One mole of an ideal gas at 300 K in thermal contact with its surroundings expands isothermally from 1 . 0 L to 2 . 0 L against a constant pressure of 3 . 0 atm . In this process, the change in entropy of the surroundings ∆ S s u r r in J K - 1 is: 1 L atm = 101 . 3 J

Step-by-step solution

From 1 s t law of thermodynamics, q s y s = ∆ U - w = 0 - - P e x t . ∆ V = 3.0   a t m × 2.0   L - 1.0   L = 3.0   L   a t m ∴ ∆ S s u r r = q r e v s u r r T = - q s y s T = - 3.0 × 101.3   J 300   K = - 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 = - 1.013   J /