Question

An ideal mixture of liquids $A$ 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 vapour pressure greater than 1 atm. When 4 moles of $C$ are added to second mixture, the vapour pressure comes down to 1 atm. Vapour pressure of $C$ in pure state i.e., $P _{\circ }^{\circ}=0.8 \,atm .$ What will be the value of $P _{A}^{\circ} \& P _{ B }^{ \circ }$

• A. $P _{ A }^{\circ}=1.2\, atm , P _{ B }^{ o }=0.7\, atm$
• B. $P _{ A }^{\circ}=1.2 \,atm , P _{ B }^{\circ}=0.6 \,atm$
• C. $P _{ A }^{\circ}=0.6 \,atm , P _{ B }^{\circ}=1.4\, atm$
• D. $P _{ A }^{\circ}=1.4 \,atm , P _{ B }^{\circ}=0.6\, atm$

Answer 1

Answer: (C)

$1=P_{A}+P_{B}$

$1=P_{A}^{\circ} \cdot \frac{2}{4}+P_{B}^{\circ} \cdot \frac{2}{4}$

$P _{ A }^{\circ}+ P _{ B }^{\circ}=2 \ldots \ldots$(1)

$x = P _{ A }^{\circ} x _{ A }+ P _{ B }^{\circ} x _{ B }$

$x = P _{ A }^{\circ} \frac{1}{4}+ P _{ B }^{\circ} \frac{3}{4}$

$P _{ A }^{\circ}+3 P _{ B }^{\circ}=4 x \dots$(2)

$1=P_{A}^{\circ} x_{A}+P_{B}^{\circ} x_{B}+P_{c}^{\circ} x_{C}$

$1=P_{A}^{\circ} \cdot \frac{1}{8}+P_{B}^{\circ} \cdot \frac{3}{8}+P_{C}^{\circ} \cdot \frac{4}{8}$

$8=P_{A}^{\circ}+3 P_{B}^{\circ}+3.2$

$P_{A}^{\circ}+3 P_{B}^{\circ}=4.8 \ldots \ldots(3)$

From 2 and $3: 4 x=4.8$

$x=1.2$

Also, $1-2=P_{A}^{\circ}+P_{B}^{\circ}-\left(P_{A}^{\circ}+3 P_{B}^{\circ}\right)=2-4 x$

$=P_{A}^{\circ}+P_{B}^{\circ}-P_{A}^{\circ}-3 P_{B}^{\circ}=2-4 x$

$=-2 P_{B}^{\circ}=2-4 x$

$P_{B}^{\circ}=4 x-2$

$P_{B}^{\circ}=4 \times 1.2-2$

$P_{B}^{\circ}=4.8-2$

$P_{B}^{\circ}=\frac{2.8}{2}=1.4$

$P_{A}^{\circ}+P_{B}^{\circ}=2$

$\Rightarrow \quad P_{A}^{\circ}+1.4=2$

$P_{A}^{\circ}=0.6$