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The temperature can be interpreted as a measure of kinetic energy per particle. In fact, denoting with $f(\vec{v}) d \vec{v}$ the number of particles with velocity in $d \vec{v}$ one obtains that the pressure exerted on the walls of the container of area $A$ along the direction $z$ is $$ p=\frac{1}{A} \int d F_A=\frac{N}{V} \int d v_x \int d v_y \int d v_z f(\vec{v}) 2 m v_z^2 $$

Assuming an isotropic distribution of velocities in a ideal gas, $f(\vec{v})=g\left(v^2\right)=g_1\left(v_x^2\right) g_2\left(v_y^2\right) g_z\left(v_z^2\right)$, this becomes $$ p V=m N \int d^3 v f(|\vec{v}|) v_z^2=m N\left\langle v_z^2\right\rangle=\frac{m N}{3}\left\langle v^2\right\rangle=\frac{2}{3} N\left\langle\varepsilon_{K I N}\right\rangle $$ ie $p V$ is proportional to the total kinetic energy of the particles. We are assuming here the particles have only 3 degrees of freedom.

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Can someone please explain me this deduction? I've alread tried so hard :/

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    – Semoi
    Jul 19, 2020 at 7:42

1 Answer 1

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I think the factor 2 in the right-hand side of the first equation is wrong.

I'm basically building on the gas model explanation in http://physics.bu.edu/~duffy/py105/Kinetictheory.html, and adding some explanations that I think weren't clear enough. I'm outlining the problem like this:

  • The container has area A and length L (along the z axis); its total volume is $V = L \cdot A$
  • The total pressure exerted on the wall will be an integral over the velocity space — the integrand is the pressure exerted only by the particles with velocity $\vec{v}$
  • Each particle has mass m

In the time period $\Delta t$ between two collisions, a particle with velocity $(v_x, v_y, v_z)$ will bounce so that $v_z$ will reverse; hence its momentum changes by $2mv_z$ (or minus that). Because the momentum change equals the impulse, the average force F exerted by that particle satisfies $F \Delta t = 2mv_z$.

In the time between two collisions, the particle reaches the same wall after going all the way to the other wall and coming back (covering a z-axis distance of 2L). That time will hence satisfy $v_z \Delta t = 2L$.

Eliminating the time from both equations we get $F = mv_z^2/L$. That's the expression for the average force exerted on the wall by a single particle which has velocity $(v_x, v_y, v_z)$.

Basically we have to integrate this over all velocities to get the total force, and divide it by the area to get the pressure. We define the infinitesimal force (for a given velocity $\vec{v}$) as $dF = F(\vec{v}) f(\vec{v}) d\vec{v}$ — the force for each particle multiplied by the number of particles.

$$ p = \int \frac{dF}{A} = \int \frac{F(\vec{v}) f(\vec{v}) d\vec{v}}{A} = \int \frac{L}{V} \frac{mv_z^2}{L} f(\vec{v}) d\vec{v} $$

which is equivalent to the first equation, without the factor 2.

To continue from there, we just rewrite $f(\vec{v})$ as a function of only the modulus, $f(|\vec{v}|)$, because of the isotropic distribution; what we get is the definition of the average squared velocity, $\langle v_z^2 \rangle = \int d^3 v f(|\vec{v}|) v_z^2$.

The isotropic assumption also leads to the average squared velocity components being equal in all three directions $\langle v_x^2 \rangle = \langle v_y^2 \rangle = \langle v_z^2 \rangle$, so that $\langle v^2 \rangle = \langle v_x^2 + v_y^2 + v_z^2 \rangle = 3 \langle v_z^2 \rangle$.

Substituting that and recalling that the kinetic energy is $mv^2/2$ leads to the final result.

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