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

$P_{O_2}=\frac{n_{O_{2}RT}}{V}$

$P_{N_2}=\frac{n_{N_{2}RT}}{V}$

$n_{O_2}=\frac{m}{M_{O_2}}=\frac{1.60}{32}$

$n_{N_2}=\frac{m}{M_{N_2}}=\frac{2.80}{28}=0.1$

Now, $P_{mix}=\left(\frac{n_{O_2+n_{N_2}}}{V}\right)$RT

$P_{mix}=\frac{(0.05+0.1)\times 8.3\times 300}{0.168}$

= 2250 $N/m^2$