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Atmospheric scientists often use mixing ratios to express the concentrations of trace compounds in air. Mixing ratios are often expressed as ppmv (parts per million volume): $${\text {ppmv of}} \ X=\frac{\text { vol of } X \text { at STP }}{\text { total vol of air at STP }} \times 10^{6}$$ On a certain November day, the concentration of carbon monoxide in the air in downtown Denver, Colorado, reached \(3.0 \times 10^{2}\) ppmv. The atmospheric pressure at that time was 628 torr and the temperature was \(0^{\circ} \mathrm{C}\) . a. What was the partial pressure of CO? b. What was the concentration of CO in molecules per cubic meter? c. What was the concentration of CO in molecules per cubic centimeter?

Short Answer

Expert verified
The partial pressure of CO is \(0.00616 \space atm\). The concentration of CO in molecules per cubic meter is \(8.63 × 10^{19}\), and the concentration of CO in molecules per cubic centimeter is \(8.63 × 10^{13}\).

Step by step solution

01

Calculate the partial pressure of CO

To find the partial pressure of CO, we first need to calculate the total moles of the gas at the given pressure and temperature using the Ideal Gas Law. The Ideal Gas Law is given as: \(PV = nRT\) where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature. Here, T is given in Celsius. To change it to Kelvin, we just add 273 to the Celsius temperature: T = \( 0 + 273 = 273K \) We also need to convert the pressure from torr to atm: Pressure in atm = Pressure in torr / 760 Pressure in atm = \(628 / 760 \approx 0.8263 \) Now, rewrite the Ideal Gas Law equation for the number of moles and rearrange as follows: Moles (n) = \(P_{air}V / RT\) We are given mixing ratio in ppmv, which we can rewrite using the given formula, ppmv of CO = \( (n_{CO}) / (n_{total}) × 10^6 \) So we can rearrange this formula to find moles (n) of CO: Moles (n_{CO}) = \( ppmv_{CO} × n_{total} / 10^6 \) Now, to find the partial pressure of CO (P_{CO}), use the Ideal Gas Law again by rearranging the equation for pressure: Partial Pressure of CO (P_{CO}) = \( n_{CO}RT / V \) Plug in the values we found for the moles of CO and the total pressure: Partial Pressure of CO = \(3.0 × 10^2 \times 0.8263 \times (0.08206 \times 273) / 10^6\) Partial Pressure of CO = \(0.00616 \space atm\)
02

Convert concentration to molecules per cubic meter

To convert concentration to molecules per cubic meter, we will first find the number of moles of CO in 1 m^3 using the Ideal Gas Law: Moles of CO in 1 m^3 (n_{CO}) = \( (P_{CO} × 10^3) / RT \) Now, to convert moles to molecules, we can use Avogadro's number (6.022 × 10^23 molecules per mole): Molecules of CO in 1 m^3 = \( n_{CO} × (6.022 × 10^23) \) Plug in the values we found: Molecules of CO in 1 m^3 = \(((0.00616 \times 10^3) / (0.08206 \times 273)) × (6.022 × 10^23)\) Molecules of CO in 1 m^3 = \(8.63 × 10^{19} \)
03

Convert concentration to molecules per cubic centimeter

To convert concentration from molecules per cubic meter to molecules per cubic centimeter, we can use the conversion factor \( (1 m^3 = 10^6 cm^3) \) by dividing the molecules per cubic meter by the conversion factor: Molecules of CO per cm^3 = Molecules of CO per m^3 / 10^6 Plug in the value we found: Molecules of CO per cm^3 = \( 8.63 × 10^{19} / 10^{6} \) Molecules of CO per cm^3 = \( 8.63 × 10^{13} \) So the results are: a. The partial pressure of CO is \(0.00616 \space atm\). b. The concentration of CO in molecules per cubic meter is \(8.63 × 10^{19}\). c. The concentration of CO in molecules per cubic centimeter is \(8.63 × 10^{13}\).

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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 fundamental principle in chemistry and physics that describes the behavior of a gas in terms of pressure, volume, and temperature. This law is usually stated as \(PV = nRT\), where \(P\) is the pressure of the gas, \(V\) is its volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin.

To use the Ideal Gas Law effectively, temperatures must be converted to Kelvin because the law is based on absolute temperatures. You just add 273 to the Celsius temperature to convert it. For example, \(0^\circ C\) becomes 273 K.

If you're working with pressure in units other than atm, such as torr, you'll need to convert it. To convert from torr to atm, divide the pressure in torr by 760, since 1 atm = 760 torr.

