Question

The vapour pressure of two pure liquids A and B, that form an ideal solution are $100$ and $900\text{torr}$ respectively at temperature T. This liquid solution of A and B is composed of $1$ mole of A and $1$ mole of B. What will be the pressure, when $1$ mole of mixture has been vaporized?

• A. $800\text{torr}$
• B. $500\text{torr}$
• C. $300\text{torr}$
• D. $200\text{torr}$

Answer 1

The correct option is C $300\text{torr}$

Number of moles of $A$ in vapour phase = $n_A$
Number of moles of $B$ in vapour phase = $n_B$
Mole fraction of $A$ in vapour phase =$y_A$
Mole fraction of $B$ in vapour phase =$y_B=\frac{n_B}{n_A+n_B}=\frac{n_B}{1}$
Mole fraction of $B$ in liquid phase =$x_B=\frac{1-n_B}{n_A+n_B}=\frac{1-n_B}{1}$
For liquid A
$P_A^o{n}_B=P_T(1-n_B)=P_A...\left(i\right)$
Where, $P_A^o$ is the vapour pressure of pure liquid $A$ and $P_T$ is the total pressure of the liquid
Similarly for liquid B
$P_B^o(1-n_B)=P_T{n}_B=P_B...\left(ii\right)$
On dividing (i) and (ii), we get
$\frac{P_A^o}{P_B^o}=\frac{(1-n_B{)}^2}{n_B^2}$
On substituting the values, we get
$\frac{100}{900}=\frac{(1-n_B{)}^2}{n_B^2}$
$\frac{1}{3}=\frac{(1-n_B)}{n_B}$
$n_B=3-3n_B$
$-3=-3n_B-n_{B}3=4n_B\frac{3}{4}=n_B$
On substituting the value of $n_B$ in equation (i), we get
$100\times \frac{3}{4}=P_T(1-\frac{3}{4})$
$25\times 3=P_T\left(0.25\right)$
$75=0.25P_T$
$\frac{75}{0.25}=P_T\Rightarrow P_T=300\text{torr}$