4
$\begingroup$

Why is there a $\frac23$ term?

I fully understand that it makes sense for pressure to be the volume weighted average kinetic energy. And the equation above shows that this is so (for an ideal gas). I also fully understand how to prove this with momentum and force. I'm wondering is there an intuitive reason why the for one cubic meter, the the pressure should be two thirds of the total internal kinetic energy, not some other value. Or is it just because the math says so?

$\endgroup$

2 Answers 2

5
$\begingroup$

The 2/3 factor is from the math of the kinetic theory of gases. Think of a cube with six faces (i.e. three orthogonal directions), the particles have equal probability of hitting a face in the x, y or z direction. The pressure is exerted equally among the three orthogonal directions, hence a factor of 1/3.

Wikipedia also shows the steps in deriving the kinetic theory.

$\endgroup$
3
$\begingroup$

The expression you've written can be obtained from the equipartition theorem

$$U = n_\mathrm{d.f.} \times \frac{k_\mathrm B T}2$$

stating that the energy is evenly distributed over all available degrees of freedom $n_\mathrm{d.f.}$, where each degree of freedom contributes an amount of energy equal to $k_\mathrm B T/2$. For a monoatomic gas of $N$ particles, each atom has only translational degrees of freedom (a total of three per particle, corresponding to three orthogonal spatial coordinates), and $n_\mathrm{d.f.} = 3N$, so that

$$U = 3N\frac{k_\mathrm B T}2$$

Combining this with the ideal gas expression $N k_\mathrm B T= pV$ leads to a constant of proportionality of 2/3.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.