Question

A mixture of \(N_{2}\) and \(A r\) gases in a cylinder contains \(7 g\) of \(N_{2}\) and \(8 g\) of \(A r\). If the total pressure of the mixture of the gases in the cylinder is 27 bar, the partial pressure of \(N_{2}\) is:
[Use atomic masses (in g \(\mathrm{mol}^{-1}\) ) : \(N=14, A r=40\) ]
A 9 bar
B 12 bar
(c) \(15 \mathrm{bar}\)
D \(18 \mathrm{bar}\)

Answer 1

Solution:
Partial presure of \(N_{2}=\) mole fraction \(\times P_{\text {Total }}\) of \(N_{2}\)
\(X_{N_{2}}=\frac{\text { molesof } N_{2}}{\text { total } \text { moles }}\)
moles of \(N_{2}=\frac{7}{28}=\frac{1}{4} ;\)
moles of \(A r=\frac{8}{40}=\frac{1}{5}\)
\(X_{N_{2}}=\frac{\left(\frac{1}{4}\right)}{\frac{1}{4}+\frac{1}{5}}=\frac{5}{9}\)
\(P_{N_{2}}=\frac{5}{9} \times 27=15\) bar