Asked by Josh
A certain substance has a heat of vaporation 56.21 kj/mol. At what kelvin temperature will the vapor pressure be 6.50 times higher than it was at 361k?
I need help in setting the problem:
I got 56.21/8.3145j/mol.k(1/T1-1/T2)?
Answers
Answered by
DrBob222
Did you substitute into the Clausius-Clapeyron equation?
ln (p2/p1) = delta Hvap/R(1/T1 - 1/T2)
I would make up a number, like 1 for the vp at 361K. Then ln(6.5/1) = 56,210J/R(1/361 - 1/T2)
delta H vap must be in J/mol and not kJ/mol. You could have substituted p for p1 and 6.5p for p2 and you end up with 6.5/1 just as above.Solve for T2. R is 8.314
ln (p2/p1) = delta Hvap/R(1/T1 - 1/T2)
I would make up a number, like 1 for the vp at 361K. Then ln(6.5/1) = 56,210J/R(1/361 - 1/T2)
delta H vap must be in J/mol and not kJ/mol. You could have substituted p for p1 and 6.5p for p2 and you end up with 6.5/1 just as above.Solve for T2. R is 8.314
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