Question

Two compounds A and B in when mixed form nearly ideal solutions. At room temperature, $P_A^0=40mmHg$ and $P_B^0=120mmHg$.
A liquid mixture is composed of $1mol$ of A and $3mol$ of B.
When the first trace of vapour is formed, choose the correct option(s) regarding composition of A and B in vapour phase :
Here,
$y_A=\text{Composition of A in vapour phase when first trace of vapour is formed}$
$y_B=\text{Composition of B in vapour phase when first trace of vapour is formed}$

• A. $y_A=0.1$
• B. $y_A=0.9$
• C. $y_B=0.1$
• D. $y_B=0.9$

Answer 1

Answer: (A), (D)

$y_B=0.9$

Initially
$x_A=\frac{1}{1+3}=0.25x_B=1-0.25=0.75$
Given:
$P_A^0=40mmHg{P}_B^0=120mmHg$
The pressure at which first vapour is formed is the bubble point pressure.
At bubble point:
When the first vapour is formed, the composition of A and B in the liquid mixture will be the same.
So, at bubble point
$x_A=0.25$
$x_B=0.75$
$P=x_A{P}_A^0+x_B{P}_B^{0}P=0.25\times 40+0.75\times 120P=100mmHg$

Calculating the vapour phase compositions at this bubble point pressure (P) :
$y_A=\frac{x_A{P}_A^0}{P}=\frac{0.25\times 40}{100}{y}_A=0.1$
$y_B=\frac{x_B{P}_B^0}{P}=\frac{0.75\times 120}{100}{y}_B=0.9$