1. Tardigrade
  2. Question
  3. Chemistry
  4. Mol fraction of the component A in vapour phase is x1 and mol fraction of component A in liquid phase is x2 (If pA°= vapour pressure of pure A ; pB°= vapour pressure of pure B ), then total vapour pressure of the liquid mixture is:

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Q. Mol fraction of the component $A$ in vapour phase is $x_{1}$ and mol fraction of component $A$ in liquid phase is $x_{2}$ (If $p_{A}^{\circ}=$ vapour pressure of pure $A ; p_{B}^{\circ}=$ vapour pressure of pure $B$ ), then total vapour pressure of the liquid mixture is:

Solutions

Solution:

$p _{ A }= p _{ A }^{\circ} \cdot x _{2} p _{ B }= p _{ B }^{\circ}=\left(1- x _{2}\right)$

$p _{\text {Total}}= p _{ A }^{\circ} \cdot x _{2}+ p _{ B }^{\circ}\left(1- x _{2}\right)...(i)$

Also $x_{1}=\frac{p_{A}}{p_{A}+p_{B}}=\frac{p_{A}}{p_{A}^{\circ} x_{2}+p_{B}^{\circ}\left(1-x_{2}\right)}$

or $p _{ A }^{\circ} \cdot x _{2}+ p _{ B }^{\circ}\left(1- x _{2}\right)=\frac{ p _{ A }}{ x _{1}}$

Put in (i) $ p_{\text {Total }}=\frac{p_{A}}{x_{1}}=\frac{p_{A}^{\circ} x_{2}}{x_{1}}$