Graph shows a hypothetical speed distribution for a sample of N gas molecules.
Given $\frac{d N}{d v} = 0 f o r v > v_{0}$

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the value of $a v_{0}$ is 2 N

the ratio $v_{a v g} / v_{0}$ is equal to $\frac{2}{3}$

The ratio $\frac{v_{r m s}}{v_{0}}$ is equal to $\frac{1}{\sqrt{2}}$

Three fourth of the total molecules has a speed between $\frac{v_{0}}{2} a n d v_{0}$

answer is A, B, C, D.

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Detailed Solution

(A) area under the curve is equal to number of molecules of the gas sample area $= \frac{a v_{0}}{2} = N$

$(B) V_{a v g} = \frac{\int V d N}{\int d N} = \frac{\int_{0}^{v_{0}} \frac{a}{V_{0}} V d V}{\int_{0}^{v_{0}} \frac{a}{V_{0}} V d V} = \frac{V^{2 / 3}}{\frac{V^{2}}{2}} |_{0}^{V_{0}} = \frac{2}{3} V_{0}$

$(C) V_{r m s}^{2} = \frac{\int V^{2} d N}{\int d N} = \frac{\int_{0}^{v_{0}} V^{2} \frac{a}{V_{0}} V d V}{\int_{0}^{v_{0}} \frac{a}{V_{0}} V d V} = \frac{V^{4 / 4}}{V^{2 / 2}} |_{0}^{V_{0}} = \frac{V_{0}^{2}}{2}$