A mixture of nitrogen, oxygen and helium contains \(0.25\), \(0.15\) and \(0.4\) mole of these gases respectively. If the pressure contribution due to oxygen was \(2.5 atm\). The partial pressure of helium is ____________
A. \(4.0 atm\) B. \(0.8 atm\) C. \(3.33 atm\) D. \(6.67 atm\)
Correct Answer: D
Explanation
\begin{aligned} &\eta N _{2}=0.25 mol , \eta O _{2}=0.15 mol , \\ &\eta He =0.4 mol \\ &\eta_{\text {Toul }}=0.25+0.15+0.4=0.8 mol \\ &\text { required equations (formulae) } \\ &\text { Mole fraction } X_{A}=\frac{\eta_{A}}{\eta_{T}} \ldots \ldots \ldots . . .(1) \end{aligned} also \(X_{A}=\frac{P_{A}}{P_{T}}\). \(X_{A}=\) mole fraction of gas \(A\) \(\eta_{1}=\) no of mole of gas \(A\) \(\eta_{T}\) - Total no of mole of the gases \(P_{A}=\) Partial pressure of \(A\) \(P_{T}=\) Total partial pressure of the gases \(P_{O_{2}}=2.5 atm\) \(X_{o_{1}}=\frac{\eta_{o_{7}}}{\eta_{T}}=\frac{0.15}{0.8}=0.1275\)rom eqn (2) \begin{aligned} P_{T} &=\frac{P_{O_{2}}}{X_{A}}=\frac{2.5}{0.1875}=13.33 atm \\ X_{H e} &=\frac{0.4}{0.8}=0.5 \\ P_{H e} &=X_{H e} \times P_{T}=0.5 \times 13.33 \\ &=6.67 atm \end{aligned}