Question

How do you calculate the partial pressure of water?

Answer 1

Hint: Partial pressure of water in a mixture is defined as the equilibrium vapour pressure of water at a specified temperature. It is denoted using ${P_{water}}$ . It changes with temperature.

Complete step by step answer:
If we take a gaseous mixture, there will be many compounds. The partial pressure of water is the pressure due to water vapour which is in equilibrium with the solid or liquid water. It is also called vapour pressure. The vapour pressure of water changes with temperature. Due to high surface tension of water, vapour pressure of water has a low value.
If we take water in a closed container, there will be movement of water molecules between liquid phase and vapour phase. At this condition water vapour remains constant.
There are many equations which helps us to calculate the vapour pressure of water. Some of them are,
A.Clausius-Clapeyron equation
$\ln \left( {\dfrac{{{P_2}}}{{{P_1}}}} \right) = - \dfrac{L}{R}\left( {\dfrac{1}{{{T_2}}} - \dfrac{1}{{{T_1}}}} \right)$
Where ${P_1}$ and ${P_2}$ are pressures at temperature ${T_1}$ and ${T_2}$ respectively. R is gas constant and L is specific latent heat of substance.
B.The Buck equation
$P = 0.61121\exp \left( {\left( {18.678 - \dfrac{T}{{234.5}}} \right)\left( {\dfrac{T}{{257.14 + T}}} \right)} \right)$
Where T is the temperature in degree Celsius and P is the pressure in kPa.
C.Tenens equation
$P = 0.61078\exp \left( {\dfrac{{17.27T}}{{T + 237.3}}} \right)$
Where T is the temperature in degree Celsius and P is the pressure in kPa.
D.Antoine equation
${\log _{10}}P = A - \dfrac{B}{{C + T}}$
Where T is the temperature in degree Celsius and P is the pressure in mmHg. A, B and C are some constants.

Note:
According to Dalton’s law of partial pressure, when two or more non-reacting gases are taken in a container at constant temperature, the total pressure exerted by the mixture will be the sum of partial pressures of individual gases.