1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Mean free path and speed as a function of pressure and temp

  1. Feb 22, 2016 #1
    1. The problem statement, all variables and given/known data

    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?




    2. Relevant 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?

    3. 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), Im 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?
     
  2. jcsd
  3. Feb 22, 2016 #2

    mfb

    User Avatar
    2017 Award

    Staff: Mentor

    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.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Loading...