NTA Abhyas JEE Main2020ChemistryThermodynamics (C)Practice
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
- A5.763
- B1.013
- C- 1.013
- 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