Free Expansion and the Meaning of Entropy Change
A rigid, insulated container is divided by a partition into two equal halves, each of volume V = 10 L. One half holds n = 0.50 mol of an ideal monatomic gas at T = 300 K; the other half is evacuated. The partition is removed and the gas expands to fill the whole container. A student computes the entropy change as Delta_S = Q/T with Q = 0 (the container is insulated) and concludes Delta_S = 0, calling the process reversible. A second student argues the gas does no work (rigid walls) so the internal energy is constant and again concludes nothing changed. The final temperature is measured to be 300 K.
1. Compute the correct entropy change for the gas and identify the flaw in the first student's reasoning.
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Free expansion of an ideal gas: Q = 0, W = 0, so Delta_U = 0 and T is constant (consistent with the 300 K measurement). Entropy is a state function, so compute it along a reversible isothermal path from V to 2V: Delta_S = nRln(V_f/V_i) = (0.50)(8.314)ln(2) = 4.1570.693 = 2.88 J/K. The first student's error is applying Delta_S = Q/T to an irreversible process. That relation holds only for a reversible path; here Q = 0 but the process is spontaneous and irreversible, so Delta_S > 0. The correct statement is Delta_S = integral(dQ_rev/T) along a reversible path connecting the same endpoints, not the actual heat divided by T.
2. Address the second student's argument that constant internal energy means nothing changed.
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Constant internal energy (Delta_U = 0) and constant temperature do not imply constant entropy. For an ideal gas U depends only on T, so Delta_U = 0 is correct, but entropy depends on both T and V. The gas now occupies twice the volume, so there are more accessible microstates; the number of spatial configurations increases, giving Delta_S = nR ln2 > 0. Macroscopically, the pressure halves (from nRT/V to nRT/2V), so the final state is measurably different even at the same T. The second student conflates 'no energy change' with 'no state change.' The process is irreversible: the reverse (gas spontaneously compressing into half the volume) is never observed, which is the operational signature of irreversibility.
3. If the gas were instead a van der Waals gas with attractive interactions, would the final temperature be exactly 300 K? Explain.
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No. For a real gas with attractive intermolecular forces, free expansion does internal work against those attractions as molecules move apart, converting kinetic energy into potential energy; the temperature therefore drops (Joule-Thomson-like cooling in free expansion for attractive-dominated gases). Formally, U depends on both T and V for a van der Waals gas: (dU/dV)_T = a*n^2/V^2 > 0, so at constant U, an increase in V requires a decrease in T. The measured 300 K would then be inconsistent with a purely attractive van der Waals model unless the gas were ideal or the effect were below resolution. This tests whether the student knows the ideal-gas result T_final = T_initial is a special case, not a general law.