Question

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.


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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?

Answer 1

The pressure is constant, and so the heat at constant pressure is qp = - dH vaporization. W=-pdV and so since p=pext=1atm and dV = (vf - vi) = vliquid - vvapor, we'll say that since the volume of the vapor is significantly greater than that of the liquid, then vf-vi is approximately equal to -volume of vapor = - nRT/pext. Solve for these and then you get dU.

Answer 2

Well, the pressure here is not constant, but the temperature is. Reversible implies that there are infinitesimal changes in certain parameters.
For isothermal, reversible systems

delta H=0
delta U=0
q=-w
and w=-nRTln(Vf/Vi)

You can find your Vf by using the density constant for liquid water which s essentially 1.0 g/mL.
I used SATP laws for an ideal gas to find the volume of water at room temp. Then I converted to the volume at higher temp using the relation V1/T1=V2/T2. Hope this works.

Answer 3

it really worked and helped me I really appreciate.