Question

The normal boiling of water is $ 373K $ (at $ 760mm $ ). Vapour pressure of water at $ 298K $ is $ 23mm. $ If enthalpy of vaporization is $ 40.656kJmo{{l}^{-1}} $ the boiling point of water at $ 23mm $ atmospheric pressure will be:
(A) $ 250K $
(B) $ 294K $
(C) $ 51.6K $
(D) $ 12.5K $

Answer 1

Hint: We know that the melting point is the temperature at which the solid and the liquid forms of a pure substance can exist in equilibrium, that is, they are equal. If heat is applied to any solid, its temperature will rise until it hits its melting point. More heat will convert the solid into the liquid. At this point, temperature will remain unchanged.

Complete step by step solution:
The term boiling point can be defined as the temperature at which the external pressure exerted on a liquid is equal to the pressure exerted by the vapour of the liquid. The boiling point of water at normal atmospheric pressure is $ 212{}^\circ F $ . Addition of the heat further results in the transformation of the liquid into its vapour.
 At this point, the temperature does not increase. At any given temperature, some part of the liquid vaporizes into the space above it until the pressure exerted by the vapour reaches a certain value. This characteristic value is termed as the vapour pressure of the liquid at the given temperature. If the temperature is raised, the vapour pressure increases consequently. At the boiling point, the bubbles of the vapour form within the liquid.
They then rise to the surface of the liquid. The boiling point of a liquid depends upon the pressure applied. At sea level, the water boils at a temperature of $ 100{}^\circ C $ , that is, $ 212{}^\circ F $ . If the height is increased, the temperature of the boiling point reduces. Therefore, the boiling point of water at normal atmospheric pressure is $ 100{}^\circ C $
But by clausius clapeyron equation:
 $ \ln \dfrac{{{P}_{2}}}{{{P}_{1}}}=\dfrac{\Delta H}{R}\left[ \dfrac{1}{{{T}_{1}}}-\dfrac{1}{{{T}_{2}}} \right] $ Substituting the values we get the value for $ {{T}_{1}} $ i.e. boiling point temperature,
  $ \ln \dfrac{760}{23}=\dfrac{40656}{8.314}\left[ \dfrac{1}{{{T}_{1}}}-\dfrac{1}{373} \right] $
 $ \Rightarrow {{T}_{1}}=293.98\approx 294K $
Thus correct answer is option B i.e. $ 294K $ .

Note:
Students get confused between the melting point and the boiling point. They must know that the melting point is the temperature at which solid and liquid phases are at equilibrium. On the other hand, the boiling point is the temperature at which the vapour pressure of a liquid is equal to the external pressure.