General Chemistry - Heat, Work, Enthelpy, Entropy Question

[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.

Homework 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

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).

 

[AJKing]

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

Since

ΔE = q_v + w = nC_vΔT + (-P_extΔV)

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

ΔE = q_v = nC_vΔT

?