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Question

# A gas is held within a closed chamber. It is undergoing a cyclic process as shown in the figure. The energy removed as heat QCA during iso-choric process CA is 20 J. No energy is transferred as heat during adiabatic process BC and net work done during the cyclic process is 15 J. Determine the energy transferred by the system as heat during iso-baric process AB.

A
10 J of heat is absorbed by the gas.
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B
10 J of heat is released by the gas.
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C
5 J of heat is absorbed by the gas.
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D
5 J of heat is released by the gas.
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Solution

## The correct option is C 5 J of heat is absorbed by the gas. We know that , as the initial and final states of the gas are same . The change in internal energy in a cyclic process is zero (ΔU=0) . So, Net Heat absorbed/rejected by the gas = Net work done during cyclic process. From first law of thermodynamics, we can say that, Q=ΔU+W ⇒Q=W ∴ we can write that , QAB+QBC+QCA=WAB+WBC+WCA ⇒QAB=Wnet−QBC−QCA (∵Wnet=WAB+WBC+WCA) From the data given in the question, QAB=−15−0−(−20) ⇒QAB=+5 J Positive sign shows that, heat has been absorbed by the gas during the iso-baric process Hence, option (c) is the correct answer

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