One mole of an ideal gas at 300K in thermal contact with its surroundings expands isothermally from 1 . 0L to 2 . 0L against a constant pressure of 3 . 0atm . In this process, the change in entropy of the surroundings

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Correct answer: C. - 1 . 013

From 1 ^ st law of thermodynamics, q _ sys = ∆ U - w = 0 - [- P _ ext . ∆ V] = 3.0atm × (2.0L - 1.0L) = 3.0Latm ∴ ∆ S _ surr = ( q _ rev ) _ surr T = - q _ sys T = - (3.0 × 101.3J)/(300K) = - 1 .013 J / K Alternate solution: ∆ S _ surr = q _ surr T = - q _ sys T = W _ sys T ∵ For the isothermal process: ∆ U = 0 q _ sys = - W _ sys ∆ S _ surr. = - P _ ext ( V _ f - V _ i ) T = (- 3 ( 2 - 1 ))/(300) × 101.3 = - 1.013J / K

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