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3 March, 15:42

A certain substance has a heat of vaporization of 37.51 kJ / mol. At what Kelvin temperature will the vapor pressure be 3.50 times higher than it was at 307 K?

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  1. 3 March, 18:22
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    T2=336K

    Explanation:

    Clausius-Clapeyron equation is used to determine the vapour pressure at different temperatures:

    where:

    In (P2/P1) = ΔvapH/R (1/T1 - 1/T2)

    p1 and p2 are the vapour pressures at temperatures

    T1 and T2

    ΔvapH = the enthalpy of vaporization of the liquid

    R = the Universal Gas Constant

    p1=p1, T1=307K

    p2=3.50p1; T2=?

    ΔvapH=37.51kJ/mol=37510J/mol

    R=8.314J. K^-1moL^-1

    In (3.50P1/P1) = (37510J/mol) / (8.314J. K^-1) * (1/307 - 1/T2)

    P1 and P1 cancelled out:

    In (3.50) = 4511.667 (T2 - 307/307T2)

    1.253=14.696 (T2 - 307/T2)

    1.253 = (14.696T2) - (14.696*307) / T2

    1.253T2=14.696T2 - 4511.672

    Therefore,

    4511.672=14.696T2 - 1.253T2

    4511.672=13.443T2

    So therefore, T2=4511.672/13.443=335.61

    Approximately, T2=336K
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