- #1

roam

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## Homework Statement

http://img191.imageshack.us/img191/3277/59596364.jpg## The Attempt at a Solution

I used the Maxwell-Boltzmann distribution equation:

[itex]KE_{avg}=\frac{1}{2}mv^2 = \frac{3}{2}k T[/itex]

[itex]\implies v = \sqrt{\frac{3kT}{m}}[/itex]

The most probable velocity for argon would be:

[itex]v = \sqrt{\frac{3\times (1.38 \times 10^{-23}) \times 293.15 \ K}{40}} = 3.034 \times 10^{-22}\ m/s[/itex]

And for water vapour at (not sure if I need to modify the equation):

[itex]v = \sqrt{\frac{3\times (1.38 \times 10^{-23}) \times 293.15 \ K}{18}} = 6.74 \times 10^{-22} \ m/s[/itex]

Is this correct? How do we get to use the info we were given about the molecular radius and the pressure (at 1 atm)?

And do we expect them to have collision rates? Personally I guess water vapour molecules have a higher collision rate because they have a higher velocity and thus collide more often. How do we make the actual calculation?

I appreciate any help and suggestions.

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