A) No molecule can have speed greater than \[\sqrt{2}{{v}_{rms}}\]
B) No molecule can have speed less than \[{{v}_{p}}/\sqrt{2}\]
C) \[{{v}_{p}}=\bar{v}<{{\bar{v}}_{rms}}\]
D) The average kinetic energy of a molecule is \[\frac{3}{4}mv_{p}^{2}\]
Correct Answer: D
Solution :
\[{{v}_{rms}}=\sqrt{\frac{3RT}{M}},\,\,{{v}_{P}}=\sqrt{\frac{2RT}{M}}=0.816\,{{v}_{rms}}\] \[\bar{v}=\sqrt{\frac{8RT}{\pi M}}=0.92{{\upsilon }_{rms}}\Rightarrow {{v}_{P}}<\bar{v}<{{\bar{v}}_{rms}}\] Further \[{{E}_{av}}=\frac{1}{2}mv_{rms}^{2}=\frac{1}{2}m\frac{3}{2}v_{P}^{3}=\frac{3}{4}mv_{P}^{2}\]
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