Chemistry - Thermodynamics - II Question with Solution | TestHub

ChemistryThermodynamics - IICarnot Cycle/Cyclic processMedium2 minQB

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Passage / Comprehension

One mole of an ideal monoatomic gas is taken round cyclic process ABCA as shown in figure. Calculate.

The net heat absorbed by the gas in the path $BC$ .

Answer:

Solution:

Also net heat absorbed $= 3 P_{0} V_{0} - \frac{5}{2} P_{0} V_{0} = \frac{P _{0} V _{0}}{2}$

$∴ q_{net} = \frac{P _{0} V _{0}}{2}$

Stream:JEESubject:ChemistryTopic:Thermodynamics - IISubtopic:Carnot Cycle/Cyclic process

⏱ 2mℹ️ Source: QB

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