Problem 36
Question
The atmosphere of Mars is mostly \(\mathrm{CO}_{2}\) (molar mass 44.0 \(\mathrm{g} / \mathrm{mol} )\) under a pressure of 650 \(\mathrm{Pa}\) , which we shall assume remains constant. In many places the temperature varies from \(0.0^{\circ} \mathrm{C}\) in summer to \(-100^{\circ} \mathrm{C}\) in winter. Over the course of a martian year, what are the ranges of \((\mathrm{a})\) the rms speeds of the \(\mathrm{CO}_{2}\) molecules, and (b) the density (in mollm') of the atmosphere?
Step-by-Step Solution
Verified Answer
Rms speeds range from 328 m/s to 393 m/s, and density ranges from 0.288 to 0.450 mol/m³.
1Step 1: Understand the Given Data
We have the following information:- Molar mass of \( \mathrm{CO}_2 \) is 44.0 g/mol.- Pressure of the \( \mathrm{CO}_2 \) is 650 Pa (constant).- Temperature ranges from \( 0.0^{\circ} \mathrm{C} \) to \( -100^{\circ} \mathrm{C} \).
2Step 2: Convert Temperatures to Kelvin
Convert both temperatures to Kelvin:- Summer: \( 0.0^{\circ}C = 273.15 \, \mathrm{K} \)- Winter: \( -100^{\circ}C = 173.15 \, \mathrm{K} \)
3Step 3: Calculate RMS Speeds
The rms speed \( v_{rms} \) of gas molecules is calculated using the formula:\[ v_{rms} = \sqrt{\frac{3kT}{m}} \]where \( k \) is the Boltzmann constant \( (1.38 \times 10^{-23} \, \mathrm{J/K}) \), \( T \) is the temperature in Kelvin, and \( m \) is the mass of one molecule.- Summer: \( m = \frac{44.0}{1000} \times \frac{1}{6.022 \times 10^{23}} \approx 7.3 \times 10^{-26} \, \mathrm{kg} \)Thus, \( v_{rms}^{\text{summer}} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 273.15}{7.3 \times 10^{-26}} } \approx 393 \, \mathrm{m/s} \)- Winter: Calculate similarly for 173.15 K.\( v_{rms}^{\text{winter}} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 173.15}{7.3 \times 10^{-26}} } \approx 328 \, \mathrm{m/s} \)
4Step 4: Use Ideal Gas Law to Find Density
The density \( \rho \) can be found using the Ideal Gas Law: \( PV = nRT \), where \( R \) is the gas constant (8.314 J/mol·K) and \( n/V \) is the molar density.Rearranging for \( \rho \) gives:\[ \rho = \frac{P}{RT} \]- Summer: \( \rho_{\text{summer}} = \frac{650}{8.314 \times 273.15} \approx 0.288 \, \mathrm{mol/m^3} \)- Winter: \( \rho_{\text{winter}} = \frac{650}{8.314 \times 173.15} \approx 0.450 \, \mathrm{mol/m^3} \)
Key Concepts
rms speed calculationideal gas lawmolar masstemperature conversion to Kelvin
rms speed calculation
The root mean square (rms) speed of gas molecules is a crucial concept in the kinetic theory of gases. It represents the average speed of gas molecules in a sample, giving us insight into their kinetic energy. The formula for calculating the rms speed, \( v_{rms} \), is given by:
\[ v_{rms} = \sqrt{\frac{3kT}{m}} \]
where:
\[ v_{rms} = \sqrt{\frac{3kT}{m}} \]
where:
- \( k \) is the Boltzmann constant, valued at \( 1.38 \times 10^{-23} \text{ J/K} \).
- \( T \) is the temperature in Kelvin.
- \( m \) is the mass of one molecule of the gas.
ideal gas law
The Ideal Gas Law is a fundamental principle in the study of gases, articulated through the equation \( PV = nRT \). It correlates the pressure, volume, and temperature of an ideal gas with its quantity. Here's a breakdown of what each symbol signifies:
- \( P \) is the pressure of the gas.
- \( V \) is the volume the gas occupies.
- \( n \) is the number of moles of the gas.
- \( R \) is the universal gas constant, fixed at \( 8.314 \text{ J/mol⋅K} \).
- \( T \) is the absolute temperature in Kelvin.
molar mass
The molar mass of a substance refers to the mass of one mole of its molecules or atoms, often expressed in grams per mole (g/mol). For carbon dioxide \( (\text{CO}_2) \), the molar mass is \( 44.0 \text{ g/mol} \), which implies that one mole of carbon dioxide molecules weighs 44 grams.
Knowing the molar mass is crucial in various calculations, such as determining the mass of a single molecule. This calculation is done by converting grams to kilograms, dividing by Avogadro's number \( (6.022 \times 10^{23} \text{ mol}^{-1}) \), which gives the mass of one \( \text{CO}_2 \) molecule as approximately \( 7.3 \times 10^{-26} \text{ kg} \).
This significant factor enters the rms speed calculation, as it enables the transition from macroscopic molar mass to the microscopic mass of individual molecules. Understanding molar mass bridges the particle world with tangible calculations, allowing us to apply theoretical concepts to real-world scenarios effectively.
Knowing the molar mass is crucial in various calculations, such as determining the mass of a single molecule. This calculation is done by converting grams to kilograms, dividing by Avogadro's number \( (6.022 \times 10^{23} \text{ mol}^{-1}) \), which gives the mass of one \( \text{CO}_2 \) molecule as approximately \( 7.3 \times 10^{-26} \text{ kg} \).
This significant factor enters the rms speed calculation, as it enables the transition from macroscopic molar mass to the microscopic mass of individual molecules. Understanding molar mass bridges the particle world with tangible calculations, allowing us to apply theoretical concepts to real-world scenarios effectively.
temperature conversion to Kelvin
Temperature plays a pivotal role in physical science, and its correct measurement is essential for precise calculations in kinetics and thermodynamics. Every temperature scale serves different purposes, but the Kelvin scale is particularly suitable for scientific purposes.
To convert from Celsius to Kelvin, simply add 273.15 to the Celsius degree value. For example, Mars' summer temperature of \( 0.0^{\circ}C \) converts to \( 273.15 \text{ K} \), and the winter temperature of \( -100^{\circ}C \) becomes \( 173.15 \text{ K} \).
This conversion is vital for using the Ideal Gas Law and calculating the rms speeds because Kelvin is the unit for thermodynamic temperature in these contexts. Ensuring temperatures are in Kelvin allows for consistent and accurate use of formulas across various gas laws and kinetic energy calculations, minimizing errors that arise from temperature discrepancies.
To convert from Celsius to Kelvin, simply add 273.15 to the Celsius degree value. For example, Mars' summer temperature of \( 0.0^{\circ}C \) converts to \( 273.15 \text{ K} \), and the winter temperature of \( -100^{\circ}C \) becomes \( 173.15 \text{ K} \).
This conversion is vital for using the Ideal Gas Law and calculating the rms speeds because Kelvin is the unit for thermodynamic temperature in these contexts. Ensuring temperatures are in Kelvin allows for consistent and accurate use of formulas across various gas laws and kinetic energy calculations, minimizing errors that arise from temperature discrepancies.
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