---

Example 9.14: The Pressure of a Mixture of Gases

A 10.0 L vessel contains \(2.50\times 10^{-3 }\)mol of \(\ce{H2}\), \(1.00\times 10^{-3 }\)mol of He, and \(3.00\times 10^{-4 }\)mol of Ne at 35 °C.
(a) What are the partial pressures of each of the gases?
(b) What is the total pressure in atmospheres?

Solution

\(V_{\mathrm{gas}}\) \(= 10.0\ \mathrm{L}\)


\(n_{\mathrm{\ce{H2}}}\) \(= 2.50\times 10^{-3}\ \mathrm{mol}\)


\(n_{\mathrm{\ce{He}}}\) \(= 1.00\times 10^{-3}\ \mathrm{mol}\)


\(n_{\mathrm{\ce{Ne}}}\) \(= 3.00\times 10^{-4}\ \mathrm{mol}\)


\(T\) \(= 35.\ \mathrm{°aC}\)


The gases behave independently, so the partial pressure of each gas can be determined from the ideal gas equation, using \(P = \dfrac{n \cdot R \cdot T}{V_{\mathrm{gas:}}}\)
\(R\) \(= 8.314\ \frac{\mathrm{J}}{\mathrm{mol}\ \mathrm{K}}\)


\(R\) \(= 8.314\ \frac{\mathrm{J}}{\mathrm{mol}\ \mathrm{K}}\)

\(\ \ \ =0.08205\ \frac{\mathrm{L}\ \mathrm{atm}}{\mathrm{mol}\ \mathrm{K}}\)


\(P_{\mathrm{\ce{H2}}}\) \(= \dfrac{n_{\mathrm{\ce{H2}}} \cdot R \cdot T}{V_{\mathrm{gas}}}\)

\(\ \ \ =\dfrac{2.50\times 10^{-3}\ \mathrm{mol} \cdot 0.08205\ \frac{\mathrm{L}\ \mathrm{atm}}{\mathrm{mol}\ \mathrm{K}} \cdot 35.\ \mathrm{°aC}}{10.0\ \mathrm{L}}\)

\(\ \ \ =\dfrac{2.0513\times 10^{-4}\ \frac{\mathrm{atm}\ \mathrm{L}}{\mathrm{K}} \cdot 308.\ \mathrm{K}}{10.0\ \mathrm{L}}\)

\(\ \ \ =\dfrac{0.06321\ \mathrm{atm}\ \mathrm{L}}{10.0\ \mathrm{L}}\)

\(\ \ \ =6.32\times 10^{-3}\ \mathrm{atm}\)


\(P_{\mathrm{\ce{He}}}\) \(= \dfrac{n_{\mathrm{\ce{He}}} \cdot R \cdot T}{V_{\mathrm{gas}}}\)

\(\ \ \ =\dfrac{1.00\times 10^{-3}\ \mathrm{mol} \cdot 0.08205\ \frac{\mathrm{L}\ \mathrm{atm}}{\mathrm{mol}\ \mathrm{K}} \cdot 35.\ \mathrm{°aC}}{10.0\ \mathrm{L}}\)

\(\ \ \ =\dfrac{8.205\times 10^{-5}\ \frac{\mathrm{atm}\ \mathrm{L}}{\mathrm{K}} \cdot 308.\ \mathrm{K}}{10.0\ \mathrm{L}}\)

\(\ \ \ =\dfrac{0.02528\ \mathrm{atm}\ \mathrm{L}}{10.0\ \mathrm{L}}\)

\(\ \ \ =2.53\times 10^{-3}\ \mathrm{atm}\)


\(P_{\mathrm{\ce{Ne}}}\) \(= \dfrac{n_{\mathrm{\ce{Ne}}} \cdot R \cdot T}{V_{\mathrm{gas}}}\)

\(\ \ \ =\dfrac{3.00\times 10^{-4}\ \mathrm{mol} \cdot 0.08205\ \frac{\mathrm{L}\ \mathrm{atm}}{\mathrm{mol}\ \mathrm{K}} \cdot 35.\ \mathrm{°aC}}{10.0\ \mathrm{L}}\)

\(\ \ \ =\dfrac{2.4616\times 10^{-5}\ \frac{\mathrm{atm}\ \mathrm{L}}{\mathrm{K}} \cdot 308.\ \mathrm{K}}{10.0\ \mathrm{L}}\)

\(\ \ \ =\dfrac{7.585\times 10^{-3}\ \mathrm{atm}\ \mathrm{L}}{10.0\ \mathrm{L}}\)

\(\ \ \ =7.59\times 10^{-4}\ \mathrm{atm}\)


Because the gases behave independently, the total pressure is the sum of the partial pressures.

\(P_{\mathrm{gas}}\) \(= \sum (P_{\mathrm{\ce{H2}}}, P_{\mathrm{\ce{He}}}, P_{\mathrm{\ce{Ne}}})\)

\(\ \ \ =\sum (6.32\times 10^{-3}\ \mathrm{atm}, 2.53\times 10^{-3}\ \mathrm{atm}, 7.59\times 10^{-4}\ \mathrm{atm})\)

\(\ \ \ =9.61\times 10^{-3}\ \mathrm{atm}\)