Heat And Thermodynamics Ques 112
- 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: 112.(c,d)
Solution:
Formula:
- $\quad v _{rms}=\sqrt{\frac{3 R T}{M}}, \bar{v}=\sqrt{\frac{8}{\pi} \cdot \frac{R T}{M}} \approx \sqrt{\frac{2.5 R T}{M}}$
$$ \text { and } \quad v _p=\sqrt{\frac{2 R T}{M}} $$
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
$$ =\frac{1}{2} m v _{rms}^{2}=\frac{1}{2} m\left(\sqrt{\frac{3}{2}} v _p\right)^{2}=\frac{3}{4} m v _p^{2} $$
BETA
