Question

A(l) ^{​​​​​​}​​​​​​<======>A(g) ∆_vapH=460.6 Cal mol^{​​​​-1} at 1atm, boiling point=50K. What is the boiling point at 10atm?
a)150K. b)75K
c)100K. d)None is correct

Answer 1

Given ∆Hvap = 860.6 cal/mol = 1927.15 Joules
R is gas constant = 8.314J/molK
P1 = 1 atm
P2 = 10 atm
T1 = 50 K
T2 = T2

We will be using
Clausius Clapeyron equation
ln(P2/P1) = ∆Hvap/R[ 1/T1-1/T2]
ln(10/1) = ^​​​1927.15/8.314[1/50-1/T2]
T2 = 99.34 K