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(D) For an ideal gas at temperature $T$,the average kinetic energy per molecule is given by $\frac{1}{2} m v_{rms}^2 = \frac{3}{2} k_B T$.For gas $A$

Vedclass · Accepted Answer

$2/3$

Vedclass · Answer

(D) For an ideal gas at temperature $T$,the average kinetic energy per molecule is given by $\frac{1}{2} m v_{rms}^2 = \frac{3}{2} k_B T$.For gas $A$ with mass $m$ at temperature $T$:The mean square velocity is $\langle v_A^2 angle = \frac{3 k_B T}{m}$.Since the motion is isotropic,the mean square of the $x$-component is $\langle v_{Ax}^2 angle = w^2 = \frac{1}{3} \langle v_A^2 angle = \frac{k_B T}{m}$.For gas $B$ with mass $2m$ at temperature $T$:The mean square velocity is $v^2 = \langle v_B^2 angle = \frac{3 k_B T}{2m}$.Now,calculating the ratio $w^2 / v^2$:$\frac{w^2}{v^2} = \frac{k_B T / m}{3 k_B T / 2m} = \frac{1}{m} 	imes \frac{2m}{3} = \frac{2}{3}$.Thus,the ratio is $2/3$.

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