A mixture of \(3.5 \mathrm{~mol}\) 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?

To determine the average translational energy, remember the formula derived from the Kinetic Molecular Theory of Gases: \(\bar{E}_{trans} = \frac{3}{2}RT\) where \(\bar{E}_{trans}\) is the average translational energy (per mole), R is the universal gas constant, and T is the temperature in Kelvin. Since both gases, Kr and He, are in the same container and at the same temperature (300 K), their translational energies will be identical. So, no gas has a larger average translational energy. (a) Neither gas has a larger average translational energy.

Partial pressure can be calculated using the mole fraction and total pressure. First, we need to find the total pressure of the gas mixture using the Ideal Gas Law: \(PV = nRT\) where P is pressure, V is volume, n is the number of moles, R is the universal gas constant, and T is the temperature in Kelvin. Let's find the total pressure of the mixture: \(P = \frac{(n_{Kr} + n_{He})RT}{V} = \frac{(3.5 + 3.9)R(300)}{10.00} = \frac{7.4\cdot R\cdot 300}{10.00}\) We won't calculate the specific value of total pressure since we are interested in comparing the partial pressures, not finding their specific values. Now, let's find the mole fractions for Kr and He: \(x_{Kr} = \frac{n_{Kr}}{n_{Kr} + n_{He}} = \frac{3.5}{(3.5+3.9)}\) \(x_{He} = \frac{n_{He}}{n_{Kr} + n_{He}} = \frac{3.9}{(3.5+3.9)}\) Now we can calculate the partial pressures for Kr and He: \(P_{Kr} = x_{Kr} \cdot P = \frac{3.5}{(3.5+3.9)} \cdot \frac{7.4\cdot R\cdot 300}{10.00}\) \(P_{He} = x_{He} \cdot P = \frac{3.9}{(3.5+3.9)} \cdot \frac{7.4\cdot R\cdot 300}{10.00}\) Since \(x_{Kr} < x_{He}\), the partial pressure for He is larger. (b) Helium (He) has the larger partial pressure.

We have already calculated the mole fractions for Kr and He in Step 2: \(x_{Kr} = \frac{3.5}{(3.5+3.9)}\) \(x_{He} = \frac{3.9}{(3.5+3.9)}\) Since \(x_{Kr} < x_{He}\), the mole fraction for He is larger. (c) Helium (He) has the larger mole fraction.

Graham's Law states that the ratio of the rates of effusion of two gases is inversely proportional to the square root of their molar masses: \(\frac{r_{1}}{r_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}\) where \(r_{1}\) and \(r_{2}\) are the effusion rates, and \(M_{1}\) and \(M_{2}\) are the molar masses of gas 1 and gas 2, respectively. In our case: \(\frac{r_{Kr}}{r_{He}} = \sqrt{\frac{M_{He}}{M_{Kr}}}\) We need to look up the molar masses for Kr and He: \(M_{Kr} \approx 83.80 \frac{g}{mole}\) \(M_{He} \approx 4.00 \frac{g}{mole}\) Now, we can plug these values into Graham's Law formula: \(\frac{r_{Kr}}{r_{He}} = \sqrt{\frac{4.00}{83.80}}\) Since the molar mass of He is much smaller than the molar mass of Kr, the effusion rate of He will be larger. (d) Helium (He) has the larger effusion rate.