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Homework Help: Question about change in internal energy

  1. Oct 5, 2012 #1
    Fair warning: This is more chemistry than physics but I know that thermodynamics and heat and work are all covered in General Physics but I'll post in the Other Science forum too.

    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 5, 2012 #2


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    When water vapor condenses the vapor releases energy to the environment. The resultant liquid has less internal energy than the original vapor even though the temperature remains the same. As the molecules crowd together when forming the liquid state, the potential energy of interaction between the molecules decreases.
  4. Oct 6, 2012 #3

    Andrew Mason

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    You are assuming that ΔE (ΔU) is a function of temperature only. ie. ΔE = nCvΔT. But this is true only for an ideal gas. Water vapour is NOT an ideal gas. Its internal energy is very much a function of temperature and volume/pressure. It takes energy to increase the separation between water molecules.

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