The efficiency of an ideal gas with adiabatic exponent ' $\gamma$ ' for the shown cyclic process would be

The efficiency of an ideal gas with adiabatic exponent ' $\gamma$ ' for the shown cyclic process would be
  1. $\frac{(2 \ln 2-1)}{\gamma /(\gamma-1)}$
  2. $\frac{(1-2 \ln 2)}{\gamma /(\gamma-1)}$
  3. $\frac{(2 \ln 2+1)}{\gamma /(\gamma-1)}$
  4. $\frac{(2 \ln 2-1)}{\gamma /(\gamma+1)}$

Solution

$W_{A B}=0, W_{B C}=P \Delta V=n R \Delta T=-n R T_{0}$
$W_{C A}=n R T \ell n \frac{V_{f}}{V_{i}}=n R\left(2 T_{0}ight) \ell n 2$
$Q_{B C}=n C_{p} \Delta T=\left(\frac{n R \gamma}{\gamma-1}ight) T_{0}$
Efficiency, $\eta=\frac{W}{Q}=\left[\frac{2 \ell n 2-1}{\gamma /(\gamma-1)}ight]$ *

Asked in: JEE-TOPICTESTS-CHEMISTRY

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