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An ideal gas is subjected to cyclic process involving four thermodynamic states, the amounts of heat (Q) and work (W) involved in each of these states are,
$\begin{array}{l}
\mathrm{Q}_{1}=6000 \mathrm{~J}, \mathrm{Q}_{2}=-5500 \mathrm{~J} ; \mathrm{Q}_{3}=-3000 \mathrm{~J} \\
\mathrm{Q}_{4}=3500 \mathrm{~J} \\
\mathrm{~W}_{1}=2500 \mathrm{~J} ; \mathrm{W}_{2}=-1000 \mathrm{~J} ; \mathrm{W}_{3}=-1200 \mathrm{~J} \\
\mathrm{~W}_{4}=\mathrm{x} \mathrm{J}
\end{array}$
The ratio of the net work done by the gas to the total heat absorbed by the gas is $\eta$ ). The values of $x$ and $\eta$ respectively are
Options:
$\begin{array}{l}
\mathrm{Q}_{1}=6000 \mathrm{~J}, \mathrm{Q}_{2}=-5500 \mathrm{~J} ; \mathrm{Q}_{3}=-3000 \mathrm{~J} \\
\mathrm{Q}_{4}=3500 \mathrm{~J} \\
\mathrm{~W}_{1}=2500 \mathrm{~J} ; \mathrm{W}_{2}=-1000 \mathrm{~J} ; \mathrm{W}_{3}=-1200 \mathrm{~J} \\
\mathrm{~W}_{4}=\mathrm{x} \mathrm{J}
\end{array}$
The ratio of the net work done by the gas to the total heat absorbed by the gas is $\eta$ ). The values of $x$ and $\eta$ respectively are
Solution:
2703 Upvotes
Verified Answer
The correct answer is:
$700 ; 10.5 \%$
From first law of thermodynamics
$$
\begin{array}{l}
Q=\Delta U+W \text { or } \Delta U=Q-W \\
\therefore \Delta U_{1}=Q_{1}-W_{1}=6000-2500=3500 \mathrm{~J} \\
\Delta U_{2}=Q_{2}-W_{2}=-5500+1000=-4500 \mathrm{~J} \\
\Delta U_{3}=Q_{3}-W_{3}=-3000+1200=-1800 \mathrm{~J} \\
\Delta U_{4}=Q_{4}-W_{4}=3500-x
\end{array}
$$
For cyclic process $\Delta U=0$
$\therefore 3500-4500-1800+3500-x=0$
or $x=700 \mathrm{~J}$
Efficiency, $\eta=\frac{\text { output }}{\text { input }} \times 100$
$$
\begin{array}{l}
=\frac{W_{1}+W_{2}+W_{3}+W_{4}}{Q_{1}+Q_{4}} \times 100 \\
=\frac{(2500-1000-1200+700)}{6000+3500} \times 100 \\
=\frac{1000}{9500} \times 100=10.5 \%
\end{array}
$$
$$
\begin{array}{l}
Q=\Delta U+W \text { or } \Delta U=Q-W \\
\therefore \Delta U_{1}=Q_{1}-W_{1}=6000-2500=3500 \mathrm{~J} \\
\Delta U_{2}=Q_{2}-W_{2}=-5500+1000=-4500 \mathrm{~J} \\
\Delta U_{3}=Q_{3}-W_{3}=-3000+1200=-1800 \mathrm{~J} \\
\Delta U_{4}=Q_{4}-W_{4}=3500-x
\end{array}
$$
For cyclic process $\Delta U=0$
$\therefore 3500-4500-1800+3500-x=0$
or $x=700 \mathrm{~J}$
Efficiency, $\eta=\frac{\text { output }}{\text { input }} \times 100$
$$
\begin{array}{l}
=\frac{W_{1}+W_{2}+W_{3}+W_{4}}{Q_{1}+Q_{4}} \times 100 \\
=\frac{(2500-1000-1200+700)}{6000+3500} \times 100 \\
=\frac{1000}{9500} \times 100=10.5 \%
\end{array}
$$
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