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Vapour Pressure of Water

  1. Mar 18, 2012 #1
    1. The problem statement, all variables and given/known data
    This is problem in Mandl's book on Stats Phys; The vapour pressure of water at 298.15 K is 23.75 mmHg. What is the vapour pressure of water at 273.16 K, given that the latent heat of evaporation of water at 298.15 and 273.16 K is 43,991 and 45,049 J/mol respectively?


    2. Relevant equations
    According to the Hints, we use this equation;

    [tex]\frac{d}{dT}\ln P = \frac{1}{P} \frac{dP}{dT} = \frac{ML_{12}}{R T^2}[/tex]

    where M is the gram-molecular weight and L_12 is the latent heat of vaporization.


    3. The attempt at a solution
    Integrating, we get
    [tex]P=A \exp\left({- \frac{M L_{12}}{RT}}\right)[/tex]

    where A is an arbitrary constant. but when i sub in the values, I get some ridiculous number. where did I go wrong??

    Btw the answer is 4.58 mmHg while i got something of 10^6 in magnitude. o_O
     
  2. jcsd
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