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

6 years ago


Answers : (1)

										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
w=−P(V2−V1)= 1(30.62 - 0.018) = -30.6 L-atm = -30.6×101.3 = -3100 J
one year ago

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