Question

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?

Short answer

Answer: (a) Neither gas has a larger average translational energy. (b) Helium (He) has the larger partial pressure. (c) Helium (He) has the larger mole fraction. (d) Helium (He) has the larger effusion rate.

Step-by-step solution

Examine average translational energy

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.

Calculate partial pressure

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.

Determine mole fraction

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.

Calculate effusion rate

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.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a cornerstone of chemistry that relates the pressure, volume, temperature, and number of moles of an ideal gas. It is expressed in the formula: \[PV = nRT\]where:

• \(P\) stands for pressure in atmospheres or pascals.
• \(V\) is the volume of the gas in liters.
• \(n\) represents the number of moles of the gas.
• \(R\) is the universal gas constant, approximately \(8.314 \, \text{J/mol·K}\).
• \(T\) is the temperature in Kelvin.

This equation allows us to predict the behavior of a gas under different conditions of pressure, volume, and temperature. In practice, since the gases Kr and He are mixed, we use the Ideal Gas Law to find the total pressure of the mixture before assessing each component's contribution towards that pressure.

Graham's Law

Graham's Law helps us understand how gases effuse through a small opening. It highlights that lighter gases move faster compared to heavier gases, given the same conditions. The law is mathematically represented as: \[\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}\]where:

• \(r_1\) and \(r_2\) are the rates of effusion for gas 1 and gas 2, respectively.
• \(M_1\) and \(M_2\) represent the molar masses of the two gases.

In this specific scenario, with Krypton (Kr) and Helium (He), the lighter helium molecules effuse more quickly than the heavier krypton due to their significantly lower molar mass, confirming that He has a larger effusion rate.

Partial Pressure

In a mixture of gases, each type of gas exerts its own pressure, known as its partial pressure. This can be calculated if the mole fraction of the gas in the mixture and the total pressure of the gas mixture are known. The partial pressure is calculated as: \[ P_i = x_i \times P_{\text{total}} \]where:

• \(P_i\) is the partial pressure of individual gas \(i\).
• \(x_i\) is the mole fraction of the gas.
• \(P_{\text{total}}\) is the overall pressure exerted by the gas mixture.

Given our exercise with Kr and He, Helium possesses a higher partial pressure primarily because a larger mole fraction of helium in the mixture correlates to a more significant contribution to the total pressure.

Mole Fraction

The mole fraction provides a measure of the concentration of a particular gas within a mixture. It is simply the ratio of the number of moles of a particular gas to the total number of moles present. The formula for mole fraction is: \[ x_i = \frac{n_i}{n_{\text{total}}} \]where:

• \(x_i\) represents the mole fraction of a gas.
• \(n_i\) is the number of moles of that specific gas.
• \(n_{\text{total}}\) is the total number of moles of gas in the mixture.

In the example of a gas mixture containing Kr and He, Helium has the higher mole fraction due to its greater number of moles, emphasizing its larger presence in the mixture compared to Krypton.