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 () in is
(1 L atm = 101.3 J)
(1 L atm = 101.3 J)
- (A)5.763
- (B)1.013
- (C)-1.013✓ Correct
- (D)-5.763
Answer: (C)
Solution
For an isothermal process, , so the heat absorbed by the gas equals the work done: L·atm J. The surroundings lose this heat, so J·K.
JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Single correct option