# Doubt about the constant 1/3 of the equation $PV = \cfrac{1}{3} Mv_{rms}^2$

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

• 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. – JAlex Jun 10 at 19:24

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).