InterviewSolution
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InterviewSolution
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A flask contains argon and chlorine in the ratio 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 of the molecules of two gases. Atomic mass of argon = 39.9 u, Molecular mass of chlorine = 70.9 u. |
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Answer» 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 lt `1//2 mv_(rms)2 `= average kinetic energy per molecule = (3/2) ) kBT where m is the mass of a molecule of the gas. Therefore, `=(V_(rms)^(2))_(Ar)/(V_(rms)^(2))_(Cl)=(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 proportion by mass of argon and chlorine would give the same answers to (i) and (ii), provided the temperature remains unaltered. |
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