JEE Mains · Physics · STD 11 - 11. thermodynamics
A sample of an ideal gas is taken through the cyclic process \(ABCA\) as shown in figure. It absorbs, \(40\,J\) of heat during the part \(A B\), no heat during \(BC\) and rejects \(60\,J\) of heat during \(CA\). \(A\) work \(50\,J\) is done on the gas during the part \(BC\). The internal energy of the gas at \(A\) is \(1560\,J\). The work done by the gas during the part \(CA\) is.............\(J\)

- A \(20\)
- B \(30\)
- C \(-30\)
- D \(-60\)
Answer & Solution
Correct Answer
(B) \(30\)
Step-by-step Solution
Detailed explanation
\(\Delta Q_{\text {cycle }}=40-60=\Delta W\) \(\Rightarrow \Delta W =-20 J = W _{ BC }+ W _{ CA }\) \(\Rightarrow W _{ CA }=-20 J - W _{ BC }\) \(=-20-(-50)\) \(=30\,J\)
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