February 21, 2017

Homework Help: chemistry

Posted by chemstudent on Thursday, February 5, 2009 at 12:01am.

A sample of 1.00 mol H2O(g) is condensed isothermally and reversibly to liquid water at 100°C. The standard enthalpy of vaporization of water at 100°C is 40.656 kJ mol−1. Find w, q, ∆U, and ∆H for this process.

I can solve for ∆H: n * standard ∆H * -1 because reverse of vaporization.
-40.656 kJ

I can solve for ∆U:
∆H = ∆U + ∆n(gas)*RT
-40.656 = ∆U + -1 * -R * 373.15K
∆U = -37.55 kJ

Now, how do I figure out w and q

I know that
∆U = q + w
but beyond that, how do I solve?

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