Problem 27 - Chemistry: Principles and Reactions

Question

The gas in the discharge cell of a laser contains (in mole percent) \(11 \% \mathrm{CO}_{2}, 5.3 \% \mathrm{~N}_{2}\), and \(84 \% \mathrm{He}\) (a) What is the molar mass of this mixture? (b) Calculate the density of this gas mixture at \(32^{\circ} \mathrm{C}\) and \(758 \mathrm{~mm} \mathrm{Hg}\). (c) What is the ratio of the density of this gas to that of air \((\mathrm{MM}=29.0 \mathrm{~g} / \mathrm{mol})\) at the same conditions?

Step-by-Step Solution

Answer

Answer: The ratio of the density of the gas mixture to that of air at the same conditions is approximately 0.361.

Step 1: Calculate individual molar masses

For the gas mixture, we require the molar masses of CO2, N2, and He. Carbon has molar mass 12.01 g/mol, Oxygen has 16.00 g/mol, and Nitrogen has 14.01 g/mol. Helium has molar mass 4.00 g/mol. Now, we can calculate: Molar mass (CO2) = 12.01 + 2 * 16.00 = 44.01 g/mol Molar mass (N2) = 2 * 14.01 = 28.02 g/mol Molar mass (He) = 4.00 g/mol

Step 2: Calculate the average molar mass

Using the mole percentages given in the exercise, we can find the average molar mass of the gas mixture: Average molar mass = (11% * 44.01) + (5.3% * 28.02) + (84% * 4.00) = 0.11 * 44.01 + 0.053 * 28.02 + 0.84 * 4.00 ≈ 10.44 g/mol

Step 3: Use ideal gas law to compute the density

Using the ideal gas law, PV=nRT, where n/V represents the density in mol/L, we will first convert the given pressure and temperature into appropriate units. Temperature in Kelvin: 32°C + 273.15 = 305.15 K Pressure in atm: 758 mmHg * (1 atm / 760 mmHg) ≈ 0.997 atm Now, we can compute the density at these conditions: Density (mol/L) = P / (R * T) = (0.997 atm) / (0.0821 L atm/mol K * 305.15 K) ≈ 0.0400 mol/L

Step 4: Obtain the density in g/L

To find the density in g/L, multiply the density in mol/L with the average molar mass of the gas mixture: Density (g/L) = Density (mol/L) * Average molar mass = 0.0400 mol/L * 10.44 g/mol ≈ 0.418 g/L

Step 5: Calculate the density ratio to air

To find the ratio of the density of the gas mixture to that of air, divide the density of the gas mixture by the density of air at the same conditions. First, compute the density of air using its molar mass and ideal gas law: Density of air (g/L) = (0.0400 mol/L) * (29.0 g/mol) ≈ 1.16 g/L Now, we can find the density ratio: Density ratio = Density of gas mixture / Density of air = (0.418 g/L) / (1.16 g/L) ≈ 0.361 The ratio of the density of the gas mixture to that of air at the same conditions is approximately 0.361.

Key Concepts

Molar Mass Calculation

Calculating the molar mass of a gas mixture is a fundamental step in understanding the properties of the mixture. In a gas mixture, different gases are combined in certain molar ratios. Each gas contributes to the overall molar mass based on its own molar mass and its percentage in the mixture.

To start, we need to determine the molar mass of each component gas:

• For carbon dioxide (\(\text{CO}_2\)), its molar mass is obtained by adding the atomic masses: Carbon (12.01 g/mol) + 2 * Oxygen (16.00 g/mol) = 44.01 g/mol.
• Nitrogen (\(\text{N}_2\)) is a diatomic molecule, so its molar mass is 2 * Nitrogen (14.01 g/mol) = 28.02 g/mol.
• Helium (\(\text{He}\)) is a monatomic gas and has a molar mass of 4.00 g/mol.

Next, using the mole percentage of each gas in the mixture, we calculate the average molar mass:\[\text{Average Molar Mass} = (0.11 \times 44.01) + (0.053 \times 28.02) + (0.84 \times 4.00) \approx 10.44 \text{ g/mol}\]This is the molar mass of the entire gas mixture.

Ideal Gas Law

The Ideal Gas Law is a powerful tool for understanding the behavior of gases. The formula \(PV = nRT\) links the pressure (\(P\)), volume (\(V\)), and temperature (\(T\)) of a gas with the quantity of gas in moles (\(n\)), where \(R\) is the gas constant. This formula helps to calculate various properties of gases when conditions like temperature and pressure are known.

In our exercise, the Ideal Gas Law is used to calculate the density of the gas mixture. Density (\(\rho\)) in mol/L can be represented as \(n/V\), which means the formula becomes:\[\rho = \frac{P}{RT}\]
First, we need to ensure temperature is in Kelvin: The given temperature 32°C is converted to Kelvin by adding 273.15, resulting in 305.15 K. Pressure is given as 758 mmHg, which needs to be converted to atmospheres: \[\text{Pressure} \approx 0.997 \text{ atm}\]
Then, the density in mol/L is computed as:\[\text{Density (mol/L)} = \frac{0.997}{0.0821 \times 305.15} \approx 0.0400\text{ mol/L}\]

Gas Density

Gas density is a measure of how much mass of the gas is contained in a given volume. Calculating the gas density involves multiplying the density obtained from the Ideal Gas Law (in mol/L) by the molar mass of the mixture.

Here’s how it’s done:

• We already calculated the density in mol/L as approximately 0.0400 mol/L.
• The average molar mass of the gas mixture is approximately 10.44 g/mol.

Using these values, we convert it to density in g/L:\[\text{Density (g/L)} = 0.0400\text{ mol/L} \times 10.44\text{ g/mol} \approx 0.418\text{ g/L}\]This tells us that each liter of the gas mixture weighs approximately 0.418 grams. It helps understand how substantial the gas feels or behaves under given conditions.

Density Ratio with Air

Comparing the density of a gas mixture to that of air provides insights into how the mixture behaves in comparison to a standard reference. Air typically acts as this reference point with a known average molar mass of 29.0 g/mol under similar conditions.

To find the density ratio, we calculate the density of air under the same conditions using its known molar mass:

• The density of air in mol/L remains the same as for the gas mixture, which is approximately 0.0400 mol/L.
• Therefore, the density of air is calculated as:

\[\text{Density of air (g/L)} = 0.0400\text{ mol/L} \times 29.0\text{ g/mol} \approx 1.16\text{ g/L}\]
Finally, the density ratio of the gas mixture to air is simply:\[\text{Density Ratio} = \frac{0.418}{1.16} \approx 0.361\]This means the gas mixture is about 36.1% as dense as air.