InterviewSolution
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InterviewSolution
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N molecules each of mass m of gas A and 2 N molecules each of mass 2m of gas B are contained in the same vessel which is maintined at a temperature T. The mean square of the velocity of the molecules of B type is denoted by `v^(2)` and the mean square of the x-component of the velocity of a tye is denoted by `omega^(2)`. What is the ratio of `omega^(2)//v^(2) = ?`A. `3:2`B. `1:3`C. `2:3`D. `1:1` |
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Answer» Correct Answer - C The mean sequare velocity of gas molecules is giben by `v^(2)=(3kT)/(m).` For gas A, `v_(A)^(2)=(3kT)/(m)" "...(i)` For a gas molecule. `v^(2)-v_(x)^(2)+v_(y)^(2)+v_(z)^(2)=3v_(x)^(2)" "(thereforev_(x)^(2)=v_(y)^(2)=v_(z)^(2))` or `v_(x)^(2)=(v^(2))/(3)` From eqn. (i), we get `w^(2)=v_(x)^(2)=[(3kT)/(m/(3))]=(kT)/(m)" "...(ii)` For gas B, `v_(B)^(2)=v^(2)=(3kT)/(2m)" "...(iii)` Dividing eqn. (ii) by eqn. (iii), we get `(w^(2))/(v^(2))=(KT)/((m/(3kT))/(2m))=2/3` |
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