Question

The vapor pressure of water at 1.0 atm is 373 K and the heat of vaporization (ΔHvap) is 47.0 kJ/mol.

Estimate the vapor pressure at 363K.

Answer 1

\[\ln (\frac{ P2 }{ 1 })=-(\frac{ 47.0 }{ 8.3145 }(\frac{ 1 }{ 363 }-\frac{ 1 }{ 373 })\]

Answer 2

I got 0.99 atm

Answer 3

ohhhh gotcha! thank you so much for helping me out! :)

Answer 4

Actually nic2131, I believe the equation supposed to be:\[\ln \left( \frac{ P_2 }{ P_1 } \right) = \frac{\Delta H}{R} \left( \frac{1}{T_1} - \frac{1}{T_2}\right) \] Or:

\[\ln \left( \frac{ P_1 }{ P_2 } \right) = \frac{\Delta H}{R} \left( \frac{1}{T_2} - \frac{1}{T_1}\right) \] Actually, now that I think about it, does your form of the equation account for the negative sign on the right?

Answer 5

Actually, I think I may have just answered that on my own. :)