Suppose in a gas sample $d N$ denotes number of gas molecules having speeds between $v$ and $v + d v$ . Then, $\frac{d N}{d v}$ versus $v$ graph for the sample is called velocity distribution graph. Such a graph for hypothetical gas sample is parabolic as shown with $v_{0}$ as maximum molecular speed.

If slope of graph at O is zero, mean square speed of the sample is

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$0.6 v_{0}^{2}$

$v_{0}^{2}$

$0.8 v_{0}^{2}$

$0.4 v_{0}^{2}$

answer is C.

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Detailed Solution

Let $\frac{d N}{d v} = k v^{2}$ ……………. (i)
Here, k is constant (as for upward concave parabolic curve with vertex at $O, y \propto x^{2}$ ).
Mean square speed is defined as
$v_{m s} = \frac{\int v^{2} d N}{\int d N} = \int_{0}^{v_{0}} \frac{v^{2} \cdot k v^{2} d v}{\int_{0}^{v_{0}} k v^{2} d v} [f r o m {Eq. (i), dN = kv}^{2} d v] = \frac{\frac{k v_{0}^{5}}{5}}{k v_{0}^{3} / 3} = \frac{3 v_{0}^{2}}{5} = 0.6 {v_{0}}^{2}$