Example 9.15: The Pressure of a Mixture of Gases
A gas mixture used for anesthesia contains
2.83 mol oxygen,
\(\ce{O2}\), and
8.41 mol nitrous oxide,
\(\ce{N2O}\). The total pressure of the mixture is
192 kPa.
(a) What are the mole fractions of
\(\ce{O2}\) and
\(\ce{N2O}\)?
(b) What are the partial pressures of
\(\ce{O2}\) and
\(\ce{N2O}\)?
Solution
\(n_{\mathrm{\ce{O2}}}\) \(= 2.83\ \mathrm{mol}\)
\(n_{\mathrm{\ce{N2O}}}\) \(= 8.41\ \mathrm{mol}\)
\(P_{\mathrm{total}}\) \(= 192.\ \mathrm{kPa}\)
\(X_{\mathrm{\ce{O2}}}\) = ?
\(X_{\mathrm{\ce{N2O}}}\) = ?
\(P_{\mathrm{\ce{O2}}}\) = ?
\(P_{\mathrm{\ce{N2O}}}\) = ?
Notice that the temperature is not given here, so we can't apply the ideal gas law. So instead, we will first calculate the mole fractions. The partial pressure of each component is equal to the total pressure times its mole fraction.
\(n_{\mathrm{total}}\) \(= n_{\mathrm{\ce{O2}}} + n_{\mathrm{\ce{N2O}}}\)
\(\ \ \ =2.83\ \mathrm{mol} + 8.41\ \mathrm{mol}\)
\(\ \ \ =11.24\ \mathrm{mol}\)
\(X_{\mathrm{\ce{O2}}}\) \(= \dfrac{n_{\mathrm{\ce{O2}}}}{n_{\mathrm{total}}}\)
\(\ \ \ =\dfrac{2.83\ \mathrm{mol}}{11.24\ \mathrm{mol}}\)
\(\ \ \ =0.2518\)
\(X_{\mathrm{\ce{N2O}}}\) \(= \dfrac{n_{\mathrm{\ce{N2O}}}}{n_{\mathrm{total}}}\)
\(\ \ \ =\dfrac{8.41\ \mathrm{mol}}{11.24\ \mathrm{mol}}\)
\(\ \ \ =0.748\)
\(P_{\mathrm{\ce{O2}}}\) \(= X_{\mathrm{\ce{O2}}} \cdot P_{\mathrm{total}}\)
\(\ \ \ =0.2518 \cdot 192.\ \mathrm{kPa}\)
\(\ \ \ =48.3\ \mathrm{kPa}\)
\(P_{\mathrm{\ce{N2O}}}\) \(= X_{\mathrm{\ce{N2O}}} \cdot P_{\mathrm{total}}\)
\(\ \ \ =0.748 \cdot 192.\ \mathrm{kPa}\)
\(\ \ \ =143.7\ \mathrm{kPa}\)
To check your result, make sure the mole fractions add up to one and the partial pressures add up to the total pressure.