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As Enthalpy means 鈥渉eat鈥� so enthalpy of the heat of vaporization can be defined as the heat which is a form of energy required to change the liquid phase into the vapour phase.

$\mathbf{Log}\frac{P1}{P2}\mathbf{=}\frac{\mathbf{-}{{\mathbf{螖贬}}^o}_{\mathrm{vap}}}{\mathbf{2}\mathbf{.}\mathbf{303}\mathbf{脳}\mathbf{R}}\mathbf{(}\frac{\mathbf{1}}{\mathbf{T}\mathbf{1}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{T}\mathbf{2}}\mathbf{)}$

Where

${\mathrm{P}}_1$= Vapour pressure at temperature${\mathrm{T}}_1$

${\mathrm{P}}_2$= Vapour pressure at temperature${\mathrm{T}}_2$

R = Gas constant at a universal level.

The relationship between pressure and temperature is directly proportional, if the temperature increases the pressure also increases and vice-versa.

Temperature,${\mathbf{T}}_1\mathbf{=}{\mathbf{25}}^o\mathbf{C}\mathbf{=}\mathbf{298}\mathbf{k}$

Vapour pressure at temperature,${\mathbf{P}}_1\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{703}\mathbf{atm}$

Temperature,${\mathbf{T}}_2\mathbf{=}{\mathbf{95}}^o\mathbf{C}\mathbf{=}\mathbf{368}\mathbf{k}$

Let vapour pressure at temperature,${\mathbf{P}}_2\mathbf{=}\mathbf{鈥�}\mathbf{x}\mathbf{鈥�}\mathbf{atm}$

Universal gas constant, $\mathbf{R}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{314}\mathbf{J}\mathbf{/}\mathbf{mol}\mathbf{/}\mathbf{k}$.

Enthalpy of vaporization, ${{\mathbf{螖贬}}^o}_{\mathrm{vap}}\mathbf{=}\mathbf{29}\mathbf{.}\mathbf{1}\mathbf{kJ}\mathbf{/}\mathbf{mole}$

Put the values in the above equation of enthalpy of vaporization:

role="math" localid="1658848549221" $\begin{array}{c}\mathbf{Log}\mathbf{\left(}\frac{\mathbf{P}\mathbf{1}}{\mathbf{P}\mathbf{2}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{-}{{\mathbf{螖贬}}^o}_{\mathrm{vap}}}{\mathbf{2}\mathbf{.}\mathbf{303}\mathbf{脳}\mathbf{R}}\mathbf{(}\frac{\mathbf{1}}{\mathbf{T}\mathbf{1}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{T}\mathbf{2}}\mathbf{)}\\ \mathbf{Log}\mathbf{\left(}\frac{\mathbf{0}\mathbf{.}\mathbf{703}}{\mathbf{X}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{-}\mathbf{29}\mathbf{.}\mathbf{1}\mathbf{KJ}\mathbf{/}\mathbf{mole}}{\mathbf{2}\mathbf{.}\mathbf{303}\mathbf{脳}\mathbf{8}\mathbf{.}\mathbf{314}\mathbf{J}\mathbf{/}\mathbf{mole}\mathbf{/}\mathbf{k}}\mathbf{(}\frac{\mathbf{1}}{\mathbf{298}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{366}}\mathbf{)}\\ \mathbf{Log}\mathbf{\left(}\frac{\mathbf{0}\mathbf{.}\mathbf{703}}{\mathbf{X}}\mathbf{\right)}\mathbf{=}\frac{-29100J/\mathrm{mole}}{\mathbf{2}\mathbf{.}\mathbf{303}\mathbf{脳}\mathbf{8}\mathbf{.}\mathbf{314}\mathbf{J}\mathbf{/}\mathbf{mole}\mathbf{/}\mathbf{k}}\mathbf{\left(}\frac{\mathbf{368}\mathbf{-}\mathbf{298}}{\mathbf{298}\mathbf{脳}\mathbf{368}}\mathbf{\right)}\\ \mathbf{Log}\mathbf{\left(}\frac{\mathbf{0}\mathbf{.}\mathbf{703}}{\mathbf{X}}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{3500}\mathbf{.}\mathbf{12}\mathbf{\left(}\frac{\mathbf{70}}{\mathbf{2}\mathbf{.}\mathbf{303}\mathbf{脳}\mathbf{109664}}\mathbf{\right)}\end{array}$

role="math" localid="1658848620605" $\begin{array}{c}\mathbf{Log}\mathbf{\left(}\frac{\mathbf{0}\mathbf{.}\mathbf{703}}{\mathbf{X}}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{97}\\ \mathbf{\left(}\frac{\mathbf{0}\mathbf{.}\mathbf{703}}{\mathbf{X}}\mathbf{\right)}\mathbf{=}\mathbf{Antilog}(\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{97})\\ \mathbf{\left(}\frac{\mathbf{0}\mathbf{.}\mathbf{703}}{\mathbf{X}}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{11}\\ \mathbf{x}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{703}}{\mathbf{0}\mathbf{.}\mathbf{11}}\\ \mathbf{x}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{4}\mathbf{atm}\end{array}$

The Vapour pressure at temperature $95掳C$ is 6.4atm.