Question

The enthalpy of vaporization of Substance X is 16.0kJ mol and its normal boiling point is −43.°C. Calculate the vapor pressure of X at −89.°C. Round your answer to 2 significant digits.

Answer 1

Answer:

The vapor pressure of X at $- 89^{\circ}$ = 0.12 atm

Given:

$\Delta H_{v} = 16.0 kJ mol$

Normal boiling point, T = $- 43^{\circ}C = 230 K$

T' = $- 89^{\circ}C = 184 K$

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:

$log_{10}\frac{P'}{P} = \frac{\Delta H_{v}}{R2.303}\frac{T' - T}{T'T}$

where

R = Rydberg's constant = 8.314 J/mol.K

Putting the values in the above formula:

$log_{10}\frac{P'}{1} = \frac{16.0\times 1000}{8.314\times 2.303}\frac{184 - 230}{184\times 230}$

$log_{10}{P'}= - 0.9083$

$P'= 10^{- 0.9083} = 0.12 atm$