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Questions and Answers
What is the mean free path of a nitrogen molecule at 0 °C and 1.0 atm pressure?
What happens to the mean free path if the pressure of a gas is reduced?
What is the molecular diameter of nitrogen, given in the problem?
Which equation is used to express the mean free path in terms of pressure and molecular diameter?
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What is the role of the universal gas constant, kB, in calculating the mean free path?
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Study Notes
Mean Free Path Calculation of Nitrogen Molecule
- Determine the mean free path (( \lambda )) for nitrogen at 0 °C and 1.0 atm.
- Conditions:
- Temperature (T): 0 °C = 273 K
- Pressure (P): 1.0 atm = 1.01 × 10⁵ Pa
- Molecular diameter (d): 324 pm = 324 × 10⁻¹² m
- Equation used for ideal gases:
- ( PV = Nk_B T )
- ( \frac{N}{V} = \frac{P}{k_B T} )
- Mean free path formula:
- ( \lambda = \frac{k_B T}{\sqrt{2 \pi d^2 P}} )
- Constants used:
- Boltzmann's constant (( k_B )): 1.38 × 10⁻²³ J/K
- Calculation progression:
- Substitute values into ( \lambda ) formula:
- ( \lambda = \frac{(1.38 × 10^{-23} J/K)(273 K)}{\sqrt{2 \pi (324 × 10^{-12} m)^2 (1.01 × 10^5 Pa)}} )
- Substitute values into ( \lambda ) formula:
- Result:
- ( \lambda \approx 0.8 × 10^{-7} m )
- Comparison to molecular diameter:
- Mean free path is approximately 247 times the molecular diameter of nitrogen.
Effects of Pressure on Mean Free Path
- Lowering gas pressure leads to reduced gas density.
- As density decreases, mean free path increases.
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Description
This quiz focuses on calculating the mean free path of nitrogen molecules at specific temperature and pressure conditions. Using the ideal gas law and provided molecular data, students will derive the mean free path. Test your understanding of gas behaviors and molecular interactions!
