Question

A mixture of 3.5 mol of \(\mathrm{Kr}\) and \(3.9 \mathrm{~mol}\) of He occupies a 10.00 - \(\mathrm{L}\) container at \(300 \mathrm{~K}\). Which gas has the larger (a) average translational energy? (b) partial pressure? (c) mole fraction? (d) effusion rate?

Short answer

Question: In a container with 3.5 moles of krypton (Kr) and 3.9 moles of helium (He) at 300 K and a volume of 10.00 L, compare the (a) average translational energy, (b) partial pressure, (c) mole fraction, and (d) effusion rate of the two gases. Answer: (a) The average translational energy for both Kr and He is the same. (b) He has the larger partial pressure. (c) He has the larger mole fraction. (d) He has a larger effusion rate than Kr.

Step-by-step solution

(a) Average translational energy

To calculate the average translational energy, we can use the formula: \(E_{avg} = \frac{3}{2} k_\mathrm{B} T\) Where \(E_{avg}\) is the average translational energy, \(k_\mathrm{B}\) is the Boltzmann constant \((1.38 \times 10^{-23} \mathrm{J/K})\), and \(T\) is the temperature in Kelvin. Since the temperature is the same (300 K) for both gases, we just need to calculate \(E_{avg}\) for each gas. \(E_{avg}(\mathrm{Kr}) = \frac{3}{2}(1.38 \times 10^{-23} \mathrm{J/K})(300 \mathrm{K})\) \(E_{avg}(\mathrm{He}) = \frac{3}{2}(1.38 \times 10^{-23} \mathrm{J/K})(300 \mathrm{K})\) Notice that the result for both calculations is the same numerical value, so the average translational energy for both Kr and He is the same.

(b) Partial pressure

To calculate the partial pressure of each gas, we use the ideal gas law equation: \(PV = nRT\) Where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant \((8.314 \mathrm{J/mol \cdot K})\), and T is the temperature. We need to calculate the partial pressure of each gas individually, and then compare the values. For Kr: \(P(\mathrm{Kr}) = \frac{n(\mathrm{Kr})RT}{V} = \frac{3.5 \mathrm{mol} (8.314 \mathrm{J/mol \cdot K})(300 \mathrm{K})}{10.00 \mathrm{L}}\) For He: \(P(\mathrm{He}) = \frac{n(\mathrm{He})RT}{V} = \frac{3.9 \mathrm{mol} (8.314 \mathrm{J/mol \cdot K})(300 \mathrm{K})}{10.00 \mathrm{L}}\) Comparing the calculations, we see that He has the larger partial pressure.

(c) Mole fraction

To calculate the mole fraction, we need to know the total number of moles (\(n_{total}\)) and the number of moles of the specific gas (\(n_i\)). The formula for mole fraction is: \(\chi_i = \frac{n_i}{n_{total}}\) Calculating the mole fraction for Kr: \(\chi_\mathrm{Kr} = \frac{3.5 \mathrm{mol}}{3.5 \mathrm{mol} + 3.9 \mathrm{mol}}\) Calculating the mole fraction for He: \(\chi_\mathrm{He} = \frac{3.9 \mathrm{mol}}{3.5 \mathrm{mol} + 3.9 \mathrm{mol}}\) Comparing the calculations, we can see that He has the larger mole fraction.

(d) Effusion rate

To determine which gas has a larger effusion rate, we use Graham's law of effusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. The form of the equation is: \(\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}\) Where r denotes the effusion rate and M represents the molar mass. For this problem, we will compare the effusion rates of Kr \((r_\mathrm{Kr})\) and He \((r_\mathrm{He})\): \(\frac{r_\mathrm{Kr}}{r_\mathrm{He}} = \sqrt{\frac{M_\mathrm{He}}{M_\mathrm{Kr}}}\) With the molar mass of Kr being 83.798 g/mol and He being 4.0026 g/mol: \(\frac{r_\mathrm{Kr}}{r_\mathrm{He}} = \sqrt{\frac{4.0026 \mathrm{g/mol}}{83.798 \mathrm{g/mol}}}\) Calculating the ratio shows that \(r_\mathrm{Kr} < r_\mathrm{He}\), meaning that He has a larger effusion rate than Kr.