Get Answers to all your Questions

header-bg qa

 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 Vq, where V is the volume of the gas.  The value of q is :       

\left ( \gamma =\frac{C_{p}}{C_{v}} \right )

Option: 1

\frac{3\gamma +5}{6}


Option: 2

\frac{3\gamma -5}{6}


Option: 3

\frac{\gamma +1}{2}


Option: 4

\frac{\gamma -1}{2}


Answers (1)

best_answer

\text {Average time of collision} : \tau=\frac{1}{n \pi \sqrt{2} v_{r m s} d^{2}}$\\ since $v_{r m s} \propto \sqrt{T}$ \\ and $n \propto \frac{1}{V}

\text {For an adiabatic process}:\\ T V^{\gamma-1}=$constant\\ $\left.\sqrt{T} \propto V^{(} \frac{1-\gamma}{2}\right)

\begin{array}{l} \therefore \tau \propto \frac{V}{\sqrt{T}} \\ \left.\therefore \tau \propto V^{(} \frac{1+\gamma}{2}\right) \end{array}

Posted by

sudhir.kumar

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks