According to Raoult's law,

$\mathrm{p}={\mathrm{p}}^0\mathrm{\chi }$

According to Dalton's law of partial pressure,

${\mathrm{p}}_{\mathrm{T}}={\mathrm{p}}_{\mathrm{x}}+{\mathrm{p}}_{\mathrm{y}}$

${\mathrm{p}}_{\mathrm{T}}={\mathrm{p}}^0{\mathrm{x\chi }}_{\mathrm{x}}+{\mathrm{p}}^0{\mathrm{y\chi }}_{\mathrm{y}}$

$={\mathrm{p}}^0{\mathrm{x\chi }}_{\mathrm{x}}+{\mathrm{p}}_{\mathrm{y}}^0\left(1-{\mathrm{\chi }}_{\mathrm{x}}\right)$

${\mathrm{p}}_{\mathrm{T}}={\mathrm{p}}^0\mathrm{Y}+\left({\mathrm{p}}^0\mathrm{x}-{\mathrm{p}}^0\mathrm{y}\right){\mathrm{\chi }}_{\mathrm{x}}$

Thus, total vapour pressure of solution varies linearly with mole fraction of component $\mathrm{X}$.

Case $\mathrm{I}$, at ${\mathrm{\chi }}_{\mathrm{X}}=0$

${\mathrm{p}}_{\mathrm{T}}={\mathrm{p}}^0\mathrm{Y}$

Case $\mathrm{II}$, at ${\mathrm{X}}_{\mathrm{X}}=1$

${\mathrm{p}}_{\mathrm{T}}={\mathrm{p}}^0\mathrm{Y}+{\mathrm{p}}^0\mathrm{X}-{\mathrm{p}}^0\mathrm{Y}$

${\mathrm{p}}_{\mathrm{T}}={\mathrm{p}}_{\mathrm{X}}^0$