In: Chemistry
A certain substance has a heat vaporization of 36.63kJ/mol. At which Kelvin temperature will the vapor pressure be 5.50 times higher than it was at 363K?
Hvap = 36.63Kj/mole = 36630J/mole
T1 = 363K
P1 = P
P2 = 5.5P
log(P2/P1) = Hvap/2.303R [1/T1 -1/T2]
log(5.5P/1P) = 36630/2.303*8.314 [ 1/363 -1/T2]
log(5.5/1) = 1913.08(0.00275-1/T2)
0.7403/1913.08 = (0.00275-1/T2)
0.000387 = 0.00275-1/T2
0.000387-0.00275 = -1/T2
-0.002363 = -1/T2
T2 = 1/0.002363 423.19K >>>>>answer