Graph shows a hypothetical speed distribution for a sample of N gas particle (for $v > v_{0}; \frac{d N}{d v} = 0$ )

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

The ratio $v_{a v g} / v_{0}$ is equal to 3/2

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

Three fourth of the total particle has a speed between 0.5 $v_{0}$ and $v_{0}$ .

answer is A, C, D.

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Detailed Solution

Area under the curve gives total number of molecules in the gas sample.
Total number of gas molecules $N = \int (\frac{d N}{d v}) d v = \frac{1}{2} a v_{0} \Rightarrow a v_{0} = 2 N$
From the graph $\frac{d N}{d v} = \frac{a}{v_{0}} v \Rightarrow d N = \frac{a}{v_{0}} v d v$
Average velocity $v_{a v g} = \frac{\int v d N}{\int d N} = \frac{\int_{0}^{v_{0}} \frac{a}{v_{0}} v^{2} d v}{\int_{0}^{v_{0}} \frac{a}{v_{0}} v d v} = \frac{v_{0}^{3} / 3}{v_{0}^{2} / 2} = \frac{2}{3} v_{0}$
$v_{r m s} = v_{a v g} = \sqrt{\frac{\int v^{2} d N}{\int d N}} = \sqrt{\frac{\int_{0}^{v_{0}} \frac{a}{v_{0}} v^{3} d v}{\int_{0}^{v_{0}} \frac{a}{v_{0}} v d v}} = \sqrt{\frac{v_{0}^{4} / 4}{v_{0}^{2} / 2}} = \frac{1}{\sqrt{2}} v_{0}$
Area under the curve from 0.5 $v_{0}$ to $v_{0}$ is $\frac{3}{4}$ of total area