Answer:
The vapor pressure of X at [tex]- 89^{\circ}[/tex] = 0.12 atm
Given:
[tex]\Delta H_{v} = 16.0 kJ mol[/tex]
Normal boiling point, T = [tex]- 43^{\circ}C = 230 K[/tex]
T' = [tex]- 89^{\circ}C = 184 K[/tex]
Solution:
At boiling point, vapor pressure = atmospheric pressure
Thus at T, P = 1 atm
Now, to calculate vapor pressure, P' at T' = 184 K, we use:
[tex]log_{10}\frac{P'}{P} = \frac{\Delta H_{v}}{R2.303}\frac{T' - T}{T'T}[/tex]
where
R = Rydberg's constant = 8.314 J/mol.K
Putting the values in the above formula:
[tex]log_{10}\frac{P'}{1} = \frac{16.0\times 1000}{8.314\times 2.303}\frac{184 - 230}{184\times 230}[/tex]
[tex]log_{10}{P'}= - 0.9083[/tex]
[tex]P'= 10^{- 0.9083} = 0.12 atm[/tex]
