Heat And Thermodynamics Ques 195
- 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: 195.(b)
Solution:
Formula:
- Average time between two collisions is given by
$$ \tau=\frac{1}{\sqrt{2} \pi n v _{rms} d^{2}} \cdots(i) $$
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,
$$ \tau \propto \frac{V}{\sqrt{T}} \cdots(ii) $$
For adiabatic process, $T V^{\gamma-1}=$ constant
substituting in Eq. (ii), we have $\tau \propto \frac{V}{\sqrt{\left(\frac{1}{V^{\gamma-1}}\right)}}$
$\text { or } \quad \tau \propto V^{1+\left(\frac{\gamma-1}{2}\right)} \quad \text { or } \tau \propto V^{\left(\frac{1+\gamma}{2}\right)}$
BETA
