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 (ΔSsurr) in JK

Engineering

Chemistry

Entropy Calculations

Question

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 (ΔS_surr) in JK^–1 is [1 L atm = 101.3 J]

–5.763

–1.013

5.763

1.013

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

${ΔS}_{surr} = \frac{- Q_{sys}}{T} = \frac{- (- W_{irr isothermal})}{T} = \frac{W_{irr isothermal}}{T}$

W_{irr isothermal} = – P_ext (V₂–V₁) = – 3 (2 –1) = – 3 atm lit

= – 3 × 101.3 J = –303.9 J

${ΔS}_{surr} = \frac{- 303 .9}{300} = - 1 .013 {JK}^{- 1}$