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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 \(\left.\mathrm{CO}_{2}\right)\). What is the mole fraction of \(\mathrm{CO}_{2}\) in the atmosphere?

Short Answer

Expert verified
The mole fraction of COâ‚‚ in the atmosphere can be calculated using the simplified equation \(n=\frac{V}{V_{atm}}\). Plugging in the given values, we get: \(n = \frac{390\,\text{L}}{10^6\,\text{L}}\). The mole fraction of COâ‚‚ is equal to the ratio of moles of COâ‚‚ to the total number of moles in the atmosphere: \(\chi_{COâ‚‚} = \frac{n_{COâ‚‚}}{n_{air} + n_{COâ‚‚}}\). Since the sum of all mole fractions is 1, we can solve for \(\chi_{COâ‚‚}\) by plugging in the values: \(\chi_{COâ‚‚} = \frac{0.00039\,\text{L}}{1\,\text{L}}\). Thus, the mole fraction of COâ‚‚ in the atmosphere is approximately \(3.9 \times 10^{-4}\).

Step by step solution

01

Convert ppm concentration to Liters

The given concentration of COâ‚‚ is 390 ppm. In this case, ppm is by volume, so we need to convert it to liters. It means that for every 10^6 L of the atmosphere, there are 390 L of COâ‚‚.
02

Calculate moles of COâ‚‚

Now use the ideal gas law to calculate the moles of COâ‚‚ from the given volume. The ideal gas law equation is: \(PV = nRT\) Where: - P = pressure of the gas - V = volume of the gas - n = moles of the gas - R = universal gas constant (\(8.314 \frac{J}{mol*K}\)) - T = temperature of the gas (in Kelvin) Considering that we only need the mole fraction, we can simplify. Since the pressure and temperature are the same in the atmosphere and COâ‚‚, we can use a simplified equation: \(n=\frac{V}{V_{atm}}\) Where: - n = moles of COâ‚‚ - V = volume of COâ‚‚ - \(V_{atm}\) = total volume of the atmosphere Therefore, \(n = \frac{390\,\text{L}}{10^6\,\text{L}}\)
03

Calculate the mole fraction of COâ‚‚

Now that we have the moles of COâ‚‚, we can find the mole fraction of COâ‚‚. The mole fraction is the ratio of the number of moles of COâ‚‚ to the total number of moles of the atmosphere. Assuming the atmosphere contains only COâ‚‚ and air, the mole fraction is: \(\chi_{COâ‚‚} = \frac{n_{COâ‚‚}}{n_{air} + n_{COâ‚‚}}\) We have calculated \(n = \frac{390\,\text{L}}{10^6\,\text{L}}\), and it is important to note that the sum of the mole fractions for all components in the atmosphere is equal to 1: \(1 = \frac{n_{COâ‚‚}}{n_{air} + n_{COâ‚‚}}\) We can now replace n with the value we found and solve for the mole fraction of COâ‚‚: \(\chi_{COâ‚‚} = \frac{\frac{390\,\text{L}}{10^6\,\text{L}}}{1}\)
04

Calculate the final mole fraction

Now simply plug the value of n into the equation and solve for \(\chi_{COâ‚‚}\): \(\chi_{COâ‚‚} = \frac{0.00039\,\text{L}}{1\,\text{L}}\) Therefore, the mole fraction of COâ‚‚ in the atmosphere is \(3.9 \times 10^{-4}\) (approximately).

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a cornerstone principle in atmospheric chemistry. This law provides a simple equation which relates several important properties of a gas: pressure (P), volume (V), moles (n), the universal gas constant (R), and temperature (T) in Kelvin. Expressed mathematically, the Ideal Gas Law is given by: \[PV = nRT\] This equation allows us to understand how gases behave under different conditions. When using this equation to find moles of a gas like COâ‚‚ from a given volume, simplifying assumptions can be made if the pressure and temperature are constant.

