The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is
- $\frac{1}{4}$
- $\frac{1}{2}$
- $\frac{2}{3}$
- $\frac{1}{3}$
Solution
$\mathrm{Q}_{1}=\mathrm{T}_{0} \mathrm{~S}_{0}+\frac{1}{2} \mathrm{~T}_{0} \mathrm{~S}_{0}=\frac{3}{2} \mathrm{~T}_{0} \mathrm{~S}_{0}$
$\mathrm{Q}_{2}=\mathrm{T}_{0}\left(2 \mathrm{~S}_{0}-\mathrm{S}_{0}ight)=\mathrm{T}_{0} \mathrm{~S}_{0}$ and $\mathrm{Q}_{3}=0$
$\eta=\frac{\mathrm{W}}{\mathrm{Q}_{1}}=\frac{\mathrm{Q}_{1}-\mathrm{Q}_{2}}{\mathrm{Q}_{1}}$
$=1-\frac{\mathrm{Q}_{2}}{\mathrm{Q}_{1}}=1-\frac{\mathrm{T}_{0} \mathrm{~S}_{0}}{\frac{3}{2} \mathrm{~T}_{0} \mathrm{~S}_{0}}=\frac{1}{3}$ ~
Asked in: JEE-TOPICTESTS-CHEMISTRY