- #1
Seydlitz
- 263
- 4
So basically I was wondering whether it's possible to get the expression of ideal gas using molecular flux equation which is ##\phi = \frac{1}{4}\bar{v}n##. The derivation should be straightforward. I need to get the expression of pressure. Because the flux by definition already gives the rate of number of particles per area. I multiplied ##\phi## by the change of momentum when a particle hits a wall and rebounds which is ##\frac{2}{3}m\bar{v}##.
Eventually I get.
$$
P = \phi \, dp = \frac{1}{2}\frac{1}{2}m\bar{v}^2\frac{2}{3}n
$$
Notice however that I get extra ##\frac{1}{2}## factor which makes the final expression smaller by a half than the correct. ##PV = NkT##
What should be fixed in this case?
Thank You
Eventually I get.
$$
P = \phi \, dp = \frac{1}{2}\frac{1}{2}m\bar{v}^2\frac{2}{3}n
$$
Notice however that I get extra ##\frac{1}{2}## factor which makes the final expression smaller by a half than the correct. ##PV = NkT##
What should be fixed in this case?
Thank You