Mean free path and speed as a function of pressure and temp

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SUMMARY

The discussion focuses on the effects of altitude on the speed distribution and mean free path (mfp) of air molecules. At 18,000 ft, the atmospheric pressure is approximately 380 Torr, which is half of sea level pressure (760 Torr), and the temperature decreases by 70°C. The speed distribution can be calculated using the Maxwell-Boltzmann distribution formula, while the mean free path can be determined using the equation λmfp = (kT)/(πd²P). The analysis concludes that the decrease in pressure increases the mean free path, with calculations suggesting a ratio of approximately 0.66 for mfp under the new conditions.

PREREQUISITES
  • Understanding of the Maxwell-Boltzmann speed distribution
  • Familiarity with the ideal gas law and its implications
  • Knowledge of the mean free path concept in kinetic theory
  • Basic proficiency in thermodynamics and gas properties
NEXT STEPS
  • Study the derivation and implications of the Maxwell-Boltzmann distribution
  • Learn how to apply the ideal gas law at varying altitudes and temperatures
  • Explore the relationship between pressure, temperature, and mean free path in gases
  • Investigate the effects of altitude on air density and molecular behavior
USEFUL FOR

This discussion is beneficial for students in physics or engineering, particularly those studying thermodynamics, atmospheric science, or kinetic theory. It is also relevant for researchers analyzing gas behavior under varying environmental conditions.

jhartc90
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Homework Statement



I know that at 18,000 ft. above mean sea level, the atm. pressure ~half of what it is at seal level (760 Torr). The temperature also decreases by 70C.

a. How would this change the speed distribution of the air molecules quanitatively?
b.How would it change the mean free path of the air molecules quantitatively?

Homework Equations


For a) all I can seem to find is this:
$$ c=\int_0^\infty vP(v)dv=(8kT/\pi m)^{1/2}$$

But since I am interested in the speed distribution, should I use:

$$P(v)=4\pi [\frac{m}{2 \pi kT}]^{3/2} (v^2) exp(\frac{-mv^2}{2kt})$$

For b) all I can seem to find is this:

$$\lambda_{mfp}=\frac{kt}{\pi d^2 P (2)^{1/2}}$$

Since the pressure is half, that would seem to double the mfp correct? and then i need to take into account the change in temp?

Thus the change would be

$$\frac{T1}{T2} \frac{P2}{P1}$$
is this correct?

The Attempt at a Solution



For b) If I plug the numbers in I get like 0.66 which to me, might be correct. This is the ratio of mfp_1 to mfp_2 (lower pressure and temperature). Thus 0.66 * mfp_1 / mfp_2 = 0.65 so mfp_2 is larger (which I would expect)

I'm not sure if my approach to b) is correct, and for a), I am not very sure how to solve it. Do I just plug the temperatures into the equation and see the difference, noting that v also depends on temperature?
 
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b: correct.

a: I'm not sure what else you can do quantitatively apart from writing down the formula. The distribution changes as given by this formula.
 

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