Answer:
The vapor pressure at temperature 363 K is 0.6970 atm
The vapor pressure at 383 K is 1.410 atm
Explanation:
To calculate
of the reaction, we use clausius claypron equation, which is:
![\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%3D%5Cfrac%7B%5CDelta%20H_%7Bvap%7D%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= vapor pressure at temperature 
= vapor pressure at temperature 
= Enthalpy of vaporization
R = Gas constant = 8.314 J/mol K
1) 
= initial temperature =363 K
= final temperature =373 K

Putting values in above equation, we get:
![\ln(\frac{1 atm}{P_1})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{363}-\frac{1}{373}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7B1%20atm%7D%7BP_1%7D%29%3D%5Cfrac%7B40680%20J%2Fmol%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B363%7D-%5Cfrac%7B1%7D%7B373%7D%5D)

The vapor pressure at temperature 363 K is 0.6970 atm
2) 
= initial temperature =373 K
= final temperature =383 K

Putting values in above equation, we get:
![\ln(\frac{P_2}{1 atm})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{383}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7B1%20atm%7D%29%3D%5Cfrac%7B40680%20J%2Fmol%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B373%7D-%5Cfrac%7B1%7D%7B383%7D%5D)

The vapor pressure at 383 K is 1.410 atm