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Byju's Answer

Standard XII

Chemistry

Work Done in Isothermal Reversible Process

An ideal gas ...

Question

An ideal gas is initially at $P_{1}, V_{1}$ is expanded to $P_{2}, V_{2}$ and then compressed adiabatically to the same volume $V_{1}$ and pressure $P_{3}$ . If $W$ is the net work done by the gas in the complete process, which of the following is true?

W > 0; P_{3} > P_{1}

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W < 0; P_{3} > P_{1}

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W > 0; P_{3} < P_{1}

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W < 0; P_{3} < P_{1}

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Solution

The correct option is B $W < 0; P_{3} > P_{1}$
In the first process $W$ is $+ v e$ as $△ V$ is $+ v e$ , in the second process $W$ is $- v e$ as $△ V$ is $- v e$ and area under the curve of second process is more. Therefore, the net work $< 0$ and also $P_{3} > P_{1}$ .

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An ideal gas is initially at P1,V1 is expanded to P2,V2 and then compressed adiabatically to the same volume V1 and pressure P3. If W is the net work done by the gas in the complete process, which of the following is true?

The correct option is B W<0;P3>P1In the first process W is +ve as △V is +ve, in the second process W is −ve as △V is −ve and area under the curve of second process is more. Therefore, the net work <0 and also P3>P1.

An ideal gas is initially at $P_{1}, V_{1}$ is expanded to $P_{2}, V_{2}$ and then compressed adiabatically to the same volume $V_{1}$ and pressure $P_{3}$ . If $W$ is the net work done by the gas in the complete process, which of the following is true?

The correct option is B $W < 0; P_{3} > P_{1}$
In the first process $W$ is $+ v e$ as $△ V$ is $+ v e$ , in the second process $W$ is $- v e$ as $△ V$ is $- v e$ and area under the curve of second process is more. Therefore, the net work $< 0$ and also $P_{3} > P_{1}$ .