The entropy of sublimation of a certain compound is 30.8 J / K mol at its normal sublimation point of 75 C. Calculate the vapour pressure in torr of the compound at 40 C. Assume that the enthalpy of sublimation is constant.
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The answer is 502 torr but I am not sure why. I got 1150 torr for some reason.
I used the following equations to get P(final):
Delta S = Delta H / T
ln(P(final)/P(initial) = Delta H / R * (1/T(final) - 1/T(initial))
I also used the following information: R = 8.314 J/Kmol
1atm = 760 Torr
I would really appreciate it if you could tell me what I am doing wrong.
Thank you very much.