For 2 mol of a van der Waals gas (a negligible), the graph of lo g P against lo g ( V − 2 b ) is a straight line with an intercept of 1.20 on the lo g P axis. P is in atm and V is in L. The temperatur

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For 2 mol of a van der Waals gas (a negligible), the graph of lo g P against lo g ( V − 2 b ) is a straight line with an intercept of 1.20 on the lo g P axis. P is in atm and V is in L. The temperature at which the graph is plotted is ( R = 0.08 L atm K − 1 mol − 1 , lo g 2 = 0.30 )

TestHub · Accepted Answer

For n mol with a negligible: P ( V − nb ) = n RT , so lo g P = − lo g ( V − nb ) + lo g n RT . The intercept is lo g n RT = 1.20 , so n RT = 1 0 1.20 = 16 (since 4 lo g 2 = 1.20 ). T = 16/ ( 2 × 0.08 ) = 100 K . Traps: taking n = 1 gives 200 K; forgetting R gives 16 K; 373 K is 100°C. Answer: (D).

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