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Isothermal reversible condensation

  1. May 24, 2010 #1
    1. The problem statement, all variables and given/known data

    A sample of 1.00 mol H20(g) is condensed isothermally and reversibly
    to liquid water at 100°C. The standard enthalpy of vaporization of water at
    100°C is 40.656 kJ mol-1. Find w, q, change in internal energy, and change in enthalpy for this process.

    2. Relevant equations


    3. The attempt at a solution

    So since this is an isothermal reversible compression, change in temperature is equal to 0, so change in internal energy should also be equal to 0.
    Because deltaH = deltaU + delta(pV), and for an ideal gas pV = nRT so deltaH = 0 + delta(nRT) = 0, change in enthalpy should also be equal to 0.

    HOWEVER the solutions were quite different. Apparently deltaH = -standard enthalpy of vapourization, q = deltaH, w = -(external pressure) x deltaV, and deltaU = q + w.

    What am I missing here? Why are the changes in internal energy and enthalpy not zero if the change in temperature is zero?

    I know this is probably a really basic thermodynamics question but I just don't get it!
     
  2. jcsd
  3. May 24, 2010 #2

    Mapes

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    Liquid water is not an ideal gas.
     
  4. May 24, 2010 #3
    okay, so what are the expressions for deltaU and deltaH for a liquid?
     
  5. May 25, 2010 #4

    Mapes

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    Exactly what is given in the answer: the change in enthalpy is the enthalpy of vaporization. Condensation releases heat, so this value is negative. And by definition, the change in energy is the enthalpy without the work performed during the process.
     
  6. May 25, 2010 #5
    Okay, i understand that now, thanks. There's one more thing that i am having trouble understanding, though. In the solutions, it says that because the condensation is done isothermally and reversibly, the external pressure is constant at 1atm.
    I thought the whole point of a reversible process was that the pressure was not constant? Shouldn't the expression for work be
    w = -nRTln(Vf/Vi),
    not w = -(external pressure)*(Vf-Vi)?
     
  7. May 25, 2010 #6

    Mapes

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    Reversibility doesn't imply anything about the pressure being constant or not constant. And work is always [itex]\int P(V)\,dV[/itex]. For ideal gases at constant temperature this expression can be reduced to the other expression you give, but again, we're not dealing with an isolated ideal gas here.
     
  8. May 25, 2010 #7
    Ah, I see. I really must learn which expressions apply only to ideal gases!
    So because this question mentions the standard enthalpy of vaporization, the external pressure is at standard conditions, which is 1atm?
    Thank you very much for your help.
     
  9. Sep 30, 2011 #8
    I am having the same problems in this question. Not sure if I did it correct.
    I said "Delta H = -Enthalpy of Vaporization"
    "W=0" because the final state (liquid) does not alter volume
    "Delta H = q " this is true for isothermal processes
    "Delta U= q + W, W=0 so Delta U=q"

    is this thought correct?
     
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