0
$\begingroup$

So when learning about the kinetic model of gases, my teacher said that in a volume $A*v_{x}*\Delta t$, exactly half the molecules will have a velocity component of $v_x$ and the other half will have a velocity component of $-v_x$. But I think they are wrong!

Because there should be some small portion of gas molecules that will have a velocity component of: $v_x = 0$

Am I correct? So is the term for the average number of collisions with the wall during the time interval $\Delta t$, just an approximation?:

$(0.5)*A*v_{x}*\Delta t*n*N_a/V$

This is clear with an example. If there are three molecules in a container, one moving left on the x axis, one moving right on the x axis, and one moving directly up the y axis, then on average, only 1/3 of the molecules is moving towards the right wall.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

You have pointed out the problem of how to handle the infinitesimal in probability theory. So yes a molecule could traveling in the $yz$ plane and have 0 $x$ velocity. However the probability of the motion being in exactly the $yz$ plane can be made as small as you like.

Let's make the problem one dimensional and choose a random angle between 0 and 359.999... degrees. So the question is what is the probability that the chosen angle will be for a vector on the x-axis?

  • To a degree if I chose 0 or 180 then the vector will be on the x-axis. Odds 2/360

  • To a tenth of a degree if I chose 0.0 or 180.0 then the vector will be on the x-axis. Odds 2/3600

  • To a hundredth of a degree if I chose 0.00 or 180.00 then the vector will be on the x-axis. Odds 2/36000

and so on...

So what probability should be assigned to a randomly chosen angle being exactly along the x-axis?

Improve this answer
$\endgroup$
2
  • $\begingroup$ Are you basically arguing that the x-velocity component is best described by a continuous distribution and therefore it having an exact value of zero would be zero? $\endgroup$
    – Nova
    Commented Feb 14, 2017 at 22:26
  • $\begingroup$ @Nova - I do like that you had your thinking cap on. You really need to look for boundary conditions on the various chemistry factoids that get thrown around. $\endgroup$
    – MaxW
    Commented Feb 14, 2017 at 22:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.