Free Expansion of an Ideal Gas: Why Work and ΔU Are Zero

Work in gas expansion is calculated as w = −p_ext × ΔV, where p_ext is the pressure pushing back from OUTSIDE. In free expansion the gas expands into a vacuum, so there is nothing outside pushing back: p_ext = 0. Even though the volume changes (ΔV is large), multiplying by zero gives w = 0. Key idea: work is not about the gas's own pressure — it is about the external pressure the gas pushes against. No opposing pressure means no work.

For an ideal gas, internal energy U depends only on temperature. In free expansion under adiabatic conditions, q = 0 and w = 0, so by the first law ΔU = q + w = 0. Since ΔU = nC_v·ΔT for an ideal gas, and ΔU = 0, it follows that ΔT = 0. So the gas ends at the same temperature it started. This is only true for an ideal gas (real gases cool slightly because of attractive forces between molecules).

It can be both at the same time for an ideal gas. In NEET problems, free expansion is usually described as adiabatic — the container is insulated, so q = 0. Because w = 0 too, ΔU = 0, which forces ΔT = 0. So the process is ALSO isothermal (constant temperature) as a result. The starting cause is adiabatic (no heat allowed); the constant temperature is the effect. Do not confuse this with a normal isothermal expansion, where heat DOES flow in to keep temperature steady.

In a normal isothermal expansion the gas pushes against a real external pressure, so it does work (w is not zero), and heat flows in from the surroundings to keep the temperature constant. In free expansion the gas expands into vacuum, so p_ext = 0, w = 0, and no heat is needed. Both can have ΔU = 0 for an ideal gas, but the reason is different: isothermal balances heat and work, while free expansion has both equal to zero separately.

Energy (ΔU) and entropy (ΔS) are different things. ΔU = 0 because no energy enters or leaves. But the gas now fills a bigger volume, so its molecules are more spread out and more disordered — this raises entropy. ΔS_system = nR ln(V2/V1) > 0. Since q = 0, the surroundings do not change, so ΔS_surroundings = 0. The total entropy of the universe increases, which is why free expansion is irreversible (it never happens in reverse on its own).

p_ext is the external pressure — the pressure of whatever is on the other side of the piston or boundary, pushing back against the gas. When gas expands into a vacuum, there is no gas or piston on the other side, so there is nothing to push back: p_ext = 0. The gas's OWN internal pressure does not appear in the irreversible work formula w = −p_ext·ΔV. That is why the gas can expand a lot yet do zero work.

Under adiabatic conditions: q = 0 (no heat), w = 0 (external pressure is zero), ΔU = q + w = 0, and ΔT = 0 (because ΔU = nC_v·ΔT for an ideal gas). All four are zero. This is the standard NEET result.

No. The gas expands into a vacuum, so the external pressure p_ext = 0. Work w = −p_ext·ΔV = 0, even though the volume increases a lot. Work depends on the pressure the gas pushes against, not on how much the volume changes.

It is irreversible. A gas will never gather itself back into a smaller volume on its own. The total entropy of the universe increases (ΔS_system > 0, ΔS_surroundings = 0), which is the sign of an irreversible process.

ΔU measures energy change, which is zero because no heat or work crosses the boundary. Entropy measures disorder, which increases because the gas fills a larger space. They are two separate quantities, so one can be zero while the other rises.

No. A real gas cools slightly (the Joule effect), because energy is used to overcome the attractive forces between molecules as they move apart. Only an IDEAL gas, which has no intermolecular forces, keeps ΔT = 0. NEET questions about free expansion assume an ideal gas.