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 I. 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 I by decreasing the applied pressure from a starting of 760 mm Hg. A pressure gauge (in mm Hg) is connected

Answer 1

Vapour pressure of pure liquid $\left(P^{o}A\right)=400mmHg$

mole fraction of $A=3/4$

Partial vapour pressure of $A\left(PA\right)=P^{o}A\times x_A$

$\Rightarrow \frac{3}{4}\times 400$

$\Rightarrow 300mmHg$

vapour pressure of pure liquid $B\left(P^{o}B\right)=800mmHg$

mole fraction of $B=1/4$

Partial vapour pressure of $B\left(P_B\right)=P_B^o\times x_B$

$=800\times \frac{1}{4}$

$=200mmHg$

Total partial pressure $P_T=300+200$

$=500mmHg$

Hence, pressure gauge is connected to $500mmHg$.