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QUESTION

# A certain substance has a heat of vaporization of 46.34 kJ/mol. At what Kelvin temperature will the vapor pressure be 5.00 times higher than it was at 323 K? Use the Clausius-Clapeyron equation.

The temperature will be 356 K.

So, you know that you must use the Clausius-Clapeyron equation. Now, you'll find this equation written is several equivalent forms, so I'll just choose one of these forms

ln(P_1/P_2) = (DeltaH_("vap"))/R * (1/T_2 - 1/T_1), where

P_1 - the measured at T_1; P_2 - the vapor pressure measured at P_2; DeltaH_("vap") - the of vaporization; R - the gas constant - expressed in Joules per mol K;

You have everything you need to solve for T_2. Since the pressure measured at this new temperature will be 5.00 times bigger than P_1, you can write it as P_2 = 5 * P_1 and use it in this form in the equation.

So, plug all in and you'll get

ln(P_1/(5 * P_1)) = (46340"J"/"mol")/(8.31446"J"/("mol" * "K")) * (1/T_2 - 1/"323 K")

ln(1/5) = "5573.4" * 1/T_2 - "5573.4" * 1/"323"

-1.6094 = "5573.4"/T_2 - 17.2252

15.646 = "5573.4"/T_2 => T_2 = 5573.4/15.646 = "356.2 K"

Rounded to three , the answer will be

T_2 = "356 K"