Use the Maxwell distribution to calculate the average value of $v^2$for the molecules of an ideal gas. Check that your answer agrees with equation 6.41.

The Maxwell speed distribution function is give by

$D\left(v\right)={\left(\frac{m}{2\pi kT}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}4\pi v^2{e}^{-\raisebox{1ex}{$m{v}^2$}\!\left/ \!\raisebox{-1ex}{$2kT$}\right.}................................\left(1\right)$

The formula to calculate the average of $v^2$is given by

Substitute the distribution function from equation (1) into equation (2) to calculate the required average value.

Substitute $x$for $\left(\sqrt{\frac{m}{2kT}}v\right)$into equation (3) and evaluate the required value.

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The rms speed is defined as

Substitute $\left(\frac{3kT}{m}\right)$for into equation (4) to obtain the rms speed.

$v_{\mathrm{rms}}=\sqrt{\frac{3kT}{m}}$

Thus, the above equation exactly matches with equation 6.41.