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1 mole of superheated water at 110 degrees Celsius and 1atm is evaporated to a vapour at the same temperature. Find the change in entropy given that i) In the pressure range from 1atm to 1.4atm the entropy of water at 110 degrees Celsius can be considered constant. ii) At 110 degrees Celsius the liquid water and vapour are in equilibrium at 1.4 atm. The latent heat of vaporization at 110 degrees Celsius is $4e4J/mol$ iii) In the pressure range 1atm to 1.4atm the vapour at 110 degrees Celsius behaves as a perfect gas.

My attempt:

At first I thought it was as simple as using $\Delta S = \int \frac{dQ}{T}$, but then that doesn't seem to account for the second given point (the one about equilibrium)

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closed as off-topic by John Duffield, Sebastian Riese, CuriousOne, ACuriousMind, Gert Mar 8 '16 at 1:55

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  • $\begingroup$ I think they expect you to take the final state as vapor at 110 C and 1.0 atm. $\endgroup$ – Chet Miller Mar 7 '16 at 21:06
  • $\begingroup$ @ChesterMiller yea I figured this one out, I'll write the answer later. This is a 3 part problem, I didn't realize it at first. $\endgroup$ – whatwhatwhat Mar 7 '16 at 22:06