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  • Under an adiabatic process, the volume of an ideal gas gets doubled. Consequently the mean collision time between the gas molecule changes from   to <img alt="\tau_{2}" src="/latex-image/?%5Cta

Under an adiabatic process, the volume of an ideal gas gets doubled. Consequently the mean collision time between the gas molecule changes from \tau_{1}  to \tau_{2} . If  \frac{C_{p}}{C_{v}}=\gamma for this gas then a good estimate for \frac{\tau_{2}}{\tau_{1}}  is 
Option: 1 \left(\frac{1}{2}\right)^{(\gamma+1)/{2}}  
  
Option: 2 \frac{1}{2}

Option: 3 \left ( \frac{1}{2} \right )^{\gamma }  

Option: 4 (2)^{{(1+\gamma})/{2}}
 

Answers (1)

best_answer

As, relaxation time= \tau \ \ \alpha \ \ \frac{V}{\sqrt{T}}

and using PV^{\gamma }=Const gives \mathrm{T} \propto \frac{1}{\mathrm{V}^{\gamma-1}}

\begin{array}{l}{\tau \propto V^{1+{(\gamma-1)}/{2}}} \\ {\tau \propto V^{{(1+\gamma)}/{2}}}\end{array}

\begin{aligned} &\frac{\tau_{2}}{\tau_{\mathrm{1}}}=\left(\frac{2 \mathrm{~V}}{\mathrm{~V}}\right)^{{(1+\gamma})/{2}} \\ &\frac{\tau_{\mathrm{2}}}{\tau_{\mathrm{1}}}=(2)^{(1+\gamma) / 2} \end{aligned}

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Ritika Jonwal

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