Heat and Thermodynamics 6 Question 2
6. Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as $V^{q}$, where $V$ is the volume of the gas. The value of $q$ is $\left(\gamma=\frac{C _p}{C _V}\right)$
(a) $\frac{3 \gamma+5}{6}$
(b) $\frac{\gamma+1}{2}$
(c) $\frac{3 \gamma-5}{6}$
(d) $\frac{\gamma-1}{2}$
(2015 Main)
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Answer:
Correct Answer: 6. (b)
Solution:
- Average time between two collisions is given by
$$ \tau=\frac{1}{\sqrt{2} \pi n v _{rms} d^{2}} $$
Here, $n=$ number of molecules per unit volume $=\frac{N}{V}$
and $\quad v _{rms}=\sqrt{\frac{3 R T}{M}}$
Substituting these values in Eq.(i) we have,
$$ \tau \propto \frac{V}{\sqrt{T}} $$
For adiabatic process, $T V^{\gamma-1}=$ constant
substituting in Eq. (ii), we have $\tau \propto \frac{V}{\sqrt{\left(\frac{1}{V^{\gamma-1}}\right)}}$
$\tau \propto V \quad$ or $\left.\tau \propto V^{(2}\right)$
