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Let V_{a v}, V_{r m s}  and V_p respectively denote the mean speed, the root mean square speed and the the most probable speed of the molecules in an ideal monoatomic gas at absolute temperature T. The mass of a molecule is m.

Option: 1

No molecules can have speed greater than V_{\text {rms }}.


Option: 2

No molecules can have speed less than V_p / \sqrt{2}.
 


Option: 3

V_p<V_{a v}<V_{\text {rms }}


Option: 4

The average K E of a molecule is \frac{3}{4} m V_p^2.


Answers (1)

best_answer

\begin{aligned} V_{\text {rms }}&=\sqrt{\frac{3 R T}{M}} \\ V_{a v} & =\sqrt{\frac{8 R T}{\pi M}} \\& =\sqrt{2.55 \frac{R T}{M}} \\V_p&=\sqrt{\frac{2 R T}{M}}\\&=\sqrt{\frac{2 R T}{N_0 M}}\\&=\sqrt{\frac{2 k T}{M}} \\ \end{aligned};where

N_0=\text{Avogadro's~number} , and

k=\frac{R}{N_0}=\text{Boltzmann~constant}  
\therefore Average kinetic energy of a molecule -
\\=\frac{3}{2} k T\\=\frac{3}{2} \times \frac{1}{2} M \frac{2 k T}{M}\\=\frac{3}{4}\left(M V_p^2\right)

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