# Pressure for ideal gas in terms of stat.

1. Jan 8, 2009

### KFC

I am trying to deduce the expression for pressure of perfect gas when the momentum distribution $$n(p)$$ is given.

Here is how I did. First we assume a box with side length $$L_x, L_y, L_z$$, when a particle , say moving a long x direction, collide with one side of the wall, the total change of momentum would be

$$\Delta p_x = -2m v_x$$

Assume it takes time t for one round-trip (from one wall to the oppsite and come back), hence

$$t = \frac{2L_x}{v_x}$$

and 1/t is the rate of colliding.

Now consider the average impact per unit time,

$$\overline{f} = \Delta p_x \times (\textnormal{rate of colliding}) = 2mv_x \frac{v_x}{2L_x} = \frac{mv_x^2}{L_x}$$

For N particles, the total average impact per unit time would be

$$\overline{F} = \sum_i^Nf_i = \sum_i^N \frac{mv_{ix}^2}{L_x}$$

Hence, the pressure on the side $$A=L_yL_z$$ woule be

$$P = \frac{\overline{F}}{A} = \frac{\overline{F}}{L_yL_z}$$

For continuous case, the average impact becomes

$$\overline{F} = \int \frac{pvn(p)}{L_x}dp$$

So, the pressure becomes

$$P = \frac{\overline{F}}{A} = \int \frac{pvn(p)}{L_xL_yL_z}dp$$

In unit volume, $$L_xL_yL_z=1$$, wehave

$$P = \frac{\overline{F}}{A} = \int pvn(p)dp$$

I know there is something wrong here. The correct answer should be

$$P = \frac{1}{3}\int pvn(p)dp$$

Well, I don't know where my reasoning is going wrong. From $$\overline{F} = \sum_i^N \frac{mv_{ix}^2}{L_x}$$ to $$\overline{F} = \int \frac{pvn(p)}{L_x}dp$$, I feel that there is something missing?

2. Jan 13, 2009

### Thaakisfox

Because when you say N particles with the side L_yL_z, you say that all of the particles move in the x direction. Whereas in average only one third of the particles move in the x direction.
So you should divide the average rate of impact by three for all three directions...