Question

Calculate the boiling point of water at $ 24torr $ pressure. The average $ \vartriangle {H_{vap}} $. Over temperature range is $ 10.12kcalmo{l^{ - 1}} $ . Will all the water go from gaseous state if placed in a closed container, if not then how can we convert all the water into vapor state?

Answer 1

Hint: The boiling point of a liquid is the temperature at which its vapor pressure is equivalent to the weight of the gas above it. The typical boiling point of a liquid is the temperature at which its vapor pressure is equivalent to one atmosphere.

Complete step by step solution:
The connection between the temperature of a liquid and its vapor pressure is certainly not a straight line. The vapor weight of water, for instance, increments altogether more quickly than the temperature of the system. This conduct can be clarified with the Clausius-Clapeyron equation.
Pressure $ = 24torr $
 $ \vartriangle {H_{vap}} = 10.12kcal/mol $
Using Clausius- Clapeyron formula:
In, $ \left( {\dfrac{{{P_2}}}{{{P_1}}}} \right) = \dfrac{{ - \vartriangle {H_{vap}}}}{R}\left( {\dfrac{1}{{{T_2}}} - \dfrac{1}{{{T_1}}}} \right) $
Where; $ {P_2} $ : vapor pressure at time $ {T_1} $
 $ {P_1} $ ; vapor pressure at time $ {T_2} $
 $ \vartriangle {H_{vap}} = $ enthalpy
 $ R: $ gas constant $ :8.314J/kmol $
Let's just assume at $ 1atm $ pressure. Vapor pressure of boiling point $ 760torr $
 $ \left( {\dfrac{{760}}{{24}}} \right) = \dfrac{{ - 10.12}}{{8.314}}\left( {\dfrac{1}{{{T_2}}} - \dfrac{1}{{273}}} \right) \\
   \Rightarrow 31.6 = - 1.21\left( {\dfrac{{273 - {T_2}}}{{273{T_2}}}} \right) \\
   \Rightarrow 31.6 = \dfrac{{ - 330.33 + 1.21{T_2}}}{{273{T_2}}} \\
   \Rightarrow 8626.8{T_2} - 1.21{T_2} = - 330.33 \\
   \Rightarrow {T_2} = \dfrac{{ - 330.33}}{{8626.59}} = - 0.03829 \\
$
Hence, the answer is $ - 0.03829 $ .

Note:
As the temperature of a liquid expands, the vapor pressure of the liquid increments until it approaches the external pressure, or the air pressure on account of an open container. Bubbles of vapor start to form all through the liquid, and the liquid starts to boil. The temperature at which a liquid boils at precisely $ 1atm $ Pressure is the ordinary boiling point of the liquid. For water, the normal liquid point is actually $ {100^ \circ }C $ .
This equation is exceptionally helpful because it gives the variation with temperature of the pressure at which water and steam are in equilibrium, that is the boiling temperature.