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Vapor Pressure of Water

  1. Oct 30, 2009 #1
    1. The problem statement, all variables and given/known data
    The vapor pressure of water at 25.0 C is 23.8 torr, and the heat of vaporization is 43.9 kJ/mol. What is the vapor pressure (atm) of water at 59.8C?


    2. Relevant equations
    Clausius-Clapeyron equation.
    ln(P2/P1)=(ΔH/R)((1/T1)-(1/T2))


    3. The attempt at a solution

    P2=?
    P1=23.8 torr = .031315789 atm
    ΔH=43.9 KJ/mol=43900 J/mol
    R= 8.31 (J)/(mol∙K)
    T1=25.0°C=298 K
    T2=59.8°C=332.8 K

    plugged into the equation and keep getting the wrong answer.

    Can someone please help.

    ln(p2) - ln(.031315789) = -(439005)/(8.31) * [(1/332.8) - (1/298)]

    HELP idk what i did wrong
     
  2. jcsd
  3. Oct 30, 2009 #2
    Atm is not KMS units
     
  4. Oct 31, 2009 #3
    That shouldn't matter as long as P1 and P2 have the same units.

    The setup looks good, but it does look like the calculations have a typo in the heat of vaporization though.
     
    Last edited: Oct 31, 2009
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