Question

An ideal mixture of liquid A and B with 2 moles of A and 2 moles of B has a total vapour pressure of 1 atm at a certain temperature. Another mixture with 1 mole of A and 3 moles of B has a vapour pressure greater than 1 atm. But if 4 moles of C are added to the second mixture, the vapour pressure comes down to 1 atm. Vapour Pressure of C, $P_c^0$ = 0.8 atm, calculate the vapour pressures of pure A.

Answer 1

$P_{\tau }=P_A^0{x}_A+P_B^0{x}_B\Rightarrow \frac{P_A^0}{2}+\frac{P_B^0}{2}=1atm\Rightarrow P_A^0+P_B^0=2atm.....\left(1\right)\frac{P_A^0}{4}+\frac{3P_B^0}{4}>1atm\Rightarrow P_A^0+3P_B^0>4atm\text{and}\frac{P_A^0}{8}+\frac{3P_B^0}{8}+\frac{4P_C^0}{8}=1atm\Rightarrow P_A^0+3P_B^0+4P_C^0=8atmAs{P}_C^0=0.8atm.So,P_A^0+3P_B^0=8-(4\times 0.8)atm=4.8atm....\left(2\right)$
Hence using (1) and (2),
$P_B^0=1.4atm{P}_A^0=0.6atm$