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Question
Calculate the ratio of the mean square speeds of molecules of a gas at 30 K and 120 K.
Solution
Data: T1 = 30 K, T2 = 120 K
The mean square speed, `bar"v"^2 = "3RT"/"M"_0`
∴ `bar"v"_1^2/bar"v"_2^2 = "T"_1/"T"_2` for a given gas
∴ `bar"v"_1^2/bar"v"_2^2 = (30"K")/(120"k") = 1/4`
This is the required ratio.
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