1
$\begingroup$

The problem I have with this proof is how to acquire the fraction $\frac{1}{3}$. In theory we were taught by taking into consideration a rectangular vessel which obviously has three paths which a particle can take when colliding between three faces.nevertheless we still apply this equation to spherical vessels. which have infinite equi-probable paths.

I'm convinced that there is a more generalized way of acquiring the constant(my guess is by integration but i'm clueless as to what is integrated )

Please describe the derivation clearly with a mathematical perspective.

$\endgroup$
  • $\begingroup$ I guess considering a 1d case with velocity $v_x$ and then generalizing it to 3d with velocity $v$ assuming uniform velocity in each direction such that $v^2=v_x^2+v_y^2+v_z^2 = 3v_x^2$ and now replacing $v_x^2 = v^2/3$ should do it. $\endgroup$ – exp ikx May 14 at 2:48
  • $\begingroup$ Elementary treatments of kinetic theory wave aside a lot of hard issues, of which this is the least. One of the biggest stinkers is showing that the averages you use in these procedures stay the same over time. It took Boltzmann about 60 pages though he was notorious for not bothering to tighten up his work, and instead shoving it at a journal as soon as he had a complete treatment. $\endgroup$ – dmckee May 14 at 17:57
2
$\begingroup$

(a) I won't reproduce a complete proof (See for example Jeans: Kinetic Theory of Gases), but will outline how the factor of 1/3 enters, even for a container of arbitrary shape.

If we consider a small 'patch' of the container wall, it is straightforward to show that the total momentum of molecules hitting that patch per unit time is proportional to $\overline {u^2},$ the molecules' mean square velocity component normal to the patch.

If $\overline {v^2}$ and $\overline {w^2}$ are the mean square velocity components in the two directions orthogonal to each other and to the normal, then we can easily show that the mean square speed of the molecules is given by $$\overline{c^2} = \overline{u^2} + \overline{v^2} + \overline{w^2}$$

Assuming no direction is favoured,$$\overline{u^2} = \overline{v^2} = \overline{w^2}$$ So $$\overline{u^2} = \tfrac 13 \overline{c^2}.$$

(b) A delightful elementary derivation of $pV= \frac13 Nm \overline{c^2}$ considers a gas in a spherical container and assumes that molecules bounce off the walls as elastic spheres. Orthogonal components are not considered; rather the factor of 1/3 arises almost magically, as the quotient of numerical coefficients in the formulae for the volume and area of a sphere!

(c) It might seem as though the factor of 1/3 arises quite differently in (a) and (b). But on a deeper level there is a common origin: the 3-dimensionality of space!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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