Give the relation between partial pressure and mole fraction.

Suppose at a constant temperature $(T)$, three gases are enclosed in a volume $(V)$, exerting partial pressures $p_{1}, p_{2}$, and $p_{3}$ respectively.
Using the ideal gas equation $pV = nRT$, we have:
$(i) \ p_{1} = \frac{n_{1}RT}{V} \quad \dots (Eq.-i)$
$(ii) \ p_{2} = \frac{n_{2}RT}{V} \quad \dots (Eq.-ii)$
$(iii) \ p_{3} = \frac{n_{3}RT}{V} \quad \dots (Eq.-iii)$
where $n_{1}, n_{2}$, and $n_{3}$ are the number of moles of these gases.
According to Dalton's Law, the total pressure $(p_{\text{total}})$ is:
$p_{\text{total}} = p_{1} + p_{2} + p_{3} = (n_{1} + n_{2} + n_{3}) \frac{RT}{V} \quad \dots (Eq.-iv)$
To find the relation, divide the partial pressure of a gas by the total pressure:
$\frac{p_{1}}{p_{\text{total}}} = \frac{n_{1}RT/V}{(n_{1} + n_{2} + n_{3})RT/V} = \frac{n_{1}}{n_{1} + n_{2} + n_{3}}$
Since the mole fraction $(\chi_{1})$ of the first gas is defined as $\chi_{1} = \frac{n_{1}}{n_{\text{total}}}$, where $n_{\text{total}} = n_{1} + n_{2} + n_{3}$, we get:
$\frac{p_{1}}{p_{\text{total}}} = \chi_{1}$
Therefore, the relation is $p_{1} = \chi_{1} p_{\text{total}}$.