Physics · Thermodynamics · NEET
Work is done by a gas only when it pushes something (a piston or the outside gas) through a distance. In free expansion the gas expands into a vacuum. A vacuum has zero pressure, so there is nothing to push against. External pressure P(ext) = 0, therefore W = P(ext) x dV = 0 x dV = 0. The gas gains volume but does no work on anything.
The gas is kept inside a container with insulated (adiabatic) walls, so no heat can flow in or out. That makes the process adiabatic by construction, which means Q = 0. Note this is set by the insulation of the box, not by the speed of the process.
No. Apply the first law: dU = Q - W = 0 - 0 = 0. So internal energy U stays constant. For an ideal gas U depends only on temperature, so if U is constant then T is constant. The gas cools or heats only if it is a real (non-ideal) gas, because of intermolecular forces.
It is both in a special way, and truly neither. It is adiabatic because Q = 0 (insulated box). For an ideal gas it is also isothermal because T does not change. But it is NOT a quasi-static reversible process, so you cannot draw it as a smooth line on a P-V graph and you cannot use PV(gamma) = constant or W = nRT ln(V2/V1). Those formulas need a slow, controlled process.
That formula assumes the process is quasi-static, so the gas has a single well-defined pressure at every step and pushes the wall gently. In free expansion the gas rushes into vacuum, is turbulent, and has no single pressure during the process. The correct work is found from the external pressure the gas pushes against, which is zero, giving W = 0.
In a slow isothermal expansion the gas pushes a piston, so W is positive, and heat Q flows in to keep T constant, so Q is positive. In free expansion there is no piston and no heat: W = 0 and Q = 0. Both keep T constant for an ideal gas, but for very different reasons.
In which of the following processes, heat is neither absorbed nor released by a system?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
A gas is kept in one half of an insulated box, the other half is empty (vacuum). When a valve is opened, the gas spreads out to fill the whole box on its own. It does no work and no heat enters or leaves, so its internal energy stays the same.
It is named after James Joule, who did the experiment. He let a gas expand into a vacuum in an insulated container and found almost no temperature change for gases that behave nearly ideally. So it is also called Joule expansion or Joule free expansion.
It is irreversible. The gas rushes into empty space in an uncontrolled, turbulent way and will never gather back into one half by itself. Because it is not quasi-static, it cannot be shown as a smooth curve on a P-V diagram.
Q = 0 (insulated box), W = 0 (expands into vacuum), and dU = Q - W = 0. Since U is constant and U depends only on T for an ideal gas, the temperature also stays constant.
Yes. Even though Q = 0, the process is irreversible, so the entropy of the gas increases: dS = nR ln(V2/V1). The entropy of the surroundings does not change because no heat is exchanged with them.
