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### homework 9.19

Posted: Wed Jan 31, 2018 10:12 pm
"Calculate the standard entropy of vaporization of water at 85 C, given that its standard entropy of vaporization at 100. C is 109.0 J K 1 mol 1 and the molar heat capacities at constant pressure of liquid water and water vapor are75.3 J K 1 mol 1 and 33.6 J K 1 mol 1, respectively, in this range."

Should we use S=q/T where q=n*C*T? do we have to know the value of n in order to get the final result? I feel a little bit confused. Thanks!

### Re: homework 9.19

Posted: Wed Jan 31, 2018 10:27 pm
To find ΔS for change in temperature (for example: in the first step of this problem, T1=358 K and T2=373 K), you would use the equation for ΔS=n*Cp*ln(T2/T1) which is derived from ΔS=q/T and nCΔT. Since this problem is asking for standard entropy, you can use the formula without n (ΔS=Cp*ln(T2/T1)) and your answer will have units of J/K*mol, which are the units for standard entropy.

### Re: homework 9.19

Posted: Thu Feb 01, 2018 4:01 pm
why in this problem do you have to heat then cool the H2O?

### Re: homework 9.19

Posted: Thu Feb 01, 2018 7:57 pm
From a previous CC post:

"The water is heated up to 100 C because that is the boiling point of water where water turns from liquid to gas. There is a change because the question asks for the standard enthalpy of vaporization of water at 85 C, which implies that it was a liquid that was heated up to 100C (accompanied by entropy change) along with the entropy of vaporization at 100C and then it implies that now the water is in gas form that needs to be cooled back down to 85C since the gas will condensate once it reaches 85 C (which is accompanied by the entropy change). Then, use the respective standard molar entropies of water at the different phases for the right accompanied steps of heating, heat of fusion, and cooling of water."

### Re: homework 9.19

Posted: Thu Feb 01, 2018 8:44 pm
Thank you!