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Let \stackrel{1}{v}, \mathrm{v}_{\text {rms }} and \mathrm{v}_{\mathrm{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 \mathrm{T}.The mass of a molecule is \mathrm{m}. Then:

Option: 1

no molecule can have a energy greater than \sqrt{2} v rms


Option: 2

no molecule can have speed less than \frac{v_{p}}{\sqrt{2}}


Option: 3

\mathrm{v}_{\mathrm{p}}<\bar{v}<\mathrm{v}_{\mathrm{rms}}


Option: 4

the average kinetic energy of a molecule is \frac{3}{2} m v_{p}^{2}


Answers (1)

best_answer

\mathrm{V}_{\mathrm{rms}} =\sqrt{\frac{3 R T}{M}}
\mathrm{ \bar{v}=\sqrt{\frac{8}{\pi} \cdot \frac{R T}{M}} \approx \sqrt{\frac{25 R T}{M}} \text { and } \mathrm{v}_{\mathrm{p}}=\sqrt{\frac{2 R T}{M}}}

From these expression we can see that

\mathrm{ \mathrm{V}_{\mathrm{P}}<\bar{v}<\mathrm{V}_{\text {rms }}}
\mathrm{ secondly, \mathbf{v}_{\text {rms }}}             \mathrm{ =\sqrt{\frac{3}{2}} \mathrm{v}_{\mathrm{p}}}

and average kinetic energy of a gas molecule

\mathrm{ =\frac{1}{2} m v_{\mathrm{rms}}^{2} }
\mathrm{ =\frac{1}{2} m\left(\sqrt{\frac{3}{2}} v_{p}\right)^{2}=\frac{3}{4} m v_{p}^{2}}
 





 

Posted by

Kuldeep Maurya

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