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Answer in brief: Show that rms velocity of an oxygen molecule is √22sqrt2 times that of a sulfur dioxide molecule at S.T.P.

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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.

योग
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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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अध्याय 3: Kinetic Theory of Gases and Radiation - Exercises [पृष्ठ ७४]

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बालभारती Physics [English] Standard 12 Maharashtra State Board
अध्याय 3 Kinetic Theory of Gases and Radiation
Exercises | Q 14 | पृष्ठ ७४

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