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 :
Average time of collisim between molecules is \[t=\frac{\lambda }{{{v}_{rms}}}\]mean free path \[\lambda \frac{1}{\pi {{d}^{2}}N/V}=\frac{V}{\pi {{d}^{2}}N}\] \[\lambda \propto V\] \[{{v}_{rms}}\propto \sqrt{T}\] So\[t\propto \frac{V}{\sqrt{T}}\] \[t\propto V.{{T}^{\frac{-1}{2}}}\]and\[V{{T}^{\gamma =1}}=\]constant \[t\propto V.{{\left( {{V}^{1-\gamma }} \right)}^{\frac{-1}{2}}}\] \[t\propto V.{{V}^{\frac{r-1}{2}}}\] \[\]So\[q=\frac{\gamma +1}{2}\]
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