Question

Pressure over ideal binary liquid mixture containing 10 moles each of liquid A and B is gradually decreased isothermally. If $P_A^{\circ }=200mmHg$ and $P_B^{\circ }=100mmHg$, the pressure at which half of the liquid is converted into vapour is:

• A. $150$ mm Hg
• B. $166.5$ mm Hg
• C. $133$ mm Hg
• D. $141.4$ mm Hg

Answer 1

The correct option is D $141.4$ mm Hg

Let $P$ be the total pressure. Let $x$ be the number of moles of $A$ in the liquid state.

The number of moles of the vapour state will be $10-x$.

Let $y$ be the number of moles of $B$ in the liquid state

The number of moles of $B$ in vapour state will be $10-y$.
Hence, $x+y=10$
$P\times \frac{(10-x)}{(10-x+10-y)}=200\times \frac{\left(x\right)}{(x+y)}$
$P\times \frac{(10-y)}{(10-x+10-y)}=\frac{100\times \left(y\right)}{(x+y)}$
These equations are solved for $x,y$ and $P$.

$x=4.142$

$y=5.858$

$P=141.4mmHg$