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Clausius Clapeyron

  1. Dec 13, 2009 #1
    It has been years scene I have taken a math class and I need help to see if I am setting the equation up correctly and if I have solved the problem correctly.

    Diethyl ether has ∆Hvap of 29.1kJ/mol and a vapor pressure of 0.703 atm at 25.0 C. What is its vapor pressure at 95.0 C?

    I converted the temp to K and kJ to J

    In P2/P1=delta H vaporization/R (1/T2- 1/T1)

    rearranged to

    P2/P1= antilog (delta H vap/R (1/T2- 1/T1))

    Then muliply by P1
    P2= P1 antilog (delta H vap/R (1/T2- 1/T1))

    antilog is also exp so...

    P2= P1 exp(delta H vap/R (1/T2- 1/T1))

    P2= .703 atm exp (29100/8.3145 (1/368-1/298)

    P2= 2.19 atm
  2. jcsd
  3. Dec 13, 2009 #2


    User Avatar

    Staff: Mentor

    Re: Clausius-Clapeyron

    Looks OK. I have not checked the final result.

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