One mole of an ideal gas at temperature T was cooled isochorically till the gas pressure fell from P to n P . Then, by an isobaric process, the gas was restored to the initial temperature. The net amount of heat absorbed by the gas in the process is

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One mole of an ideal gas at temperature $T$ was cooled isochorically till the gas pressure fell from $P$ to $\frac{P}{n}$ . Then, by an isobaric process, the gas was restored to the initial temperature. The net amount of heat absorbed by the gas in the process is

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Answer:

Solution:

Since $T_{A} = T_{C} = T, Δ U = 0$ over the whole process: net heat = net work.

$W_{A B} = 0 (isochoric); W_{BC} = \frac{P}{n} (nV - V) = P V (1 - n^{- 1}) = RT (1 - n^{- 1}) .$

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:Thermodynamics

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