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Change in Internal Energy of 1kg Water to Steam?

  1. Nov 17, 2015 #1
    1. The problem statement, all variables and given/known data
    Hvap = 22.6E5 J/Kg. R=8.315E3 Jkmol^-1K^-1. 1kmol water = 18kg
    Keeping in mind that latent heats are enthalpies, what is the change in internal energy when 1kg water is converted to steam at pressure of 1atm and temperature of 100C.

    2. Relevant equations
    dU = dQ - PdV.

    3. The attempt at a solution
    dQ = mHvap
    PdV = 1atm (Vfinal - Vinitial) Vfinal >>Vinitial -> PdV = 1atm(Vfinal).
    Assume ideal gas-> V = nRT/P =18kilomoles * R*373 / 1atm?
     
  2. jcsd
  3. Nov 17, 2015 #2

    SteamKing

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    Is there an organized question here, or is it just stream of consciousness which you are writing down?
     
  4. Nov 17, 2015 #3
    You have 1 kg of liquid water in a closed container with a piston at 100 C and 1 atm. You add heat until all the water has vaporized to water vapor at 100 C and 1 atm (holding the pressure constant). What is the change in enthalpy ΔH for the water?

    Chet
     
  5. Nov 17, 2015 #4
    The H is given as the heat of vaporization which is the change in enthalpy. How do I get internal energy here? I'm uncertain of the final volume.
     
  6. Nov 17, 2015 #5
    I have to calculate the change in internal energy of the steam. My main problem is obtaining the final volume of steam.
     
  7. Nov 17, 2015 #6
    What is the definition of ΔH in terms of ΔU and Δ(PV)?
     
  8. Nov 17, 2015 #7
    Lvap = H = U + PV => dH = dU + PdV. I just don't see how the given information will allow me to compute this. I feel that I need the density of steam or specific volume. I don't see what I'm missing here. This is actually an old exam problem from another school so I'm just doing it for practice for my exam tomorrow.
     
  9. Nov 17, 2015 #8
    From the ideal gas law, what is the volume of 1 kg of water vapor at 1 atm and 100 C?
    What is the volume of 1 kg of liquid water at 1 atm and 100 C?
     
  10. Nov 17, 2015 #9
    density of water = 1000kg/m^3 => V = 1E-3m^3.
    vapor: V = nRT/P. n = 1kg/16kg kilomoles.
    Then dU = Lm - P(Vvapor - Vwater)?
     
  11. Nov 17, 2015 #10
    Sure. That's right.

    Incidentally, I don't understand your answer for the volume of the vapor. It should have units of m^3. I comes out to about 1.67 m^3. Notice that the volume of the liquid water is negligible compared to the volume of the water vapor.

    Chet
     
  12. Nov 17, 2015 #11
    I think I didn't make it too clear. I was just specifying what n would be; not volume. Thanks.
     
  13. Nov 17, 2015 #12
    n = 1000gm/18 = 55.6 moles
     
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