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प्रश्न
Answer in brief:
Show that rms velocity of an oxygen molecule is `sqrt2` times that of a sulfur dioxide molecule at S.T.P.
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उत्तर
`("M"_0 ("SO"_2))/("M"_0 ("O"_2)) = (64 "kg/mol")/(32 "kg/mol") = 2`
The rms speed, `"v"_"rms" = sqrt("3RT"/"M"_0)`
∴ `"v"_"rms" ∝ 1/sqrt"M"_0` at constant T
∴ `("v"_"rms"("O"_2))/("v"_"rms"("SO"_2)) = sqrt(("M"_0 ("SO"_2))/("M"_0 ("O"_2))) = sqrt2`
Thus, `"v"_"rms"("O"_2) = sqrt2 "v"_"rms"("SO"_2)`
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