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