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At what temperature is the root mean square (rms) speed of Neon gas atoms is equal to the rms speed of Helium gas atom at $-33^{\circ} \mathrm{C}$ ?
(atomic mass of $\mathrm{Ne}=2.02 \mathrm{u}$, and that of $\mathrm{He}=4.0 \mathrm{u}$ )
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(atomic mass of $\mathrm{Ne}=2.02 \mathrm{u}$, and that of $\mathrm{He}=4.0 \mathrm{u}$ )
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The correct answer is:
1212 K
$\begin{aligned} & \left(\mathrm{V}_{\mathrm{rms}}\right)_{\mathrm{Neon}}=\left(\mathrm{V}_{\mathrm{rms}}\right)_{\mathrm{He}} \\ & \Rightarrow \sqrt{\frac{3 \mathrm{R} \times \mathrm{T}}{20.2}}=\sqrt{\frac{3 \mathrm{R} \times 240}{4}} \\ & \Rightarrow \frac{\mathrm{T}}{20.2}=60 \Rightarrow \mathrm{T}=1212 \mathrm{~K}\end{aligned}$
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