Consider a gas kept in a closed container. ‘ $N$ ’ is the no of molecules present in the container and $v$ is the speed of molecules. Molecular speed distribution graph is shown in the figure with suitable S.I. units of $\frac{d N}{d v}$ and $v$ . The $v_{r m s}^{2}$ of molecules in S.I. units is $\frac{x}{2},$ then $x =___$

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answer is 3.

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Detailed Solution

$v_{r m s}^{2} = \frac{\int v^{2} d N}{\int d N} = \frac{\int v^{2} \frac{d N}{d v} d v}{\int \frac{d N}{d v} d v} = \frac{\int_{0}^{3} v^{2} (- \frac{5}{3} v + 5) d v}{\int_{0}^{3} (- \frac{5}{3} v + 5) d v} = 1.50$

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