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From the equation $ PV = \cfrac{1}{3} Mv_{rms}^2 $, Could someone explain to me where the 1/3 comes from?

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  • $\begingroup$ Unrelated: I don't why this bothers me but try \tfrac{1}{3} instead of \frac{1}{3} for fractional coefficients. It looks a bit more streamlined. $\endgroup$ – JAlex Jun 10 at 19:24
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For a monatomic ideal gas, the average kinetic energy per particle is given by $\frac{1}{2}mv_{rms}^2= \frac{3}{2} k_BT$ via the equipartition theorem. The number 3 in the numerator of the right-hand side is due to the fact that there are three quadratic degrees of freedom in the Hamiltonian - namely, $p_x,p_y,$ and $p_z$.

Comparing this with the ideal gas law $PV=Nk_BT$, we see that $PV=N\left(\frac{1}{3}mv_{rms}^2\right)= \frac{1}{3}Mv_{rms}^2$, where $M=Nm$ is the total mass of the system (the number of particles times the mass of each particle).

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