- #1

Ethereal

Can someone explain why the pressure due to an ideal gas in a container is given by [tex]P = \frac{1}{3}\rho<c^2>[/tex]? (rho is the density of the gas and <c^2> is the mean square speed)

I happen to have the derivation of the eqn at hand, but somehow, there is this particular step in it I didn't understand; it's the step in which they calculated the change of momentum for a single gas molecule to be 2mv. Since force exerted on container walls is given by [tex]F = \frac {\triangle p}{\triangle t}[/tex] which means the change of momentum (p) divided by the time taken for the momentum to change.

However, in the case of the gas particle, [tex]\triangle t[/tex] was taken to be [tex]\frac {2l}{v}[/tex]! (l is the length of the cuboid container, and v is velocity of the gas molecule). Why?