163. Derive a relationship between mole fraction and vapour pressure of a compound of an ideal solution in the liquid phase and vapour phase.

Answer:

for any solution the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction.
p_A ∝ x_A

Where only solvent is volatile
p_A a x_A where p A is vapour pressure of solvent having mole fractionx_A,
But x_A + x_B = 1
x_A = 1 – xB where x_B is mole fraction of non-volatile solute B
p_A = p⁰A (1 – x_B)
= p⁰_A – p⁰_A x B
Total vapour pressure
$$\begin{gathered} \mathrm{p} = {\mathrm{p}}_{\mathrm{A}}+{\mathrm{p}}_{\mathrm{B}} \\ = {\mathrm{p}}_{\mathrm{A}} = {{\mathrm{p}}^0}_{\mathrm{A}}+{{\mathrm{p}}^0}_{\mathrm{A}}\times \mathrm{B} \\ {\mathrm{x}}_{\mathrm{B}} = \frac{{{\mathrm{p}}^0}_{\mathrm{A}}-{\mathrm{p}}_{\mathrm{A}}}{{{\mathrm{p}}^0}_{\mathrm{A}}} \end{gathered}$$
Solution containing non-volatile solute : For a solution of non-volatile solid in a liquid the vapour pressure contribution by the non-volatile solute is negligible. Therefore the partial vapour pressure of a solution containing a non-volatile solute is equal to the product of the vapour pressure of the pure liquid (solvent p⁰_A) and its mole fraction in solution.
PA = P⁰_A x x_B ....(i)
x_B is the mole fraction of the non-volatile solute
B, then x_A + x_B = 1
x_A = 1 – x_B ....(ii)
Substituting the value of x_A fromeq. (ii) into eq. (i), we get, p_A = p⁰_A (1 – x_B) = p⁰_A – p⁰_A x B
$\frac{{{\mathrm{p}}^0}_{\mathrm{A}}-{\mathrm{p}}_{\mathrm{A}}}{{{\mathrm{p}}^0}_{\mathrm{A}}}={\mathrm{x}}_{\mathrm{B}}.$