Question

Find the vapour pressure of the solution when $\text{ }{{\dfrac{3}{4}}^{\text{th}}}\text{ }$ of its liquid is vaporized.(Given : $\text{ P}_{\text{A}}^{\text{0}}\text{ = 200 mm of Hg }$,$\text{ P}_{\text{B}}^{\text{0}}\text{ = 300 mm of Hg }$and A = 1 mole and B = 1 mole)

Answer 1

Hint: The partial pressure of any volatile component of a solution at any temperature is equal to the vapour pressure of the pure component multiplied by the mole fraction of the component. According to Dalton’s law, Raoult's law for the binary solution is written as, \[\text{ P = }{{\text{X}}_{\text{A}}}\text{P}_{\text{A}}^{\text{0}}\text{ + }{{\text{X}}_{\text{B}}}\text{P}_{\text{B}}^{\text{0}}\,\]
Where \[{{\text{X}}_{\text{A}}}\] and \[{{\text{X}}_{\text{B }}}\] are mole fractions of components A and B.

Complete Solution :
We are provided with the following data:
The partial vapour pressure of component A is $\text{ P}_{\text{A}}^{\text{0}}\text{ = 200 mm of Hg }$
The number of moles of component A is 1 mole.
The partial vapour pressure of component B is $\text{ P}_{\text{B}}^{\text{0}}\text{ = 300 mm of Hg }$
The number of moles of component B is 1 mole.
We are interested to determine the vapour pressure of the solution when $\text{ }{{\dfrac{3}{4}}^{\text{th}}}\text{ }$ of the liquid is vaporized.

- Let's calculate the mole fraction of component A.mole fraction is a ratio of the number of moles of a component to the total number of the component present in the solution. Thus mole fraction for component A is,
$\text{ }{{\text{X}}_{\text{A}}}\text{ =}\dfrac{{{\text{n}}_{\text{A}}}}{{{\text{n}}_{\text{A}}}\text{+}{{\text{n}}_{\text{B}}}}\text{ = }\dfrac{\text{1}}{\text{1+1}}\text{ = 0}\text{.5 }$
The mole fraction of component B is,
$\text{ }{{\text{X}}_{\text{B}}}\text{ =}\dfrac{{{\text{n}}_{\text{B}}}}{{{\text{n}}_{\text{A}}}\text{+}{{\text{n}}_{\text{B}}}}\text{ = }\dfrac{\text{1}}{\text{1+1}}\text{ = 0}\text{.5 }$

- According to Raoult's law, the total vapour pressure P of the solution is given by,
$\text{ P = }{{\text{p}}_{\text{A}}}\text{ + }{{\text{p}}_{\text{B }}}\text{= }{{\text{X}}_{\text{A}}}\text{P}_{\text{A}}^{\text{0}}\text{ + }{{\text{X}}_{\text{B}}}\text{P}_{\text{B}}^{\text{0}}\,$
Then let's substitute the values of the partial pressure of component A, B, and their mole fraction .we have,
$\begin{align}
  & \text{ P = }\left( 0.5 \right)\times \left( 200 \right)\text{ + }\left( 0.5 \right)\times \left( 300 \right)\, \\
 & \Rightarrow \text{ P = 250 mm of Hg } \\
\end{align}$

- Thus vapour pressure solution is 250 mm of Hg. Now we have given that $\text{ }{{\dfrac{3}{4}}^{\text{th}}}\text{ }$of it liquid is vaporized .then volume of the solution remains behind is$\text{ }\dfrac{1}{4}\text{ }$. let the mole's solution decreases from 2 moles to $\text{ 0}\text{.5 }$ . Thus there is an increase in the vapour pressure of the solution. the increase in vapour pressure .The pressure on the solution will increase the $\text{ }{{\dfrac{3}{4}}^{\text{th}}}\text{ }$ of the original vapour pressure of the solution. Thus the vapour pressure on the solution when its liquid is vaporized is equal to,
$\text{ }{{\text{P}}_{\text{T}}}\text{ = P + P }\times \dfrac{3}{4}\text{ = 250 + 250}\times \dfrac{3}{4}\text{ = 437}\text{.5 mm of Hg }$

Thus, final vapour pressure increases to $\text{437}\text{.5}$ mm of Hg.

Note: Note that vaporization is a process by which the molecules escape from the liquid state and exerts pressure on the liquid. Thus final vapour pressure is more compared to the vapour pressure of the solution.