This law helps us calculate unknown properties by using the known ones, making it essential for understanding gas behavior under different conditions.
Mixing Ratios
Mixing ratios are commonly used in atmospheric science to express the concentration of a particular component, like a trace gas, within a mixture. For gases, mixing ratios are often expressed in parts per million volume (ppmv). This unit tells us how many parts of a gas are present for every million parts of the total air volume.

The calculation for ppmv is:
  • \({\text {ppmv of}} \ X=\frac{\text {vol of } X \text { at STP }}{\text { total vol of air at STP }} \times 10^{6}\)
Where STP refers to standard temperature and pressure. This method is a concise way to refer to very small quantities of a substance, like pollutants or greenhouse gases, in the air.

Mixing ratios allow scientists to easily compare the concentration of gases despite changes in pressure and temperature. They're particularly useful when assessing air quality and pollution levels.
Partial Pressure
Partial pressure refers to the pressure that a specific gas in a mixture would exert if it occupied the entire volume alone at the same temperature. It's based on Dalton's Law of Partial Pressures, which states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of each individual component.

To find the partial pressure of a gas, you can use the formula:
  • \(P_{\text{gas}} = \frac{n_{\text{gas}}}{n_{\text{total}}} \times P_{\text{total}}\)
However, when using mixing ratios given in ppmv, you can integrate this ratio in your calculation as well.

Knowing a gas's partial pressure is crucial for understanding its impact on chemical reactions and physiological effects, like oxygen's role in respiration. It's also vital for industries using gas mixtures to ensure safety and efficiency.
Concentration Conversion
Converting gas concentration between different units is often necessary for standardization and interpretation. When converting from ppmv to molecules per cubic meter or cubic centimeter, you typically start with the Ideal Gas Law to find moles per unit volume.

Using Avogadro's number (\(6.022 \times 10^{23}\) molecules per mole), you can convert this to molecules. For instance, to convert concentration of CO at a certain partial pressure \(P\) and temperature \(T\):
  • Calculate moles in a cubic meter using \((P_{\text{gas}} \times 10^3) / RT\)
  • Multiply by Avogadro's number to find the number of molecules
For conversions to smaller units like cubic centimeters, simply divide by \(10^6\) since \(1 \space m^3 = 10^6 \space cm^3\).

This ensures compatibility of the data used in various fields like environmental science, where precise gas concentrations are crucial.

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Most popular questions from this chapter

Cyclopropane, a gas that when mixed with oxygen is used as a general anesthetic, is composed of 85.7\(\% \mathrm{C}\) and 14.3\(\% \mathrm{H}\) by mass. If the density of cyclopropane is 1.88 \(\mathrm{g} / \mathrm{L}\) at STP, what is the molecular formula of cyclopropane?

Consider a sample of a hydrocarbon (a compound consisting of only carbon and hydrogen) at 0.959 atm and 298 \(\mathrm{K}\). Upon combusting the entire sample in oxygen, you collect a mixture of gaseous carbon dioxide and water vapor at 1.51 atm and 375 \(\mathrm{K}\). This mixture has a density of 1.391 g/L and occupies a volume four times as large as that of the pure hydrocarbon. Determine the molecular formula of the hydrocarbon

The rate of effusion of a particular gas was measured and found to be 24.0 mL/min. Under the same conditions, the rate of effusion of pure methane \(\left(\mathrm{CH}_{4}\right)\) gas is 47.8 mL/min. What is the molar mass of the unknown gas?

A compound containing only \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{N}\) yields the following data: i. Complete combustion of 35.0 \(\mathrm{mg}\) of the compound produced 33.5 \(\mathrm{mg}\) of \(\mathrm{CO}_{2}\) and 41.1 \(\mathrm{mg}\) of \(\mathrm{H}_{2} \mathrm{O} .\) ii. A 65.2 -mg sample of the compound was analyzed for nitrogen by the Dumas method (see Exercise \(137 ),\) giving 35.6 \(\mathrm{mL}\) of dry \(\mathrm{N}_{2}\) at \(740 .\) torr and \(25^{\circ} \mathrm{C}\) . iii. The effusion rate of the compound as a gas was measured and found to be \(24.6 \mathrm{mL} / \mathrm{min}\). The effusion rate of argon gas, under identical conditions, is \(24.6 \mathrm{mL} / \mathrm{min}\). What is the molecular formula of the compound?

Consider a 1.0-L container of neon gas at STP. Will the average kinetic energy, average velocity, and frequency of collisions of gas molecules with the walls of the container increase, decrease, or remain the same under each of the following conditions? a. The temperature is increased to \(100^{\circ} \mathrm{C}\) . b. The temperature is decreased to \(-50^{\circ} \mathrm{C}\) c. The volume is decreased to 0.5 L . d. The number of moles of neon is doubled.

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