turn

Common mistakes

Try these

$Q = W = Δ U = 0$

$P_{1} V_{1} = P_{2} V_{2}$