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 pressure is reduced further at room temperature, at what pressure does the last trace of liquid disappear?

• A. $40mmHg$
• B. $80mmHg$
• C. $120mmHg$
• D. $160mmHg$

Answer 1

The correct option is B $80mmHg$

The pressure at which last trace of liquid disappears is the dew point pressure.
Initially
$x_A=\frac{1}{1+3}=0.25x_B=1-0.25=0.75$
But when all the liquid converts to vapour then vapour phase composition is :
$y_A=0.25y_B=0.75$
Since,

$x_A=\frac{y_A\times P_T}{p_A^{\circ }}$

$x_B=\frac{y_B\times P_T}{p_B^{\circ }}$

Also $x_A+x_B=1$

So, $\frac{1}{P_T}=\frac{y_A}{p_A^{\circ }}+\frac{y_B}{p_B^{\circ }}$
Putting the values :
$\frac{1}{P}=\frac{0.25}{40}+\frac{0.75}{120}\frac{1}{P}=\frac{1}{80}P=80mmHg$
So, the dew point pressure i.e. the pressure at which all the liquid disappears is $80mmHg$