Question

The normal boiling point of cyclohexane (enthalpy of vaporization: $30.08kJmo{l^{ - 1}}$ ) is ${80.75^ \circ }C$ . What will be its vapour pressure at ${25^ \circ }C$ ?

Answer 1

We can calculate the vapour pressure using the enthalpy of vapourization, temperature, and initial vapour pressure. We have to substitute the values of the enthalpy of vapourization, temperature, and initial vapour pressure in the expression below to get the vapour pressure at the given temperature.
Formula used: The formula to calculate the vapour pressure is,
$\log \dfrac{{{P_2}}}{{{P_1}}} = \dfrac{{{\Delta _{vap}}H}}{{2.303R}}\left( {\dfrac{1}{{{T_2}}} - \dfrac{1}{{{T_1}}}} \right)$
Here ${P_1}$ represents the initial vapour pressure
${P_2}$ represents the vapour pressure at the given temperature
${T_2}$ represents the final temperature
${T_1}$ represents the initial temperature
$\Delta {H_{vap}}$ represents the enthalpy of vapourization

Complete answer:
The given data contains,
The value of ${T_1}$ is ${80.75^ \circ }C$ .
The value of ${T_2}$ is ${25^ \circ }C$ .
The value of ${P_1}$ is $760mmHg$ .
The value of $\Delta {H_{vap}}$ is $30.08kJmo{l^{ - 1}}$ .
Let us now convert the degree Celsius into Kelvin by adding $273.15$ to the given temperature. The formula is written as,
$K{ = ^ \circ }C + 273.15$
Let us now substitute the given value of Celsius in the expression.
$K{ = ^ \circ }C + 273.15$
$K = {80.75^ \circ }C + 273.15$
On adding we get,
$\Rightarrow K = 353.9K$
The value of ${T_1}$ in Kelvin is $353.9K$ .
$K{ = ^ \circ }C + 273.15$
$\Rightarrow K = {25^ \circ }C + 273.15$
$\Rightarrow K = 298.15K$
The value of ${T_2}$ in Kelvin is $298.15K$.
Let us now use the expression to calculate the vapour pressure at the given temperature.
The formula to calculate the vapour pressure is,
$\log \dfrac{{{P_2}}}{{{P_1}}} = \dfrac{{{\Delta _{vap}}H}}{{2.303R}}\left( {\dfrac{1}{{{T_2}}} - \dfrac{1}{{{T_1}}}} \right)$
Here ${P_1}$ represents the initial vapour pressure
${P_2}$ represents the vapour pressure at the given temperature
${T_2}$ represents the final temperature
${T_1}$ represents the initial temperature
$\Delta {H_{vap}}$ represents the enthalpy of vapourization
Let us now substitute the values of enthalpy of vapourization, temperatures, and pressure to get the vapour pressure at ${25^ \circ }C$.
The vapour pressure at ${25^ \circ }C$ is calculated as,
$\log \dfrac{{{P_2}}}{{{P_1}}} = \dfrac{{{\Delta _{vap}}H}}{{2.303R}}\left( {\dfrac{1}{{{T_2}}} - \dfrac{1}{{{T_1}}}} \right)$
On substituting the known values we get,
$\Rightarrow \log \left( {\dfrac{{{P_2}}}{{760mmHg}}} \right) = \dfrac{{30.08 \times {{10}^3}Jmo{l^{ - 1}}}}{{2.303\left( {8.314J{K^{ - 1}}mo{l^{ - 1}}} \right)}}\left( {\dfrac{1}{{298.15K}} - \dfrac{1}{{353.9K}}} \right)$
We get,
$\log \left( {\dfrac{{{P_2}}}{{mmHg}}} \right) = \log \left( {760} \right) - \dfrac{{30.08 \times {{10}^{ - 3}}J\,mo{l^{ - 1}}}}{{\left( {2.303} \right)\left( {8.314J{K^{ - 1}}mo{l^{ - 1}}} \right)}}\left[ {\dfrac{1}{{298.15K}} - \dfrac{1}{{353.9K}}} \right]$
On simplification we get,
${P_2} = 112.4mmHg$
We have calculated the vapour pressure at ${25^ \circ }C$ is $112.4mmHg$.

Note:
Based on the values and given variables we have to calculate the vapour pressure. We can also calculate the vapour pressure of the solution from mole fraction of the solvent and vapour pressure of the pure solvent. While calculating the vapour pressure from enthalpy of vapourization, we have to remember to convert the degree Celsius to Kelvin for temperatures failing which the exact vapour pressure could not be obtained.