Problem 55 - Chemistry The Molecular Science

Question

The density of air at \(20.0 \mathrm{~km}\) above Earth's surface is \(92 \mathrm{~g} / \mathrm{m}^{3}\). The pressure is \(42 \mathrm{mmHg}\) and the temperature is \(-63^{\circ}\) C. Assuming the atmosphere contains only \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2},\) calculate (a) the average molar mass of the air at \(20.0 \mathrm{~km}\). (b) the mole fraction of each gas.

Step-by-Step Solution

Answer

(a) Average molar mass is 28.70 g/mol. (b) Mole fractions: \( \mathrm{O}_2 = 0.231 \), \( \mathrm{N}_2 = 0.769 \).

Step 1: Convert Temperature to Kelvin

Temperature needs to be in Kelvin for calculations involving gases. The conversion formula is \[ T(K) = T(°C) + 273.15 \]For this problem:\[ T = -63 + 273.15 = 210.15 \text{ K} \]

Step 2: Convert Pressure to Atmospheres

Convert the pressure from mmHg to atm using:\[ 1 \text{ atm} = 760 \text{ mmHg} \]So, \[ P = \frac{42 \text{ mmHg}}{760 \text{ mmHg/atm}} = 0.05526 \text{ atm} \]

Step 3: Use Ideal Gas Law to find molar volume

The ideal gas law is \( PV = nRT \). We rearrange it to find the molar volume \( V_m \) (volume per mole) using \[ V_m = \frac{RT}{P} \]where \(R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1} \). Substituting the values:\[ V_m = \frac{0.0821 \cdot 210.15}{0.05526} = 312.04 \text{ L/mol} \]

Step 4: Find the Molar Mass of Air

The density \( \rho \) is given by \( \rho = \frac{m}{V} \), and rearranging gives \( m = \rho V \). Using it together with the molar mass \( M \), we use the relation:\[ M = \rho \cdot V_m \]Given \( \rho = 0.092 \text{ g/L} \), we find:\[ M = 0.092 \cdot 312.04 = 28.70 \text{ g/mol} \]

Step 5: Assume Composition of Air is 21% \(O_2\) and 79% \(N_2\)

Using known composition ratios of air, we assume 21% oxygen and 79% nitrogen in terms of their natural abundance at Earth's surface. While this can change at altitude, without additional data we proceed with these ratios.

Step 6: Find Molar Mass of Each Gas

The molar masses are:- \( \mathrm{O}_2: 32.00 \text{ g/mol} \)- \( \mathrm{N}_2: 28.02 \text{ g/mol} \)

Step 7: Use Weighted Average Formula

To find the average molar mass, use the formula:\[ \text{Average Molar Mass} = 0.21 \times 32.00 + 0.79 \times 28.02 \]Calculating gives:\[ 28.84 \text{ g/mol} \]

Step 8: Check Consistency with Calculated Molar Mass

Compare the weighted average molar mass of \(28.84\, \text{g/mol}\) with the calculated molar mass of \(28.70\, \text{g/mol}\) to ensure the results are coherent.

Step 9: Calculate Mole Fractions

First, we determine the number of moles for each component present in 1 mole of air. Let \(x_o\) and \(x_n\) be the moles for \( \mathrm{O}_2 \) and \( \mathrm{N}_2 \) respectively, as follows:- \(0.21x_o \times 32.00 \text{ g/mol} = 0.21 \times 32\)- \(0.79x_n \times 28.02 \text{ g/mol} = 0.79 \times 28.02 \)For mole fraction:\[ \text{Mole fraction of } \mathrm{O}_2 = \frac{\text{moles of } \mathrm{O}_2}{\text{total moles}} \approx \frac{32.00}{32.00 + 28.02} \approx 0.231 \]\[ \text{Mole fraction of } \mathrm{N}_2 = \frac{\text{moles of } \mathrm{N}_2}{\text{total moles}} \approx \frac{28.02}{32.00 + 28.02} \approx 0.769 \]

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry and physics that relates the pressure, volume, number of moles, and temperature of a gas. It is mathematically expressed as \( PV = nRT \), where:

• \( P \) is the pressure of the gas
• \( V \) is the volume of the gas
• \( n \) is the number of moles of gas
• \( R \) is the ideal gas constant (0.0821 L atm K\(^{-1}\) mol\(^{-1}\) is commonly used)
• \( T \) is the temperature in Kelvin

To make sure the calculations with the Ideal Gas Law are accurate, always convert the temperature into Kelvin using the formula: \( T(K) = T(°C) + 273.15 \). Similarly, when working with different units of pressure, ensure a conversion to atm for consistency, such as converting from mmHg to atm using \( 1 \, \text{atm} = 760 \, \text{mmHg} \).

In this exercise, the Ideal Gas Law is used to determine the molar volume of air, which is an important step to find the average molar mass and involves rearranging the equation to \( V_m = \frac{RT}{P} \). This helps calculate the amount of space one mole of the gas occupies under given conditions of pressure and temperature.

Mole Fraction

The concept of mole fraction deals with the proportion of individual gas components in a mixture. Mole fraction is a useful way to describe the composition of a gas mixture because it represents the ratio of the moles of one component to the total number of moles in the entire mixture. The mole fraction is symbolized by \( x \) and can be calculated using the formula:

• \( x_i = \frac{n_i}{n_{\text{total}}} \)

where \( n_i \) is the moles of a specific component and \( n_{\text{total}} \) is the total moles of all components.

In the context of the air mixture assumed, made up of 21% \( \mathrm{O}_2 \) and 79% \( \mathrm{N}_2 \), we calculated the mole fraction for each gas. It is essential because it provides the ability to understand the behavior of each gas in a mixture. While oxygen and nitrogen naturally have different properties, knowing their mole fraction helps predict their combined behavior at high altitudes.

The calculated mole fraction values \( \approx 0.231 \) for \( \mathrm{O}_2 \) and \( \approx 0.769 \) for \( \mathrm{N}_2 \), show the prevalence of nitrogen in air even at higher altitudes.

Molar Mass

Molar mass is a measure of the mass of one mole of a substance. It is expressed in g/mol and is crucial for converting between the mass of a substance and the amount in moles. The molar mass of a compound is determined by summing the molar masses of its constituent elements, based on the periodic table.

For instance, in calculating the molar mass of air, an important parameter is to use the weighted average of the molar masses of its components, taking into account their proportion in the mixture. The exercise exploits the known approximate composition of air:\( \mathrm{O}_2 = 32.00 \text{ g/mol} \) and \( \mathrm{N}_2 = 28.02 \text{ g/mol} \).

Using reasonable assumptions about the mixture, the average molar mass is calculated with a formula:

• \( \text{Average Molar Mass} = 0.21 \times 32.00 \text{ + } 0.79 \times 28.02 \)

The molar mass is then approximately \( 28.84 \text{ g/mol} \), indicating how closely the calculated value reflects the real situation at 20 km above the Earth's surface. This accurate estimation is key in diverse applications, such as in atmospheric science and engineering calculations.