Problem 62
Question
Suppose that on another planet the atmosphere consists of \(10 \% \mathrm{Kr}, 40 \% \mathrm{CH}_{4},\) and \(50 \% \mathrm{O}_{2}\). What is the average molar mass at the surface? What is the average molar mass at an altitude at which all the \(\mathrm{O}_{2}\) is photodissociated?
Step-by-Step Solution
Verified Answer
The average molar mass at the surface is 30.78 g/mol; at altitude, it is 29.56 g/mol.
1Step 1: Understanding the Components at Surface
The atmosphere on the planet consists of Krypton (Kr), Methane (CH\(_4\)), and Oxygen (O\(_2\)). By percentage, the composition is 10% Kr, 40% CH\(_4\), and 50% O\(_2\).
2Step 2: Determine the Molar Mass of Each Component
The molar mass of Kr is approximately 83.8 g/mol, CH\(_4\) is approximately 16.0 g/mol, and O\(_2\) is 32.0 g/mol.
3Step 3: Calculate Average Molar Mass at the Surface
The average molar mass at the surface is calculated using the formula: \[ \text{Average Molar Mass} = (0.1 \times 83.8) + (0.4 \times 16.0) + (0.5 \times 32.0) \] After calculation: \[ = 8.38 + 6.4 + 16.0 = 30.78 \text{ g/mol} \]
4Step 4: Atmospheric Composition at Altitude
At the altitude where all the O\(_2\) is photodissociated, we are left with only Kr and CH\(_4\) in their previous proportions of 20% Kr and 80% CH\(_4\) (as these take over the percentage previously held by O\(_2\)).
5Step 5: Molar Mass Calculation at Altitude
The average molar mass at this altitude can be calculated by: \[ \text{Average Molar Mass} = (0.2 \times 83.8) + (0.8 \times 16.0) \] After calculation: \[ = 16.76 + 12.8 = 29.56 \text{ g/mol} \]
6Step 6: Conclusion: Results Interpretation
The average molar mass at the surface is 30.78 g/mol, while at the altitude where all O\(_2\) is photodissociated, it is 29.56 g/mol.
Key Concepts
Atmospheric CompositionMolar Mass CalculationPhotodissociation
Atmospheric Composition
The atmosphere of a planet is made up of various gases. The composition and proportions of these gases influence the properties of the atmosphere like its molar mass. For example, the planet mentioned in the exercise has an atmosphere composed of
- 10% Krypton (Kr)
- 40% Methane (CH4)
- 50% Oxygen (O2)
Molar Mass Calculation
Molar mass calculation is the process of determining the mass of one mole of a substance. It's an essential concept when studying chemistry and helps understand the properties of gases in an atmosphere. To calculate the average molar mass of a mixture, the formula used is:\[ \text{Average Molar Mass} = \sum (\text{Fraction of Gas} \times \text{Molar Mass of Gas}) \]In the example provided, you learn how this calculation works with a mixture of Kr, CH4, and O2. At the planet's surface, the average molar mass is calculated by:
- Multiplying the percentage of each gas by its molar mass
- Summing up these values
Photodissociation
Photodissociation refers to the process in which molecules are broken down into smaller components due to absorption of light. In atmospheres, this process commonly affects gases like O2. The exercise illustrates how O2 can be photodissociated at certain altitudes, altering the atmospheric composition. As the O2 is dissociated, only Kr and CH4 are left, taking up its previous percentages. Photodissociation is influenced by wavelengths of light, as certain components absorb specific wavelengths, leading to more or less photodissociation. The change from 50% O2 to none affects the molar mass which decreases to 29.56 g/mol at that altitude. Comprehending photodissociation helps in understanding the dynamic nature of atmospheres and how solar radiation impacts planetary environments.
Other exercises in this chapter
Problem 60
In the following three instances, which choice is greener in a chemical process? Explain. (a) A reaction that can be run at \(350 \mathrm{~K}\) for \(12 \mathrm
View solution Problem 61
A friend of yours has seen each of the following items in newspaper articles and would like an explanation: (a) acid rain, \((\mathbf{b})\) greenhouse gas, \((\
View solution Problem 65
What properties of CFCs make them ideal for various commercial applications but also make them a long-term problem in the stratosphere?
View solution Problem 68
Explain, using Le Châtelier's principle, why the equilibrium constant for the formation of \(\mathrm{NO}\) from \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) increa
View solution