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## Q. 9.5

Use the shape of the Maxwell speed distribution curve (Figure 9.7) to explain why $v_{ rms }>\bar{v}>v^{*}$. ## Verified Solution

The speed distribution curve shown in Figure 9.7 is not symmetric. It could best be described as lopsided, with the wider part of the curve to the right of $v^{*}$. This means that there are more molecules with speeds greater than $v^{*}$ than molecules with speeds less than $v^{*}$. The computation of the mean speed $\bar{v}$ is a weighted average. With more molecules having speeds above $v^{*}$ than below, this result is $\bar{v}>v^{*}$.

The root-mean-square speed $v_{ r ms }$ comes from a similar average, but one that uses the average square of the speed. In such a computation, higher speeds are weighted even more heavily, because the squares of numbers rise more rapidly than the numbers themselves. For example, the number 11 is 1.1 times larger than the number 10, but comparing the squares of these numbers, $121\left(11^{2}\right) \text { is } 1.21$ times larger than $100\left(10^{2}\right) . \text { This is why } v_{ rms }>\bar{v}$.