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Q.

$U_{rms} o {⨍ CH}_{4}$ TK is 6 times of $U_{mp} o ⨍ {SO}_{2}$ at $T_{1} K .$ The temperature of ${CH}_{4}$ gas is $x$ times of ${SO}_{2} . x$ is $\dots \dots$

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answer is 6.

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Detailed Solution

$\underset{{CH}_{4}}{\underset{⏟}{\sqrt{\frac{3 RT}{16}}}} = \underset{{SO}_{2}}{\underset{⏟}{6 \sqrt{\frac{2 {RT}_{1}}{64}}}}$

Squaring both sides

$\frac{3 RT}{16} = \frac{36 \times 2 {RT}_{1}}{64} T = 6 T_{1} x = 6$

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