Question

Find the boiling point temperature at 760 torr of an isomer of octane, c8h18, if its enthalpy of vaporization is 38,210 j mol-1 and its vapor pressure at 110.0°c is 638.43 torr.

Answer 1

According to Clausius-Clayperon equation,

$ln(\frac{P_{1} }{P_{2} }) = \frac{delta H_{vap} }{R} (\frac{1}{T_{2} }-\frac{1}{T_{1} } )$

$P_{1}$is the vapor pressure at boiling point = 760 torr

$P_{2}$ is the vapor pressure at T_{2} =638.43 torr

Temperature$T_{2} = 110.0^{0}C + 273 = 383 K$

Δ$H_{vap} = 38210 J/mol$

Plugging in the values, we get

$ln(\frac{P_{1} }{P_{2} }) = \frac{delta H_{vap} }{R} (\frac{1}{T_{2} }-\frac{1}{T_{1} } )$

ln$\frac{760 torr}{638.43 torr} =\frac{38210 J/mol}{8.314 J/(mol.K)} (\frac{1}{383 K}-\frac{1}{T_{1} }$

$T_{1} = 389 K$

Therefore, the boiling point of octane = 389 K - 273 = $116^{0}C$