I am trying to deduce the expression for pressure of perfect gas when the momentum distribution [tex]n(p)[/tex] is given.(adsbygoogle = window.adsbygoogle || []).push({});

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

[tex]\Delta p_x = -2m v_x[/tex]

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

[tex]t = \frac{2L_x}{v_x}[/tex]

and 1/t is the rate of colliding.

Now consider the average impact per unit time,

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

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

[tex]\overline{F} = \sum_i^Nf_i = \sum_i^N \frac{mv_{ix}^2}{L_x}[/tex]

Hence, the pressure on the side [tex]A=L_yL_z[/tex] woule be

[tex]P = \frac{\overline{F}}{A} = \frac{\overline{F}}{L_yL_z}[/tex]

For continuous case, the average impact becomes

[tex]\overline{F} = \int \frac{pvn(p)}{L_x}dp[/tex]

So, the pressure becomes

[tex]P = \frac{\overline{F}}{A} = \int \frac{pvn(p)}{L_xL_yL_z}dp[/tex]

In unit volume, [tex]L_xL_yL_z=1[/tex], wehave

[tex]P = \frac{\overline{F}}{A} = \int pvn(p)dp[/tex]

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

[tex]P = \frac{1}{3}\int pvn(p)dp[/tex]

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

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# Pressure for ideal gas in terms of stat.

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