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One mole of an ideal gas at 300 K in thermal contact with 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 surroundings ( Δ S surr ) in J K - 1 is (1 L atm = 101.3 J)

Options

  1. A5.763
  2. B1.013
  3. C- 1.013
  4. D- 5.763

Correct answer

C. - 1.013

Step-by-step solution

Δ E = q + w 0 = q - P e x t Δ V q = P e x t Δ V = 3 atm (2 - 1) L = 3 atm L = ( 3 × 101.3 ) Joule Δ S s u r r = - q T = 3 × 101.3 300 = - 1.013 Joule/K

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