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Byju's Answer

Standard XII

Physics

Work Done

An ideal gas ...

Question

An ideal gas expands isothermally from a volume $V_{1}$ to $V_{2}$ to and then compressed to original volume $V_{1}$ adiabatically. Initial pressure is $P_{1}$ and final pressure is $P_{3}$ . The total work done is W. Then

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Solution

The correct option is C

From graph it is clear that $P_{3} > P_{1}$ .

Since area under adiabatic process (BCED) is greater than that of isothermal

process (ABDE). Therefore net work done

$W = W_{i} + (- W_{A}) ∵ W_{A} > W_{i} \Rightarrow W < 0$

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An ideal gas expands isothermally from a volume V1 to V2 to and then compressed to original volume V1 adiabatically. Initial pressure is P1 and final pressure is P3. The total work done is W. Then

The correct option is C From graph it is clear that P3>P1. Since area under adiabatic process (BCED) is greater than that of isothermal process (ABDE). Therefore net work done W=Wi+(−WA)∵WA>Wi⇒W<0

An ideal gas expands isothermally from a volume $V_{1}$ to $V_{2}$ to and then compressed to original volume $V_{1}$ adiabatically. Initial pressure is $P_{1}$ and final pressure is $P_{3}$ . The total work done is W. Then

The correct option is C

From graph it is clear that $P_{3} > P_{1}$ .

Since area under adiabatic process (BCED) is greater than that of isothermal

process (ABDE). Therefore net work done

$W = W_{i} + (- W_{A}) ∵ W_{A} > W_{i} \Rightarrow W < 0$