Benzene has a heat of vaporization of 30.72kj/mol and a normal boiling point of 80.1 C . At what temperature does benzene boil when the external pressure is 410 ? This is how I am solving this problem,

Question

Benzene has a heat of vaporization of 30.72kj/mol and a normal boiling point of 80.1 C . At what temperature does benzene boil when the external pressure is 410 ? This is how I am solving this problem, 
ln(415/760)= (30720/8.314)*(1/80.1 - 1/T2)

It is asking to answer using two significant figures and it keeps telling me I'm incorrect.  Any help would be appreciated.

Thanks.

Lost · Answer

ops I meant to put 410 in the equation.

DrBob222 · Answer

Did you and I work on this problem yesterday. If not, then let me tell you quickly where your problem(s) is/are. First you MUST use Kelvin for T1. That is 80.1 + 273.16. I think the problem asks for T in C BUT you must solve the problem first, using Kelvin, then convert back to C. Second, I see you typed the problem as 410 but substituted 415. Take care that you are solving the problem with the right numbers.