Find boiling point using Clausius-Clapeyron Equation.

The normal boiling point of benzene is 80.1°C, and the heat of vaporization is Δ Hvap = 30.7 kJ/mol. What is the boiling point of benzene in °C on top of Mt. Everest, where P = 260 mm Hg?

I tried using the following formula: ln(p1/p2) = (∆Hvap/R) (1/T2 - 1/T1)

I plugged in: P1=760, P2=260, Hvap=30.7, R=8.3145, T1=80.1, and T2=x

The correct answer is supposed to be 47°C, but I seem to be getting every answer but that one. No matter how many times I try I can't seem to get it right. Please help!

You need to convert T1 from °C to K . You also might not be multiplying Δ Hvap x 10^3 because the value is in kJ

ln(p2/p1) = -(Δ Hvap/R)*[1/T2 - 1/T1]

{ln (76/26)*8.31 }/{ 30.7 x 10^3} = 1/T2 - 1/353

Solving this we get T2 = 320 K approx or 47 °C

Check for unit consistency...

Your temperatures need to be in kelvin so
T1 = 80.1 + 273.15

And R should be 0.008314 kJ/mol K because your dH is kJ/mol... 8.314 is J/mol K

Your final answer will be in kelvin so subtract 273.15 to get it in celsius