- #1
modx07
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1. Assume that one mole of H2O(g) condenses to H2O(l) at 1.00atm and 95 celsius. 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.
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
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 celsius, 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).
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 celsius, 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).