Heat and Thermodynamics 4 Question 33

35. Let $\bar{v}, v _{\text {rms }}$ and $v _p$ 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,

$(1998,2 M)$

(a) no molecule can have a speed greater than $\sqrt{2} v _{\text {rms }}$

(b) no molecule can have speed less than $v _p / \sqrt{2}$

(c) $v _p<\bar{v}<v _{\text {rms }}$

(d) the average kinetic energy of a molecule is $\frac{3}{4} m v _p^{2}$

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Answer:

Correct Answer: 35. (c,d)

Solution:

  1. $\quad v _{rms}=\sqrt{\frac{3 R T}{M}}, \bar{v}=\sqrt{\frac{8}{\pi} \cdot \frac{R T}{M}} \approx \sqrt{\frac{2.5 R T}{M}}$

$$ \text { and } \quad v _p=\sqrt{\frac{2 R T}{M}} $$

From these expressions we can see that,

Secondly, $\quad v _{rms}=\sqrt{\frac{3}{2}} v _p$

and average kinetic energy of a gas molecule

$$ =\frac{1}{2} m v _{rms}^{2}=\frac{1}{2} m\left(\sqrt{\frac{3}{2}} v _p\right)^{2}=\frac{3}{4} m v _p^{2} $$



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