Example 9.16: Pressure of a Gas Collected Over Water
If
0.200 L of argon is collected over water at a temperature of
26 °C and a pressure of
750 torr in a system like that shown in
Figure 21, what is the partial pressure of argon?
Solution
\(P_{\mathrm{\ce{Ar}}}\) = ?
\(V_{\mathrm{\ce{Ar}}}\) \(= 0.200\ \mathrm{L}\)
\(P_{\mathrm{total}}\) \(= 750.\ \mathrm{torr}\)
\(T\) \(= 26.\ \mathrm{°aC}\)
According to Dalton’s law, the total pressure in the bottle (
750 torr) is the sum of the partial pressure of argon and the partial pressure of gaseous water:
\(P_{\mathrm{total}} = P_{\mathrm{\ce{Ar}}} + P_{\mathrm{\ce{H2O}}}\)
The pressure of water vapor above a sample of liquid water at
26 °C is
25.2 torr (
Appendix E):
\(P_{\mathrm{\ce{H2O}}}\) \(= 25.2\ \mathrm{torr}\)
Now we can calculate the partial pressure:
\(P_{\mathrm{\ce{Ar}}}\) \(= P_{\mathrm{total}} - P_{\mathrm{\ce{H2O}}}\)
\(\ \ \ =750.\ \mathrm{torr} - 25.2\ \mathrm{torr}\)
\(\ \ \ =725.\ \mathrm{torr}\)
Notice that we did not make use of the volume information (we can't use the ideal gas law to calculate the pressure because the chemical amount of argon is unknown) and we made use of the temperture information indirectly when we looked up the partial pressure of water at this temperature.
