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A temperature at which rms speed of $S O_{2}$ molecules is half of that of helium molecules at 300 K

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Detailed Solution

$r_{{SO}_{2}} = \sqrt{\frac{3 RT}{64}}; r_{He} = \sqrt{\frac{3 R \times 300}{4}} ∴ \sqrt{\frac{3 R \times 300}{4}} = 2 \sqrt{\frac{3 RT}{64}} \frac{3 R x 300}{4} = 4 x \frac{3 RT}{64} T = \frac{64}{14} x 300 = 1200 K$

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