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2015 Midterm Q4&5 part D

Posted: Tue Feb 14, 2017 9:10 pm
by Chris_Rudewicz_3H
Why do we do step 3? Why do we have to account for the entropy of the water vapor cooling from 373 K back down to 298 K? I understand why you do steps 1 and 2, but why step 3?

Re: 2015 Midterm Q4&5 part D

Posted: Tue Feb 14, 2017 9:14 pm
by Tara Ostad 1H
We have to do step 3 because it wants the entropy of vaporization of water at 298k, so we need to bring it back down to room temperature in order to do that because the first two steps just calculate it for boiling

Re: 2015 Midterm Q4&5 part D

Posted: Tue Feb 14, 2017 9:30 pm
by Chris_Rudewicz_3H
Tara Ostad 1H wrote:We have to do step 3 because it wants the entropy of vaporization of water at 298k, so we need to bring it back down to room temperature in order to do that because the first two steps just calculate it for boiling


But wouldn't the vapor turn to water again at that point? Why don't we calculate the entropy of condensation then?

Re: 2015 Midterm Q4&5 part D

Posted: Wed Feb 15, 2017 1:59 pm
by Matt Goff 1F
Water vapor will not necessarily turn to liquid water when brought back down to room temp, which is the tricky part of this question. If you look at a phase diagram for water you can see that at a low enough pressure, water can be in the vapor phase in the whole range of temps for this question.

Logically you can think that you have to do something after you do step 2 because the last calculation you made was for boiling water at 373K, and the question asks for a at 298K. So check the equation sheet and find something that relates entropy and the two temperatures you're dealing with, then plug in the given molar heat capacity for water vapor.