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The velocities of sound in an ideal gas at temperature $T_{1} K$ and $T 2 K$ are found to be $V 1$ and $V 2$ respectively. If the r.m.s velocities of the molecules of the same gas at the same temperatures $T 1$ and $T 2$ are $v 1 and v 2$ respectively, then

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v 2 = v 1 V 1 V 2

$v_{2} = v_{1} (\frac{V_{1}}{V_{2}})$

v 2 = v 1 V 2 V 1

v 2 = v 1 V 2 V 1

answer is B.

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Detailed Solution

C = r R T M

V r m s = 3 R T M V r m s α C → V 2 V 1 = C 2 C 1

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