Answer
(i) 0.375 of $H_2$ and 0.625 of $N_2$
(ii) 1.5 bar $H_2$ and 2.5 bar $N_2$
(iii) 4.0 bar
Work Step by Step
1. Calculate the total amount of moles and determine each fraction:
Total = 1.5 mol + 2.5 mol = 4.0 mol
$$\frac{1.5 \space mol}{4.0 \space mol} = 0.375 (H_2)$$
$$\frac{2.5 \space mol}{4.0 \space mol} = 0.625 (N_2)$$
2. Use PV = nRT to calculate their partial pressures:
$H_2$:
$$P = \frac{nRT}{V} = \frac{1.5 \space mol \times 8.31447×10^{−2} \space dm^3 \space bar \space K^{-1} \space mol^{-1} \times 273.15 \space K}{22.4 \space dm^3} = 1.5 \space bar$$
$N_2$:
$$P = \frac{nRT}{V} = \frac{2.5 \space mol \times 8.31447×10^{−2} \space dm^3 \space bar \space K^{-1} \space mol^{-1} \times 273.15 \space K}{22.4 \space dm^3} = 2.5 \space bar$$
3. Sum their partial pressures to get the total pressure.
$$P = 1.5 \space bar + 2.5 \space bar = 4.0 \space bar$$
