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One mole of water vapourises at 373K under atmospheric pressure. What is the work done?

One mole of water vapourises at 373K under atmospheric pressure. What is the work done?

Grade:12

1 Answers

Gaurav
askIITians Faculty 164 Points
6 years ago
Volume of 1 mol of water at 373 K, 1 atm (V1) =18cm3= 0.018 L
Volume of 1 mol of water in vapour state at 373 K,
1 =RTP =0.0821cm3atmK−1mol−1×373K1 atm = 30.62L
Now
w=−P(V2−V1)= 1(30.62 - 0.018) = -30.6 L-atm = -30.6×101.3 = -3100 J

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