Question 12.30: Calculating the Partial Pressure of a Gas Using the Ideal Ga......
Calculating the Partial Pressure of a Gas Using the Ideal Gas Law
Calculate the partial pressure of O _2 , in atmospheres, in a gaseous mixture with a volume of 2.50 L at 20°C, given that the mixture composition is 0.50 mole O _2 and 0.75 mole N _2 .
This problem differs from Example 12.29 in that both the temperature and volume of the gaseous mixture are given. This added information will enable us to calculate the partial pressure of O _2 using two different methods: (1) mole fractions, and (2) the ideal gas law.
Using the mole fraction method, we calculate first the total pressure of the gaseous mixture based on the presence of 1.25 total moles of gas (0.50 mole + 0.75 mole):
P_{\textrm{total}} = \frac{n_{\textrm{total}} RT}{V} = \frac{(1.25 \cancel{\textrm{moles}}) \Bigl(0.08206 \frac{\textrm{atm} \cdot \cancel{\textrm{L}}}{\cancel{\textrm{mole}} \cdot \cancel{\textrm{K}}} \Bigr) (293 \cancel{\textrm{K}})}{2.50 \cancel{\textrm{L}}}= 12.02179 atm (calculator answer)
= 12.0 atm (correct answer)
Next we calculate the mole fraction of O _2 :
X_{\textrm{O}_2} = \frac{n_{\textrm{O}_2}}{n_{\textrm{total}}} = \frac{0.50 \cancel{\textrm{mole}}}{(0.50 + 0.75) \cancel{\textrm{mole}}} = 0.4 (calculator answer)
= 0.40 (correct answer)
The partial pressure of the O _2 is obtained using the mole fraction O _2 and the total pressure:
P_{\textrm{O}_2} = X_{\textrm{O}_2} P_{\textrm{total}} = 0.40 × 12.0 atm = 4.8 atm (calculator and correct answer)
This problem can also be solved without the use of mole fractions by using the concept that the partial pressure exerted by the O _2 is determined by the number of moles of O _2 present. With this approach, we substitute directly into the ideal gas law the quantities given in the problem statement.
P_{\textrm{O}_2} = \frac{n_{\textrm{O}_2} RT}{V} = \frac{(0.50 \cancel{\textrm{mole}}) \Bigl(0.08206 \frac{\textrm{atm} \cdot \cancel{\textrm{L}}}{\cancel{\textrm{mole}} \cdot \cancel{\textrm{K}}} \Bigr) (293 \cancel{\textrm{K}})}{2.50 \cancel{\textrm{L}}}= 4.808716 atm (calculator answer)
= 4.8 atm (correct answer)
Answer Double Check: The partial pressure of the O _2 gas should be 40% of the total pressure, based on its mole fraction of 0.40. Such is the case; 0.40 × 12.0 atm (total pressure) = 4.8 atm.