Question

What is the boiling point of methanol at $ 25mmHg $ if its normal boiling point is $ {64.70^ \circ }C $ and $ {\Delta _{vap}}H = 35.21kJmo{l^{ - 1}} $ ?

Answer 1

Hint: The normal boiling point of any liquid is the temperature at which the vapour pressure associated with the liquid becomes equal to the standard sea-level pressure of the atmosphere. The equation in thermodynamics that deals with pressure and temperature changes is the Clausius-Clapeyron equation.

Complete answer:
The boiling point is defined as the temperature at which a liquid attains a vapour pressure exactly equal to that of the atmosphere around it. Thus, when the pressure of the surroundings changes, the boiling point also changes. A lower pressure of the surroundings would mean that the liquid quickly achieves that vapour pressure at lower temperatures.
The enthalpy of vaporization is the heat involved in the boiling process. The value of enthalpy of vapourization plays an active role in determining the temperature at which the liquid boils at a pressure different from the normal atmospheric pressure.
The normal sea-level atmospheric pressure is $ 760mmHg $ which is very high as compared to the given pressure.
The relationship between pressure changes and associated temperature changes that include the enthalpies of transformation is given by the Clausius-Clapeyron equation. The integrated form of this equation can be written as follows:
 $ \ln \left( {\dfrac{{{P_2}}}{{{P_1}}}} \right) = \dfrac{{{\Delta _{vap}}H}}{R}\left( {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right) $
The equation represents the ratio of the atmospheric and the given pressure. Thus, $ {T_1} $ is the unknown boiling point that needs to be determined.
The temperature needs to be written in Kelvin scale and the enthalpy has to be converted in joules.
 $ {T_2} = {64.70^ \circ }C + 273.15 = 337.85K $
 $ {\Delta _{vap}}H = 35.21kJmo{l^{ - 1}} = 35210Jmo{l^{ - 1}} $
Inserting the values of pressures, normal boiling point and enthalpy of vapourization in the above equation we get,
 $ \ln \left( {\dfrac{{760mmHg}}{{25mmHg}}} \right) = \dfrac{{35210Jmo{l^{ - 1}}}}{{8.314J{K^{ - 1}}mo{l^{ - 1}}}}\left( {\dfrac{1}{{{T_1}}} - \dfrac{1}{{337.85K}}} \right) $
On solving for $ {T_1} $ we get :
 $ {T_1} = 265.5K = - {7.6^ \circ }C $
 $ \Rightarrow $ Hence, the boiling point of methanol at $ 25mmHg $ is $ - {7.6^ \circ }C $

Note:
The boiling point of methanol under normal conditions of pressure is already very low as it is a very volatile organic liquid and on lowering the pressure to such an extent, it actually becomes possible to boil methanol at negative temperatures.