Posted by **Chris** on Wednesday, March 7, 2007 at 5:24pm.

When atmospheric pressure is 700. mmHg, water boils at 97.714 degrees C. Calculate entropy change of water if 50.0 grams is vaporized under these conditions. The heat of vaporization of water can be found in almost any reference text, and Cp H2O= 33.6 J/molK.

I found the heat of vaporization to be 9.72 kcal/mol at 100. degrees C in my text book, and I know that I can find change in entropy with the equation:

delta-G= delta-H - T(delta-S)

I'm just having trouble putting all the pieces together.

I was thinking something like this:

50.0 g H20 x 1 mol/18.02 g= 2.774 mol H2O

I was going to then multiply this value by 9.72 kcal/mol, but then I remembered that this value was for 100. degrees C, not 97.714 degrees C like stated in the problem. Is my thought process on the right track here? How do I manipulate the value that I have for this problem?

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