A mixture of $n_{\text{He}}=2\text{ mol}$ and $n_{{\text{O}}_2}=4\text{ mol}$ is in a container at total pressure $P_{\text{total}}=5\text{ atm}$. What is the partial pressure $P_{\text{He}}$?

Solution

1. Calculate total moles: $$\begin{array}{rl}{n}_{\text{total}}& =2+4\\ & =6\text{ mol}\end{array}$$.
2. Calculate mole fraction of He: $$\begin{array}{rl}{x}_{\text{He}}& =\frac{2}{6}\\ & =1/3\end{array}$$.
3. Calculate partial pressure: $P_{\text{He}}=x_{\text{He}}\times P_{\text{total}}$.
4. $$\begin{array}{rl}{P}_{\text{He}}& =(1/3)\times 5\\ & =\mathrm{1.666...}\text{ atm}\end{array}$$.
5. Hence the answer is (A).