Question

The density of air \(20 \mathrm{km}\) above Earth's surface is \(92 \mathrm{g} / \mathrm{m}^{3} .\) The pressure of the atmosphere is \(42 \mathrm{mm} \mathrm{Hg}\) and the temperature is \(-63^{\circ} \mathrm{C}\) (a) What is the average molar mass of the atmosphere at this altitude? (b) If the atmosphere at this altitude consists of only \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2},\) what is the mole fraction of each gas?

Short answer

(a) Molar mass is approximately 28.95 g/mol. (b) Mole fraction of \( O_2 \) is 0.2375; \( N_2 \) is 0.7625.

Step-by-step solution

Convert Temperature to Kelvin

The temperature needs to be converted to Kelvin before using it in calculations. The formula to convert from Celsius to Kelvin is: \( T(K) = T(°C) + 273.15 \). So, \( T = -63 + 273.15 = 210.15 \ K \).

Convert Pressure to Pascals

Pressure in mm Hg needs to be converted to Pascals for SI units consistency. \( 1 \, mm \, Hg = 133.322 \, Pa \). Thus, \( 42 \, mm \, Hg = 42 \, \times 133.322 \, Pa = 5600.524 \, Pa \).

Use Ideal Gas Law for Molar Mass

Apply the ideal gas law rearranged for molar mass \( M \): \( P = \frac{{dRT}}{M} \) rearranges to \( M = \frac{{dRT}}{P} \). Substitute \( d = 92 \, g/m^3 \), \( R = 8.3145 \, J/(mol\, K) \), \( T = 210.15 \, K \), \( P = 5600.524 \, Pa \). Calculate \( M = \frac{{92 \times 8.3145 \times 210.15}}{5600.524} \approx 28.95 \, g/mol \).

Calculate Mole Fraction of Each Gas

Assume the atmosphere is composed of \(x\) moles of \(O_2\) \((32 \, g/mol)\) and \(1-x\) moles of \(N_2\) \((28 \, g/mol)\) such that \( M = 28.95 \, g/mol = x \times 32 + (1-x)\times 28 \). Solving gives \( 28.95 = 32x + 28 - 28x \), simplifying to \( 4x = 0.95 \) yields \( x = 0.2375 \). Thus, the mole fraction of \( O_2 = 0.2375 \) and \( N_2 = 0.7625 \).

Validate Solution

Check that the mole fractions sum to 1: \( 0.2375 + 0.7625 = 1 \). This confirms the correctness of the mole fraction calculations.

Key concepts

Understanding Mole Fraction

The concept of mole fraction is useful when we want to determine the proportion of a particular gas in a mixture. Simply put, mole fraction is a way to express how many moles of a particular substance you have relative to the total number of moles in the mixture.
To calculate the mole fraction of a gas, divide the number of moles of that gas by the total number of moles of all gases present.

• For example, if you have a mixture containing 2 moles of oxygen and 3 moles of nitrogen, the mole fraction of oxygen would be the number of moles of oxygen divided by the total moles—i.e., 2/(2+3) = 0.4.
• The sum of all mole fractions in a mixture equals 1, making it a useful tool for checking calculations.

Molar Mass in Ideal Gases

Molar mass is an essential concept when dealing with gases, particularly through the ideal gas law. The molar mass is the mass of one mole of a substance and is expressed in grams per mole (g/mol).
When a gas is a mixture, as is the case in our air, calculating an average molar mass involves considering the molar masses of all individual gases, weighted by their respective mole fractions.
For example, if a gas mixture is composed of oxygen (32 g/mol) and nitrogen (28 g/mol), and their respective mole fractions are known, we can find the average molar mass.

• The formula used here is: \( M = x imes M_{O_2} + (1-x) imes M_{N_2} \), where \( x \) is the mole fraction of \( O_2 \).
• Assuming different mole fractions provides an average molar mass that suits precise calculations in the context, such as atmospheric layers.

Converting Temperature for Calculations

Temperature conversion is often necessary in chemistry, particularly when dealing with gas laws that require temperature inputs in Kelvin. Most temperatures encountered are in degrees Celsius, necessitating a conversion to Kelvin using the formula: \[ T(K) = T(^\circ C) + 273.15 \]
This conversion ensures that the temperature is in an absolute scale appropriate for thermodynamic calculations.
Using Kelvin is crucial because it follows an absolute scale where 0 K is absolute zero, the theoretical point where particles have minimum thermal motion.

• For example, to convert -63°C to Kelvin, simply add 273.15 to the Celsius temperature, resulting in 210.15 K.
• Always ensure the conversion step is part of any chemical calculation involving temperature to avoid errors.

Pressure Conversion in Gas Calculations

Converting pressure units is a common necessity in scientific calculations, especially when applying the ideal gas law. Pressure is often measured in mm Hg (millimeters of mercury) when dealing with atmospheric studies, but SI units require Pascals (Pa).
The conversion between these units can be achieved using the relation: \[ 1 ext{ mm Hg} = 133.322 ext{ Pa} \] Therefore, to convert 42 mm Hg to Pascals, multiply 42 by 133.322, resulting in 5600.524 Pa.

• Always adhere to consistent units when performing ideal gas law calculations to ensure accuracy.
• Being consistent with SI units allows seamless transition between different scientific disciplines and formulas.