A vessel of volume $24.6 \mathrm{~L}$ contains 1.5 moles of $\mathrm{H}_2$ and 2.5 moles of $\mathrm{N}_2$ at $300 \mathrm{~K}$. Calculate the partial pressure of $\mathrm{N}_2$ in the vessel.

Partial pressure In a mixture of gases, each constituent gas has a partial pressure, which is the notional pressure of that constituent gas, if it alone occupied the entire volume of the original mixture at the same temperature. Dalton's law Total pressure of an ideal gas mixture is the sum of the partial pressure of the gases in the mixture. Partial pressure due to $n_1=\left(p^{\circ}{ }_1=\frac{n_1 R T}{V}\right)$ Similarly, due to $n_2=\left(p_2^{\circ}=\frac{n_2 R T}{V}\right)$ Total pressure $p=p_1^{\circ}+p_2^{\circ}=\frac{\left(n_1+n_2\right) R T}{V}$ Given, total volume $=24.6 \mathrm{~L}$ Moles of $\mathrm{H}_2=1.5 \mathrm{~mol}$ Moles of $\mathrm{N}_2=2.5 \mathrm{~mol}$ Temperature $=300 \mathrm{~K}$ Partial pressure of $\mathrm{N}_2=$ ? Partial pressure of $\mathrm{N}_2=\frac{\text { Moles of } \mathrm{N}_2 \times R T}{V}$ $=\frac{2.5 \text { moles } \times 8.314 \times 10^3 \mathrm{~L} \mathrm{~Pa} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \times 300 \mathrm{~K}}{24.6 \mathrm{~L}}$ $=253475.61 \mathrm{~Pa}($ convert into atm $)$ $(1 \mathrm{~atm}=101325 \mathrm{~Pa})$ $=2.5 \mathrm{~atm}(2.50160)$