A) \[\frac{\gamma +1}{2}\]
B) \[\frac{\gamma -1}{2}\]
C) \[\frac{3\gamma +5}{6}\]
D) \[\frac{3\gamma -5}{6}\]
Correct Answer: A
Solution :
[a] : Average time of collision between molecules, \[\tau =\frac{Mean\,free\,path\,(\lambda )}{Mean\,speed\,(\overline{v})}=\frac{1}{\left( \sqrt{2}\pi {{d}^{2}}\frac{N}{V} \right)\left( \sqrt{\frac{8{{k}_{B}}T}{m\pi }} \right)}\] \[\therefore \,\,\,\,\tau \propto \frac{V}{\sqrt{T}}\,\,\,\,or\,\,\,\,\,T\propto \frac{{{V}^{2}}}{{{\tau }^{2}}}\] For adiabatic expansion, \[T{{V}^{\lambda -1}}=\] constant or\[\frac{{{V}^{2}}}{{{\tau }^{2}}}{{V}^{\gamma -1}}=\]constant or \[\tau \propto {{V}^{\frac{(\gamma +1)}{2}}}\] Comparing it with \[\tau \propto {{V}^{q}}\], we get \[q=\frac{\gamma +1}{2}\]You need to login to perform this action.
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