General Chemistry - Heat, Work, Enthalpy, Entropy Question

1. Oct 5, 2012

modx07

1. Assume that one mole of H2O(g) condenses to H2O(l) at 1.00atm and 95 Celcius. Calculate q, w, ΔH, ΔS of the system, ΔS of surroundings. BUT I AM NOT ASKING HOW TO CALCULATE THESE VALUES, SEE LAST SENTENCE OF POST.

2. Relevant equations
q = nCΔT
ΔH = n(Cp)ΔT
W = -PΔV
ΔS = [q_reversible] / T

Cp of H2O(l) = 75.3 J / K*mol
Cp of H2O(g) = 36.4 J / K*mol
ΔH-vaporization of H2O @ 100 = 40.7 kJ / K*mol

3. The attempt at a solution

Basically, I have the solution but I don't understand why it's the case. When I asked my TA, he said that the condensation is actually occurring at 95 Celcius, which confused me right off the bat since I knew that at 1atm, the condensation point should be 100 degrees.

Regardless, I thought that the step that was occurring should be simply:

H2O (g) -> H2O (l) @ 95 degrees.

Thus, I thought that since the system stays at the same temperature, then ΔE = 0 and that q = -w. However, this assumption is wrong. I am simply asking why this is wrong (and not to calculate the values).

2. Oct 6, 2012

AJKing

Take this a point of discuss and not a solution, because I'm learning the same thing as we speak.

Since

ΔE = qv + w = nCvΔT + (-PextΔV)

and there is no change in volume, isn't it true that

ΔE = qv = nCvΔT

?