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General Chemistry - Heat, Work, Enthalpy, Entropy Question

  1. Oct 5, 2012 #1
    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. jcsd
  3. Oct 6, 2012 #2
    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

    ?
     
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