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A flask contains argon and chlorine in t...

A flask contains argon and chlorine in the ratio of 2:1 by mass. The temperature of the mixture is `27^(@)C`. Obtain the ratio of (i) average kinetic energy per molecule, and (ii) root mean square speed `v_("rms")` of the molecules of the two gases. Atomic mass of argon =39.9 u, Molecular mass of chlorine = 70.9 u.

Text Solution

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The important point to remember is that the average kinetic energy (per molecule) of any (ideal) gas (be it monatomic like argon, diatomic like chlorine or polyatomic) is always equal to `(3//2) k_(B)T`. It depends only on temperature, and is independent of the nature of the gas.
(i) Since argon and chlorine both have the same temperature in the flask, the ratio of average kinetic energy (per molecule) of the two gases is 1:1.
(ii) Now `(1)/(2)v_("rms")^(2)=` average kinetic energy per molecule = `(3//2) k_(B)T` where m is the mass of a molecule of the gas. Therefore,
`(V_("rms")^(2))_(Ar)/(V_("rms")^(2))_(Cl)= (m)_(Cl)/(m)_(Ar)= (M)_(Cl)/(M)_(Ar)=(70.9)/(39.9)= 1.77`
where M denotes the molecular mass of the gas. (For argon, a molecule is just an atom of argon.) Taking square root of both sides,
`(V_("rms"))_(Ar)/(V_("rms"))_(Cl)= 1.33`
You should note that the composition of the mixture by mass is quite irrelevant to the above Calculation . any other propotion by mass of argon and chlorine would give the same answers to (i) and (ii) provided the temperature remains unaltered.
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