These constants enable us to rearrange the equation to find moles in terms of volume. For example, when finding the mole fraction of COâ‚‚ in the atmosphere, we consider COâ‚‚ as a fraction of the total volume of air. This highlights why understanding the Ideal Gas Law is crucial for analyzing atmospheric components and their concentrations.
Mole Fraction
The mole fraction is a way to express the composition of a mixture of gases. It tells us the proportion of one particular gas in comparison to others in a mixture. In the context of atmospheric chemistry, the mole fraction of a gas like carbon dioxide (COâ‚‚) allows us to quantify its presence in the air.

The formula for calculating the mole fraction (\chi) of a particular gas in a mixture is: \[\chi_{\text{COâ‚‚}} = \frac{n_{\text{COâ‚‚}}}{n_{\text{total}}}\] where n_{\text{COâ‚‚}} and n_{\text{total}} are the moles of COâ‚‚ and the total moles of all gases in the atmosphere, respectively. The sum of all mole fractions in a mixture equals 1, reflecting that they make up the complete mixture. Understanding mole fractions is critical as it provides a clear, ratio-based method to compare the presence of individual gases with the total atmospheric composition.
Carbon Dioxide Concentration
Understanding carbon dioxide (COâ‚‚) concentration in the atmosphere is vital due to its influence on climate change and global warming. The concentration of COâ‚‚ is often expressed in parts per million (ppm), which means the number of COâ‚‚ molecules in one million molecules of air. In the exercise, 390 ppm suggests that there are 390 molecules of COâ‚‚ for every million molecules of the atmosphere.

The ppm measurement is a useful way to express low concentrations for atmospheric studies. Still, converting ppm to the mole fraction can provide more insights into the role of COâ‚‚ in air. For instance, to calculate the mole fraction based on a ppm value, one assumes that the atmosphere behaves ideally, simplifying the conversion using the given ppm values and total volume assumptions. This conversion helps to model and predict the behavior of COâ‚‚ under different atmospheric conditions, aiding in the analysis of its impact on environmental and climate dynamics.

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Most popular questions from this chapter

Consider a mixture of two gases, \(A\) and \(B\), confined in a closed vessel. A quantity of a third gas, \(C,\) is added to the same vessel at the same temperature. How does the addition of gas \(\mathrm{C}\) affect the following: (a) the partial pressure of gas \(A,\) (b) the total pressure in the vessel, \((\mathbf{c})\) the mole fraction of gas \(\mathrm{B} ?\)

Table 10.3 shows that the van der Waals \(b\) parameter has units of \(\mathrm{L} / \mathrm{mol}\). This implies that we can calculate the size of atoms or molecules from \(b\). Using the value of \(b\) for Xe, calculate the radius of a Xe atom and compare it to the value found in Figure 7.6, \(1.30 \AA\) A. Recall that the volume of a sphere is \((4 / 3) \pi r^{3}\).

The temperature of a 5.00-L container of \(\mathrm{N}_{2}\) gas is increased from \(20^{\circ} \mathrm{C}\) to \(250^{\circ} \mathrm{C}\). If the volume is held constant, predict qualitatively how this change affects the following: (a) the average kinetic energy of the molecules; (b) the root-mean-square speed of the molecules; (c) the strength of the impact of an average molecule with the container walls; (d) the total number of collisions of molecules with walls ner second.

On a single plot, qualitatively sketch the distribution of molecular speeds for \((\mathbf{a}) \operatorname{Kr}(g)\) at \(-50^{\circ} \mathrm{C},(\mathbf{b}) \operatorname{Kr}(g)\) at \(0^{\circ} \mathrm{C},(\mathbf{c}) \operatorname{Ar}(g)\) at \(0{ }^{\circ} \mathrm{C}\). [Section \(\left.10.7\right]\)

How does a gas compare with a liquid for each of the following properties: (a) density, (b) compressibility, (c) ability to mix with other substances of the same phase to form homogeneous mixtures, \((\mathrm{d})\) ability to conform to the shape of its container?

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