Chemistry Vaporization and Vapor Pressure help!?

Question

Ethanol has a heat of vaporization of 38.56 kJ/mol and a normal boiling point of 78.4 degrees C. What is the vapor pressure of ethanol at 19 degrees C.

HINT: At boiling, vapor pressure = external pressure (use 760 torr)

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2. Benzene has a heat of vaporization of 30.72 kJ/mol and a normal boiling point of 80.1 degrees C. At what temperature does benzene boil when the external pressure is 465 torr?

HINT: At boiling, vapor pressure = external pressure (use 760 torr)

Answer 1

Use the Clausius-Clapeyron equation.

ln(P2/P1)=(Δ/R)((1/T1)-(1/T2))

Note: Convert KJ to J by multiplying by 10^3 so 38.56 KJ/mol=38.56x10^3  J/mol. Convert °C to K by adding 273.

ln(P2/760 torr)=(38.56x10^3 J/mol/8.31 J/mol•K)((1/351.4 K)-(1/292 K))

ln(P2/760 torr)=-2.686

ln(P2)-ln(760 torr)=-2.686

ln(P2)=-2.686+ln(760 torr)

ln(P2)=3.947 torr

P2=e^3.947=51.786 torr=52 torr(2 sig figs)

You can do the second one.