
How do you find vapor pressure given boiling point and heat of vaporization?
Answer
492.6k+ views
Hint: We need to understand the relationship between vapor pressure, boiling point and heat of vaporization. This relationship is given by the Clausius-Clapeyron equation. The relationship between the temperature of a liquid and its vapour pressure is not linear. For example, in some systems the vapour pressure increases significantly more rapidly than the temperature of the system. This is explained by the Clausius-Clapeyron equation which will help us find vapor pressure given boiling point and heat of vaporization.
Complete step by step answer:
As we know that the Clausius-Clapeyron equation was brought into picture by experiments which proved that Vapor pressure \[P\] and temperature $ T $ are related. If $ {P_1} $ and $ {P_2} $ are the vapor pressures at two different temperatures $ {T_1} $ and $ {T_2} $ respectively, then the Clausius-Clapeyron equation is:
$ \ln \left( {\dfrac{{{P_1}}}{{{P_2}}}} \right) = - \dfrac{{\Delta {H_{vap}}}}{R}\left( {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right) $
Where,
$ \Delta {H_{vap}} $ = heat of vaporization and $ R $ = gas constant whose value is $ 8.3145Jmo{l^{ - 1}}{K^{ - 1}} $ .
This equation will be able to calculate the vapor pressure at one temperature provided that the vapor pressure at another temperature and the heat of vaporization is known. The trick to this question is the understanding of the concept of boiling point and relating it to vapor pressure. The definition of boiling point says that it is the temperature at which the vapor pressure is equal to the atmospheric pressure.
Note: It must be noted that the unit of $ \Delta {H_{vap}} $ should be in $ J/mol $ since the unit of $ R $ is $ Jmo{l^{ - 1}}{K^{ - 1}} $ and the units of temperatures should be in kelvin. If the temperatures are given in the Celsius scale, adding $ 273 $ to the value converts it into the Kelvin scale. Although the values of $ {P_1} $ , $ {P_2} $ and $ {T_1} $ , $ {T_2} $ are taken arbitrarily, the value of $ {P_1} $ should correspond to $ {T_1} $ and $ {P_2} $ should correspond to $ {T_2} $ .
Complete step by step answer:
As we know that the Clausius-Clapeyron equation was brought into picture by experiments which proved that Vapor pressure \[P\] and temperature $ T $ are related. If $ {P_1} $ and $ {P_2} $ are the vapor pressures at two different temperatures $ {T_1} $ and $ {T_2} $ respectively, then the Clausius-Clapeyron equation is:
$ \ln \left( {\dfrac{{{P_1}}}{{{P_2}}}} \right) = - \dfrac{{\Delta {H_{vap}}}}{R}\left( {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right) $
Where,
$ \Delta {H_{vap}} $ = heat of vaporization and $ R $ = gas constant whose value is $ 8.3145Jmo{l^{ - 1}}{K^{ - 1}} $ .
This equation will be able to calculate the vapor pressure at one temperature provided that the vapor pressure at another temperature and the heat of vaporization is known. The trick to this question is the understanding of the concept of boiling point and relating it to vapor pressure. The definition of boiling point says that it is the temperature at which the vapor pressure is equal to the atmospheric pressure.
Note: It must be noted that the unit of $ \Delta {H_{vap}} $ should be in $ J/mol $ since the unit of $ R $ is $ Jmo{l^{ - 1}}{K^{ - 1}} $ and the units of temperatures should be in kelvin. If the temperatures are given in the Celsius scale, adding $ 273 $ to the value converts it into the Kelvin scale. Although the values of $ {P_1} $ , $ {P_2} $ and $ {T_1} $ , $ {T_2} $ are taken arbitrarily, the value of $ {P_1} $ should correspond to $ {T_1} $ and $ {P_2} $ should correspond to $ {T_2} $ .
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