Question

The vapor pressure of benzene between 10°C and 30°C fits the equation 1780 log(p/Torr) - 7.960 - T/K Calculate (a) the enthalpy of vaporization and (b) the normal boiling point of benzene.

Answer 1

(a)

The given equation is $log_{10}\left ( P/Torr \right )=7.960-\frac{1780}{T/K}$

The clausius clapeyron equation is

Η logio (P/Torr) k - 2.303R(T/K)

Here, P is the vapor pressure, k is integration constant, R is ideal gas constant, T is absolute temperature and $\Delta H$ is enthalpy of vaporisation.

Comparing above two equations, $\frac{\Delta H}{2.303R}=1780$

$\Delta H=1780 \times 2.303 \times R$

$\Delta H=1780 \ K \times 2.303 \times 8.314 \ \frac{J}{mol \cdot K}$

$\Delta H=34082 \ \frac{J}{mol }$

Hence, the enthalpy of vaporization $\Delta H=34082 \ \frac{J}{mol }$

(b)

The given equation is $log_{10}\left ( P/Torr \right )=7.960-\frac{1780}{T/K}$

At normal boiling point, the vapor pressure is equal to atmospheric pressure which is 1 atm or 760 torr.

$log_{10}\left ( 760 \ Torr/Torr \right )=7.960-\frac{1780}{T/K}$

$log_{10}760=7.960-\frac{1780}{T/K}$

$2.88=7.960-\frac{1780}{T/K}$

$\frac{1780}{T/K}=7.960- 2.88$

$\frac{1780}{T/K}=5.079$

$T/K=\frac{1780}{ 5.079}$

Hence, the normal boiling point of benzene is 350.5 K.