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”##