AP EAMCET · PHYSICS · Thermodynamics
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
\(\begin{aligned}
& Q_1=6000 \mathrm{~J}, \quad Q_2=-5500 \mathrm{~J} ; Q_3=-3000 \mathrm{~J} ; \\
& Q_4=3500 \mathrm{~J} \\
& W_1=2500 \mathrm{~J} ; \quad W_2=-1000 \mathrm{~J} ; \quad W_3=-1200 \mathrm{~J} ; \\
& W_4=x \mathrm{~J} .
\end{aligned}\)
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
- A \(500 ; 7.5 \%\)
- B \(700 ; 10.5 \%\)
- C \(1000 ; 21 \%\)
- D \(1500 ; 15 \%\)
Answer & Solution
Correct Answer
(C) \(1000 ; 21 \%\)
Step-by-step Solution
Detailed explanation
From first law of thermodynamics \(\begin{aligned} & \qquad Q=\Delta U+W \\ & \text { or } \quad \Delta U=Q-W \\ & \therefore \Delta U_1=Q_1-W_1=6000-2500=3500 \mathrm{~J} \\ & \Delta U_2=Q_2-W_2=-5500+1000=-4500 \mathrm{~J} \end{aligned}\)…
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