Question

How do you calculate the vapour pressure of ethanol?

Answer 1

Hint: The answer is based on the physical chemistry concept which tells about the calculation of vapour pressure based on the Clausius – Clapeyron equation which is given by $\ln \left( \dfrac{{{P}_{2}}}{{{P}_{1}}} \right)=\dfrac{\Delta {{H}_{vap}}}{R}\left( \dfrac{1}{{{T}_{1}}}-\dfrac{1}{{{T}_{2}}} \right)$

Complete answer:
> The concept of the physical chemistry which deals with the concept of determination of the vapour pressure, boiling point and several other factors by using various equations derived is familiar to us.
We shall solve the above given problem by using the Clausius – Clapyeron equation.
- Clausius – Clapyeron equation relates vapour pressure P, enthalpy of vaporisation $\Delta {{H}_{vap}}$ and tt two different temperatures and pressures by its equation that is given by, $\ln \left( \dfrac{{{P}_{2}}}{{{P}_{1}}} \right)=\dfrac{\Delta {{H}_{vap}}}{R}\left( \dfrac{1}{{{T}_{1}}}-\dfrac{1}{{{T}_{2}}} \right)$
- This equation pertains to the relationship between the pressure and temperature for conditions of equilibrium between two phases.
Now for ethanol, the vapour pressure can be calculated by taking a simple problem as an instance.
- We can assume the data which is given in the form of a question that has the contents as the heat of vaporisation of ethanol is $38.56kJmo{{l}^{-1}}$ and has a normal boiling point of ${{78.4}^{0}}C$ and then calculate the vapour pressure of ethanol at ${{50}^{0}}C$.
The answer would be approached as described below,
${{T}_{1}}=50+273=323K$ and ${{P}_{1}}=x$
\[{{T}_{2}}=78.4+73=351.4K\] and ${{P}_{2}}=760torr$
By substituting these values in the Clausius-Clapeyron equation, we have
$\ln \left( \dfrac{760}{x} \right)=\dfrac{38560Jmo{{l}^{-1}}}{8.314J{{K}^{-1}}mo{{l}^{-1}}}\left( \dfrac{1}{323}-\dfrac{1}{351.4} \right)$
\[\Rightarrow \ln \left( \dfrac{760}{x} \right)=4638\times 2.5\times {{10}^{-4}}=1.159\]
\[\Rightarrow \dfrac{760}{x}={{e}^{1.159}}=3.188\]
\[\Rightarrow x=\dfrac{760}{3.188}=238.3torr\]
Thus, the vapour pressure of ethanol will be \[238.3torr\]

Note: Note that vapour pressure of a solution can also be calculated using the formula which relates the vapour pressure of pure solvent and mole fraction of the solvent which is given by the formula, ${{P}_{solution}}={{\chi }_{solvent}}{{P}_{solvent}}$