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, $W_{CB}=p\Delta V=nR\Delta T$
$=-nR\left(2T_0-T_0\right)=-nR{T}_0$
$W_{BA}=p\Delta V=0(\because \Delta V=0)$
and $W_{AC}=nRT\ln \left(\frac{V_2}{V_1}\right)=2nR{T}_0\ln \left(\frac{2V_0}{V_0}\right)$
$=+2nR{T}_0\ln 2$
Work done $=W_{AC}+W_{CB}+W_{BA}$
$=2nR{T}_0\ln 2-nR{T}_0+0$
$=nR{T}_0(2\ln 2-1)$
Also, input heat, $\Delta Q_{BC}=n{C}_p\Delta T=\frac{nR\gamma T_0}{\gamma -1}$
$\therefore$ Efficiency $=\frac{\text{ Work done }}{\text{ Input heat }}=\frac{(2\ln 2-1)}{\gamma /(\gamma -1)}$