The root mean square velocity of the molecules in a sample of helium is 5/7th that of the molecules in a sample of hydrogen. If the temperature of hydrogen sample is 0°C, then the temperature of the helium sample is about

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100°C

0°C

273°

answer is A.

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Detailed Solution

$\begin{array}{l} {(V_{rms})}_{He} = \frac{5}{7} {(V_{rms})}_{H} \\ \sqrt{\frac{3 {RT}_{He}}{M_{He}}} = \frac{5}{7} \sqrt{\frac{3 {RT}_{H}}{M_{H}}} \\ \Rightarrow \frac{T_{He}}{4} = \frac{5 \times 5}{7 \times 7} \times \frac{273}{2} \\ T_{He} = \frac{50}{49} (273) nearly 0^{\circ} c \end{array}$

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