Question

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

• A. $\frac{(2 \ln 2-1)}{\gamma /(\gamma-1)}$
• B. $\frac{(1-1 \ln 2)}{\gamma /(\gamma-1)}$
• C. $\frac{(2 \ln 2+1)}{\gamma /(\gamma-1)}$
• D. $\frac{(2 \ln 2-1)}{\gamma /(\gamma+1)}$

Answer 1

Answer: (A)

As, $ W_{C B} =p \Delta V=n R \Delta T$
$=-n R\left(2 T_{0}-T_{0}\right)=-n R T_{0} $
$ W_{B A} =p \Delta V=0 (\because \Delta V=0) $
and $ W_{A C} =n R T \ln \left(\frac{V_{2}}{V_{1}}\right)=2 n R T_{0} \ln \left(\frac{2 V_{0}}{V_{0}}\right) $
$=+2 n R T_{0} \ln 2 $
Work done $=W_{A C}+W_{C B}+W_{B A} $
$= 2 n R T_{0} \ln 2-n R T_{0}+0 $
$=n R T_{0}(2 \ln 2-1)$
Also, input heat, $\Delta Q_{B C}=n C_{p} \Delta T=\frac{n R \gamma T_{0}}{\gamma-1}$
$\therefore$ Efficiency $=\frac{\text { Work done }}{\text { Input heat }}=\frac{(2 \ln 2-1)}{\gamma /(\gamma-1)}$