# Kinetic energy of an ideal gas

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 :/

• Please consider typing the text. Images are not searchable. Therefore, the answer will not help other people, but only you. This is not the goal of this website. Commented Jul 19, 2020 at 7:42

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.