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((P_(1) )/(P_(2) )) = (delta H_(vap) )/(R) ((1)/(T_(2) )-(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((P_(1) )/(P_(2) )) = (delta H_(vap) )/(R) ((1)/(T_(2) )-(1)/(T_(1) )   )`

ln
`(760 torr)/(638.43 torr) =(38210 J/mol)/(8.314 J/(mol.K)) ((1)/(383 K)-(1)/(T_(1) )`

`T_(1) = 389 K`

Therefore, the boiling point of octane = 389 K - 273 =
`116^(0)C`