Question

A Carnot engine uses first an ideal monoatomic gas $(\gamma =\frac{5}{3})$ and then a diatomic gas $(\gamma =\frac{7}{5})$ as its working substance. The source and sink are at temperatures ${411}^{o}C$ and ${69}^{o}C$ respectively. The engine extracts $1000J$ of heat from source in each cycle. Then

• A. the efficiencies of engine in the two cases $m$ in the ratio $21:25$
• B. the area enclosed by PV diagram in the first case only is $500J$
• C. the area enclosed by PV dlagram in both cases is $500J$
• D. the heat energy rejected by engine in the first case is only $600J$ while that in second case is $714.3J$

Answer 1

Answer: (C)

the area enclosed by PV dlagram in both cases is $500J$

Efficiency of carnot engine= $\frac{T_2}{T_1}$ and it does not depend upon the nature of gas where $T_1$ is the temp of the source and $T_2$ is the temp of the sink.
Hence, values of $\gamma$ are irrelevant for the problem.
$\eta =1-\frac{T_2}{T_1}=1-\frac{Q_2}{Q_1}$
Given $T_1={411}^{0}C={684}^{0}K$
$T_2={69}^{0}C={342}^{0}K$
$\therefore \eta =1-\frac{Q_2}{Q_1}=1-\frac{342}{684}=1-\frac{1}{2}=\frac{1}{2}$
Given, $Q_1=1000J$
$\therefore 1-\frac{Q_2}{Q_1}=\frac{1}{2}$
$\therefore \frac{Q_2}{Q_1}=\frac{1}{2}$
$Q_2=500J$
$\therefore Q_1-Q_2=1000-500=500J$
But, $Q_1-Q_2$ is the work done which is same as the area under P-V diagram.