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Two gases A and B have their P vs V isotherms shown at the same temperature T. The critical temperatures of A and B are T_A and T_B respectively. From the graph, gas A shows a liquid-gas phase transition loop (below critical temperature) while gas B shows a smooth hyperbolic isotherm (above critical temperature). Which statements are correct? (I) The pressure correction term is more negligible for gas B at temperature T. (II) The P-V curve for gas B will have the same shape as gas A if T > T_B. (III) Gas A will show the same P-V curve as gas B if T > T_A.

1. III only
2. II and III
3. II only
4. All

Correct answer: III only

Solution

Gas A shows phase transition loop => T < T_A. Gas B shows smooth isotherm => T > T_B. Statement I: pressure correction (a/V²) is more negligible when attractive interactions are less significant — gas B at T > T_B is further from condensation, so this is plausible but the problem context is about the shape. Statement II: B already has T > T_B and shows smooth curve — saying it 'will have same shape as A' if T > T_B makes no sense because B is already above T_B. Statement III: if T > T_A, gas A would also have T above its critical temperature and show a smooth isotherm like B's current curve — TRUE.

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