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Consider an ideal gas continued in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increase as $V9,$ Where $V$ in the volume of the gas.
The value of $q$ is $r=\frac{C_p}{C_v}$
Option: 1
$\frac{3r+5}{6}$

Option: 2
$\frac{3r-5}{6}$

Option: 3
$\frac{r+1}{2}$

Option: 4
$\frac{r-1}{2}$

Answers (1)

best_answer

The rms spped of the molecule, $c=\sqrt{3 R T / M}$

Mean free path, $\lambda=\frac{1}{\sigma_n^2}=\frac{1}{\sigma^2 \frac{N}{V}}$

$(N=$ Number of molecules, $\sigma=$ diameter of the molecules $)$

Average time of colloision between the molecules,

$t=\frac{\lambda}{c}=\frac{v}{\sigma^2 N} \cdot \frac{1}{\sqrt{3 R T / 1 M}} \text { or, } t \propto \frac{V}{\sqrt{T}}$

or $T \propto \frac{v^2}{t^2}$

Again in the adiabatic process,

$\begin{aligned} & T V \lambda-1=\text { constant or, } T \propto V^1-r \\ & \frac{v^2}{t^2} \propto V^{\prime}-r \quad \text { or, } t^2 \propto V^r+1 \text { or, } t \propto \frac{r+1}{2} \end{aligned}$

$\text { Hence, } q=\frac{r+1}{2}$

Posted by

Deependra Verma

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