Question

A vessel contains 1.60 g of oxygen and 2.80 g of nitrogen. The temperature is maintained at 300 K and the volume of the vessel is 0.166 $m^3$. Find the pressure of the mixture.

Answer 1

Given data,

$V=0.166m^3$

$T=300K$

$Massof{O}_2=1.60g$

$M_O=32g$

$n_O=\frac{1.60}{32}=0.05$

$Massof{N}_2=2.80g$

$M_N=28g$

$n_N=\frac{2.80}{28}=0.1$

Partial pressure of $O_2$ is given by

$P_O=\frac{n_{o}RT}{V}=\frac{0.05\times 8.3\times 300}{0.166}=750$

partial pressure of $N_2$ is given by

$P_N=\frac{n_{N}RT}{V}=\frac{0.1\times 8.3\times 300}{0.166}=1500$

Total pressure is sum of the partial pressures.

$\Rightarrow P=P_N+P_O=750+1500=2250Pa$