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Homework Help: Work done on steam during phase change?

  1. Aug 30, 2012 #1
    1. The problem statement, all variables and given/known data
    (a) How much work is done on the steam when 3.89 mol of water at 100°C boils and becomes 3.89 mol of steam at 100°C at 1.00 atm pressure? (Assume the latent heat of vaporization of water is 2.26 106 J/kg.)

    (b) Assume the steam to behave as an ideal gas. Determine the change in internal energy of the system of the water and steam as the water vaporizes.

    2. Relevant equations

    Q + W = ΔE
    Q = m * Lv
    W = -P * ΔV = -P * (V2 - V1)
    P * V = n * R * T

    3. The attempt at a solution

    Convert moles of water to volume (multiply by molar mass, then use density of water)
    Volume of water: 0.07002 m^3.

    Then find the volume of steam using ideal gas law: V = ((3.89 mol) * (8.31 J / mol K) * (373.15K)) / (101325 Pa)
    Volume of steam: 0.1190 m^3

    So work should be -P * (V2 - V1) -> -(101325 Pa) * ((.1190 m^3) - (.07002 m^3))

    W = -4963 J = -4.963 kJ.

    But my web assign homework tells me that I am over 10% away from the correct answer.

    I can't do part (b) until I can complete part (a) because W + Q = ΔE, so without knowing the work, I can't find the change in internal energy (at least that's what my prof said when I asked him).

    Thanks in advance for any help :)
  2. jcsd
  3. Aug 30, 2012 #2


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    Check where the decimal point should go here.
  4. Aug 30, 2012 #3
    (3.89 mol) * (18 g / mol) = 70.02 g = 70.02 * 10^(-3) kg

    move the decimal over three places: 0.07002 kg, then density of water is 1 kg/ m^3, so:

    .07002 m^3, right?
  5. Aug 30, 2012 #4


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    It would be very hard to pick up a m3 of water! So, is the density of water 1 kg/m3?
  6. Aug 30, 2012 #5
    Oh my gosh, I can't believe I did that XD! Thanks 1000 kg/m^3
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