Question

The vapour pressure of two pure liquids $A$ and $B$ forming an ideal solution are 400 mm Hg and 800 mm Hg respectively at temperature $T$. A liquid mixture containing $3:1$ molar composition of $A$ and $B$ is present in a cylinder closed by a piston so that the pressure can be varied. The solution is slowly vaporised at temperature $T$ by decreasing the applied pressure from a starting pressure of 760 mm Hg. A pressure gauge (in mm Hg) is connected which gives the reading of pressure applied.

The reading of pressure gauge at which only vapour phase exists is:

• A. $501$
• B. $457.14$
• C. $425$
• D. $525$

Answer 1

The correct option is B $457.14$

The mole fraction of $A$ is $X_A=0.75$

The mole fraction of $B$ is $X_B=0.25$

When pressure is greater than 500 mm Hg, only liquid phase exist.

The expression for the pressure at bubble point is $P_{bubblepoint}=X_A{P}_A^o+X_B{P}_B^o=0.75\times 400+0.25\times 800=500mm$

The mole fractions in vapour phase are $y_A=0.75,y_B=0.25$

At dew point, $\frac{1}{P_T}=\frac{y_A}{P_A^o}+\frac{y_B}{P_B^o}$
$\frac{1}{P_T}=\frac{0.75}{400}+\frac{0.25}{800}=\frac{1.75}{800}$
Hence, $P_T=\frac{800}{1.75}=457.14mmHg$
Below dew point only vapour phase exists.

Therefore, the reading of pressure Gauge at which only vapour phase exists is 457.14 mm of Hg

Hence, the correct option is $B$