Q.Two cylinders $A$ and $B$ of equal capacity are connected to each other via a stopcock. $A$ contains a gas at standard temperature and pressure. $B$ is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following:

Concept: Adiabatic Free Expansion (Joule Expansion) — the gas expands into a vacuum without doing work and without heat exchange.

(a) Since the gas expands into a vacuum, $W = 0$. The system is thermally insulated, so $Q = 0$. By the first law, $\Delta U = 0$. For an ideal gas, internal energy depends only on temperature, so temperature remains constant. The gas now occupies double the volume at the same temperature. Using Boyle’s law ($P_1 V_1 = P_2 V_2$), with initial pressure $P_0$ (1 atm at STP) and final volume $2V_0$:

(b) $\Delta U = 0$ (as reasoned above).

(c) $\Delta T = 0$ — temperature remains unchanged. …

This is a classic free expansion problem. The key is to recognise that when the stopcock is opened, the gas rushes into the evacuated cylinder $B$ without any opposing pressure — it does no work, and because the system is thermally insulated, no heat enters or leaves. That combination forces the internal energy to stay the same, and for an ideal gas, that means the temperature doesn't change either.

Let’s walk through it carefully.

1. Identify the process

   The gas in $A$ is initially at STP (standard temperature $T_0 = 273\ \text{K}$ and pressure $P_0 = 1\ \text{atm}$). Cylinder $B$ is empty. When the stopcock is opened, the gas expands to fill both cylinders. This is free expansion — the gas expands into a vacuum, so it does no work against an external pressure. The system is thermally insulated, so $Q = 0$.

2. First law of thermodynamics

   The first law says $\Delta U = Q - W$. Here $Q = 0$ and $W = 0$ (no work done), so

Internal energy does not change.

3. Temperature remains constant

   For an ideal gas, internal energy depends only on temperature: $U = n C_V T$. Since $\Delta U = 0$, we get $\Delta T = 0$.

   Important

   In free expansion of an ideal gas, temperature is constant — even though the process is neither isothermal nor reversible. This is a special case.

4. Final pressure

   The gas now occupies twice the original volume (both cylinders have equal capacity). The amount of gas $n$ is unchanged, and $T$ is unchanged. Using the ideal gas law:

Dividing the second equation by the first gives

So the final pressure in each cylinder is $0.5\ \text{atm}$.

5. Intermediate states and the $P$-$V$-$T$ surface …

Think about it from the molecules' point of view rather than the macroscopic first law: in free (Joule) expansion the gas rushes into a vacuum, so no molecule has to push against anything or lose kinetic energy doing work, and an ideal gas has no intermolecular attractions to work against either — so the molecules' average speed, and hence the temperature, simply can't change. That's the physical reason $\Delta U=0$ and $\Delta T=0$ fall out immediately, without formally invoking $Q=0,\ W=0$ first. It's also why part (d) is …