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Question

(D) According to the equipartition theorem,the average kinetic energy associated with each degree of freedom is $\frac{1}{2} kT$.For the $x$-component

Vedclass · Accepted Answer

$2 : 1$

Vedclass · Answer

(D) According to the equipartition theorem,the average kinetic energy associated with each degree of freedom is $\frac{1}{2} kT$.For the $x$-component of velocity,we have $\frac{1}{2} m \langle v_x^2 \rangle = \frac{1}{2} kT$,which implies $\langle v_x^2 \rangle = \frac{kT}{m}$.For gas $A$,the mass of each molecule is $m$,so the mean square velocity of the $x$-component is $u_x^2 = \frac{kT}{m}$.For gas $B$,the mass of each molecule is $2m$,so the mean square velocity of the $x$-component is $v^2 = \frac{kT}{2m}$.Taking the ratio,we get $\frac{u_x^2}{v^2} = \frac{kT/m}{kT/2m} = \frac{2}{1} = 2:1$.

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