Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm3 at 27°C.
Question:

Calculate the total pressure in a mixture of $8 \mathrm{~g}$ of dioxygen and $4 \mathrm{~g}$ of dihydrogen confined in a vessel of $1 \mathrm{dm}^{3}$ at $27^{\circ} \mathrm{Cl} \mathrm{R}=0.083 \mathrm{bar} \mathrm{dm}^{3} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$.

Solution:

Given,

Mass of dioxygen $\left(\mathrm{O}_{2}\right)=8 \mathrm{~g}$

Thus, number of moles of $\mathrm{O}_{2}=\frac{8}{32}=0.25$ mole

Mass of dihydrogen $\left(\mathrm{H}_{2}\right)=4 \mathrm{~g}$

Thus, number of moles of $\mathrm{H}_{2}=\frac{4}{2}=2$ mole

Therefore, total number of moles in the mixture $=0.25+2=2.25$ mole

Given,

V = 1 dm3

n = 2.25 mol

R = 0.083 bar dm3 K–1 mol–1

T = 27°C = 300 K

Total pressure (p) can be calculated as:

$p V=n \mathrm{R} T$

$\Rightarrow p=\frac{n \mathrm{R} T}{V}$

$=\frac{225 \times 0.083 \times 300}{1}$

$=56.025 \mathrm{bar}$

Hence, the total pressure of the mixture is 56.025 bar.