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Chapter 10: Problem 65

The atmospheric concentration of \(\mathrm{CO}_{2}\) gas is presently 390 ppm (parts per million, by volume; that is, \(390 \mathrm{~L}\) of every \(10^{6} \mathrm{~L}\) of the atmosphere are \(\mathrm{CO}_{2}\) ). What is the mole fraction of \(\mathrm{CO}_{2}\) in the atmosphere?

Short Answer

Expert verified
The mole fraction of CO鈧� in the atmosphere is \(3.90 \times 10^{-4}\).

Step by step solution

01

Understanding the concept of ppm

Ppm stands for parts per million, which represents the concentration of a substance (CO鈧� in this case) in volume. Given that 390 L of every 1,000,000 L of the atmosphere is CO鈧�, we can write this concentration as a ratio: \[\frac{390\,\mathrm{L}}{1,000,000\,\mathrm{L}}\]
02

Converting the concentration ratio into moles

One mole of any gas occupies 22.4 L under standard conditions. To convert the concentration ratio into moles, divide each volume by the molar volume of 22.4 L: \[\frac{390\,\mathrm{L}}{22.4\,\mathrm{L/mol}} = 17.41\,\mathrm{mol}\] \[\frac{1,000,000\,\mathrm{L}}{22.4\,\mathrm{L/mol}} = 44,643.3\,\mathrm{mol}\]
03

Calculating the mole fraction

Now that we have the moles of CO鈧� and the total moles of the atmosphere, we can calculate the mole fraction, which is the ratio of the moles of CO鈧� to the total moles of the atmosphere: \[Mole\,fraction\,of\,CO_2 = \frac{17.41\,\mathrm{mol\,CO_2}}{44,643.3\,\mathrm{mol\,of\,atmosphere}}\]
04

Simplifying the expression

Divide 17.41 mol CO鈧� by 44,643.3 mol of atmosphere to calculate the mole fraction: \[Mole\,fraction\,of\,CO_2 = \frac{17.41}{44,643.3} = 3.90 \times 10^{-4}\] The mole fraction of CO鈧� in the atmosphere is \(3.90 \times 10^{-4}\).

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Key Concepts

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

ppm (parts per million)
Ppm, or parts per million, is a unit of measurement commonly used to express small concentrations of substances. It is especially useful in environmental science where such measurements are essential for monitoring pollution levels. Imagine ppm as a fraction of a whole: 1 ppm indicates that one unit of a substance is present in a million units of the total mixture. For instance, if the concentration of carbon dioxide (CO鈧�) in the atmosphere is 390 ppm, this means that out of every million liters of atmospheric air, there are 390 liters of CO鈧�.
This concept simplifies the expression of concentrations without using cumbersome fractions or percentages, which would not effectively capture the tiny amounts counted in many chemical or environmental measurements.
  • Easy to use for small quantities.
  • Commonly used in environmental measures to track pollutants.
Understanding ppm helps in quantifying the influence of gases like CO鈧� in the atmosphere, important for studying climate change impacts.
carbon dioxide concentration
Carbon dioxide concentration in the atmosphere is an important metric in determining the effects of this greenhouse gas on global climate. The concentration of CO鈧� is expressed in parts per million (ppm), which allows scientists to monitor even slight increases that may have significant impacts on global temperatures. Presently, a concentration like 390 ppm indicates that, out of every one million liters of air, 390 liters are CO鈧�.
This concentration can affect a number of environmental factors, such as how much heat from the sun is trapped in the Earth's atmosphere, contributing to global warming. It's crucial to understand, track, and predict these changes in CO鈧� concentration as they play a major role in climate change and efforts towards its mitigation.
  • Measured in ppm for precision in environmental assessments.
  • Indicator of the potential greenhouse effect and climate shifts.
By studying CO鈧� concentrations, we understand how human activities, such as burning fossil fuels, contribute to atmospheric changes.
gas molar volume
The concept of gas molar volume is central to understanding how the volume of a gas relates to the number of molecules present, under standard conditions. One mole of any gas occupies approximately 22.4 liters at standard temperature and pressure (STP), which are 0掳C and 1 atm pressure. This constant makes it easier to calculate the amount of gas present in a given volume, which is particularly useful in chemistry and environmental science.
For example, to convert atmospheric carbon dioxide from a volume (390 liters in 1 million liters of air) to a molar quantity, you divide the volume of CO鈧� by the molar volume (22.4 liters/mole).
  • 22.4 liters/mole simplifies calculations under STP.
  • Facilitates conversion from volume to moles, aiding in detailed analysis.
This conversion is key in determining the mole fraction, which further helps understand the share of CO鈧� in the atmosphere, improving our predictions and responses to environmental changes.

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Most popular questions from this chapter

A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required \(105 \mathrm{~s}\) for \(1.0 \mathrm{~L}\) of the gas to effuse. Under identical experimental conditions it required \(31 \mathrm{~s}\) for \(1.0 \mathrm{~L}\) of \(\mathrm{O}_{2}\) gas to effuse. Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of \(1.0 \mathrm{~L}\); in other words, rate is the amount that diffuses over the time it takes to diffuse.)

Which gas is most dense at \(1.00\) atm and \(298 \mathrm{~K}: \mathrm{CO}_{2}, \mathrm{~N}_{2} \mathrm{O}\), or \(\mathrm{Cl}_{2}\) ? Explain.

A scuba diver's tank contains \(0.29 \mathrm{~kg}\) of \(\mathrm{O}_{2}\) compressed into a volume of \(2.3 \mathrm{~L}\) (a) Calculate the gas pressure inside the tank at \(9^{\circ} \mathrm{C}\). (b) What volume would this oxygen occupy at \(26^{\circ} \mathrm{C}\) and \(0.95 \mathrm{~atm}\) ?

Chlorine is widely used to purify municipal water supplies and to treat swimming pool waters. Suppose that the volume of a particular sample of \(\mathrm{Cl}_{2}\) gas is \(8.70 \mathrm{~L}\) at 895 torr and \(24^{\circ} \mathrm{C}\). (a) How many grams of \(\mathrm{Cl}_{2}\) are in the sample? (b) What volume will the \(\mathrm{Cl}_{2}\) occupy at STP? (c) At what temperature will the volume be \(15.00 \mathrm{~L}\) if the pressure is \(8.76 \times 10^{2}\) torr? (d) At what pressure will the volume equal \(5.00 \mathrm{~L}\) if the temperature is \(58^{\circ} \mathrm{C}\) ?

Consider the sample of gas depicted here. What would the drawing look like if the volume and temperature remained constant while you removed enough of the gas to decrease the pressure by a factor of 2\(?[\) Section 10.3\(]\) \begin{equation}\begin{array}{l}{\text { (a) It would contain the same number of molecules. }} \\ {\text { (b) It would contain half as many molecules. }} \\\ {\text { (c) It would contain twice as many molecules. }} \\ {\text { (d) There is insufficient data to say. }}\end{array}\end{equation}
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