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Chapter 11: Problem 69

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

Expert verified
(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

01

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: \( T(K) = T(掳C) + 273.15 \). So, \( T = -63 + 273.15 = 210.15 \ K \).
02

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, \( 42 \, mm \, Hg = 42 \, \times 133.322 \, Pa = 5600.524 \, Pa \).
03

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 \( M = \frac{{92 \times 8.3145 \times 210.15}}{5600.524} \approx 28.95 \, g/mol \).
04

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 \( N_2 = 0.7625 \).
05

Validate Solution

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈥攊.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掳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.

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Most popular questions from this chapter

Chlorine trifluoride, \(\mathrm{CIF}_{3}\), is a valuable reagent because it can be used to convert metal oxides to metal fluorides: \(6 \mathrm{NiO}(\mathrm{s})+4 \mathrm{ClF}_{3}(\mathrm{g}) \rightarrow 6 \mathrm{NiF}_{2}(\mathrm{s})+2 \mathrm{Cl}_{2}(\mathrm{g})+3 \mathrm{O}_{2}(\mathrm{g})\) (a) What mass of NiO will react with CIF \(_{3}\) gas if the gas has a pressure of \(250 \mathrm{mm} \mathrm{Hg}\) at \(20^{\circ} \mathrm{C}\) in a \(2.5-\mathrm{L}\) flask? (b) If the CIF a described in part (a) is completely consumed, what are the partial pressures of \(\mathrm{Cl}_{2}\) and of \(\mathrm{O}_{2}\) in the 2.5 -I. flask at \(20^{\circ} \mathrm{C}\) (in \(\mathrm{mm}\) Hg)? What is the total pressure in the flask?

If you have a sample of water in a closed container, some of the water will evaporate until the pressure of the water vapor, at \(25^{\circ} \mathrm{C},\) is \(23.8 \mathrm{mm}\) Hg. How many molecules of water per cubic centimeter exist in the vapor phase?

Forty miles above Earth's surface, the temperature is \(250 \mathrm{K},\) and the pressure is only \(0.20 \mathrm{mm}\) Hg. What is the density of air (in grams per liter) at this altitude? (Assume the molar mass of air is \(28.96 \mathrm{g} / \mathrm{mol}\) )

One of the cylinders of an automobile engine has a volume of \(400 . \mathrm{cm}^{3} .\) The engine takes in air at a pressure of 1.00 atm and a temperature of \(15^{\circ} \mathrm{C}\) and compresses the air to a volume of \(50.0 \mathrm{cm}^{3}\) at \(77^{\circ} \mathrm{C}\). What is the final pressure of the gas in the cylinder? (The ratio of before and after volumes-in this case, 400: 50 or \(8: 1-\) is called the compresion ratio.)

A steel cylinder holds \(1.50 \mathrm{g}\) of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\). What is the pressure of the ethanol vapor if the cylinder has a volume of \(251 \mathrm{cm}^{3}\) and the temperature is \(250^{\circ} \mathrm{C} ?\) (Assume all of the ethanol is in the vapor phase at this temperature.)
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