Question 10.4.1: A gas mixture is made up of O2 (0.136 g), CO2 (0.230 g), and......
A gas mixture is made up of O_{2} (0.136 g), CO_{2} (0.230 g), and Xe (1.35 g). The mixture has a volume of 1.82 L at 22.0 ºC. Calculate the partial pressure of each gas in the mixture and the total pressure of the gas mixture.
You are asked to calculate the partial pressure of each gas in a mixture of gases and the total pressure of the gas mixture.
You are given the identity and mass of each gas in the sample and the volume and temperature of the gas mixture.
The partial pressure of each gas is calculated from the ideal gas equation.
P_{O_{2}} = \frac{nRT}{V} = \frac{\left(0.136\text{ g} \times \frac{1\text{ mol o}_{2}}{32.00\text{ g}} \right) \left(0.082057\text{ L}\cdot \text{atm/K}\cdot \text{mol}\right) \left(22.0 + 273.15\text{ K}\right) }{1.82\text{ L}} = 0.0566 atm
P_{CO_{2}} = \frac{nRT}{V} = \frac{\left(0.230\text{ g} \times \frac{1\text{ mol CO}_{2}}{44.01\text{ g}} \right) \left(0.082057\text{ L}\cdot \text{atm/K}\cdot \text{mol}\right) \left(22.0 + 273.15\text{ K}\right) }{1.82\text{ L}} = 0.0695 atm
P_{Xe} = \frac{nRT}{V} = \frac{\left(1.35\text{ g} \times \frac{1\text{ mol Xe}}{131.3\text{ g}} \right) \left(0.082057\text{ L}\cdot \text{atm/K}\cdot \text{mol}\right) \left(22.0 + 273.15\text{ K}\right) }{1.82\text{ L}} = 0.137 atm
Notice that the three gases have the same volume and temperature but different pressures. The total pressure is the sum of the partial pressures for the gases in the mixture
P_{total} = P_{O_{2}} + P_{CO_{2}} + P_{Xe} = 0.0566 atm + 0.0695 atm + 0.137 atm = 0.263 atm