Question 6.11: Applying the Ideal Gas Equation to a Mixture of Gases What i...
Applying the Ideal Gas Equation to a Mixture of Gases
What is the pressure, in bar, exerted by a mixture of 1.0 g H_2 and 5.00 g He when the mixture is confined to a volume of 5.0 L at 20 °C?
Analyze
For fixed T and V, the total pressure of a mixture of gases is determined by the total number of moles of gas: P_{tot} = n_{tot}RT/V.
n_{\text {tot }} = \left(1.0 g H _2 \times \frac{1 mol H _2}{2.02 g H _2}\right) + \left(5.00 g He \times \frac{1 mol He }{4.003 g He }\right)
= 0.50 mol H_2 + 1.25 mol He = 1.75 mol gas
P = \frac{1.75 mol \times 0.0831 \text { bar } L mol ^{-1} K ^{-1} \times 293 K }{5.0 L } = 8.5 bar
Assess
It is also possible to solve this problem by starting from equation (6.12). Because 1 mol of ideal gas occupies 22.7 L at 0 °C and 1 bar, the pressure exerted by 1.75 mol of gas in a 5.0 L vessel at 293 K is (1.75 mol/1.00 mol) × (293 K/273 K) × (22.7 L/5.0 L) × 1.0 bar = 8.5 bar.
\frac{P_i V_i}{n_i T_i} = \frac{P_f V_f}{n_f T_f} (6.12)