Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
Let\[\overline{v},{{v}_{rms}}\]and\[{{v}_{mp}}\]respectively denote the mean speed, root mean square speed, and most probable speed of the molecules in an ideal monoatomic gas at absolute temperature T. The mass of a molecule is m. Then (1) \[{{v}_{mp}}<\overline{v}<{{v}_{rms}}\] (2) no molecule can have a speed greater than \[\sqrt{2}{{v}_{rms}}\] (3) the average kinetic energy of a molecule is \[\frac{3}{4}mv_{mp}^{2}\] (4) no molecule can have a speed less than \[{{v}_{mp}}/\sqrt{2}\]A) 1, 2 and 3 are correct
B) 1 and 2 are correct
C) 2 and 4 are correct
D) 1 and 3 are correct
Correct Answer: D
Solution :
\[\overline{v}=\sqrt{\frac{8{{k}_{B}}T}{m\pi }},\] \[{{v}_{rms}}=\sqrt{\frac{3{{k}_{B}}T}{m}},{{v}_{mp}}=\sqrt{\frac{2{{k}_{B}}T}{m}}\] where\[{{k}_{B}}\]is the Boltzmann's constant \[\therefore \] \[{{v}_{rms}}>\overline{v}>{{v}_{mp}}\] or \[{{v}_{mp}}>\overline{v}>{{v}_{rms}}\] Option (1) is correct. Average kinetic energy of a molecule is \[E=\frac{1}{2}mv_{rms}^{2}\] Or \[E=\frac{1}{2}m\frac{3{{k}_{B}}T}{m}\] Or \[E=\frac{3}{4}m\frac{2{{k}_{B}}T}{m}\] Or \[E=\frac{3}{4}m_{p}^{2}\] Option (3) is correct. Again, according to kinetic theory of gases, a molecule of a gas can have speed such that it lies between 0 and\[\infty \]. ie,\[0<v<\infty \]. Hence the options (2) and (4) can never be correct.
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