Pressure for ideal gas in terms of stat.

[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?

 

[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...

 

[astrokreng]

I appreciate your effort to deduce the expression for pressure of an ideal gas in terms of the momentum distribution n(p). Your reasoning is mostly correct, but there are a few things that need clarification.

Firstly, your assumption of a box with side lengths L_x, L_y, L_z is a good starting point. However, it is important to note that this is a simplified model and in reality, the particles in a gas are constantly moving in all directions.

Secondly, in your calculation of the average impact per unit time, you have assumed that all particles are moving in the x-direction, which is not the case. This assumption only holds true for a one-dimensional gas. In a three-dimensional gas, the average impact per unit time would be calculated by considering the velocity components in all directions.

Additionally, your calculation of the total average impact per unit time for N particles is incorrect. It should be the sum of the average impacts of all particles, not the sum of the individual average impacts. This means that the correct expression for \overline{F} would be:

\overline{F} = \sum_i^N \frac{m}{L_x}(v_{ix}^2 + v_{iy}^2 + v_{iz}^2)

Furthermore, in your continuous case, the expression for the average impact per unit time should be:

\overline{F} = \int \frac{pv_xn(p_x)}{L_x}dp_x \int \frac{pv_yn(p_y)}{L_y}dp_y \int \frac{pv_zn(p_z)}{L_z}dp_z

where n(p_x), n(p_y), and n(p_z) are the momentum distributions in the x, y, and z directions respectively.

Finally, the correct expression for pressure would be:

P = \frac{1}{3}\int p(v_xn(p_x) + v_yn(p_y) + v_zn(p_z))dp_xdp_ydp_z

This takes into account the contributions from all three velocity components.

In conclusion, your reasoning is on the right track but there are some key assumptions and calculations that need to be revised. I hope this helps in your understanding of the expression for pressure of an ideal gas in terms of the momentum distribution. Keep up the good work!