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An ideal gas undergoes the cyclic process abca as shown in the given $\mathrm{P}-\mathrm{V}$ diagram. It rejects $50 \mathrm{~J}$ of heat during $a b$ and absorbs $80 \mathrm{~J}$ of heat during ca. During $b c$, there is no transfer of heat and $40 \mathrm{~J}$ of work is done by the gas. What should be the area of the closed curve abca?
Solution:
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Verified Answer
The correct answer is:
$30 \mathrm{~J}$
Hint:
Area of cycle $=\mathrm{W}_{\text {total }}$
$=\Delta Q_{\text {net }}=\Delta Q_{A B}+\Delta Q_{B C}+\Delta Q_{C A}=-50+0+80=30 \mathrm{~J}$
NOTE: In process $\mathrm{BC}, \mathrm{W}_{\mathrm{BC}}>0$ and $\Delta \mathrm{U}_{\mathrm{BC}}>0$
$\therefore \Delta \mathrm{Q}_{\mathrm{BC}}=\Delta \mathrm{U}_{\mathrm{BC}}+\mathrm{W}_{\mathrm{BC}}>0$
Thus $\Delta \mathrm{Q}_{\mathrm{eC}}=0$ as given in question isn't possible. But we have neglected this technical mistake while answering.
Area of cycle $=\mathrm{W}_{\text {total }}$
$=\Delta Q_{\text {net }}=\Delta Q_{A B}+\Delta Q_{B C}+\Delta Q_{C A}=-50+0+80=30 \mathrm{~J}$
NOTE: In process $\mathrm{BC}, \mathrm{W}_{\mathrm{BC}}>0$ and $\Delta \mathrm{U}_{\mathrm{BC}}>0$
$\therefore \Delta \mathrm{Q}_{\mathrm{BC}}=\Delta \mathrm{U}_{\mathrm{BC}}+\mathrm{W}_{\mathrm{BC}}>0$
Thus $\Delta \mathrm{Q}_{\mathrm{eC}}=0$ as given in question isn't possible. But we have neglected this technical mistake while answering.
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