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Question:

Figure shows the P−V plot of an ideal gas taken through a cycle ABCDA. The part ABC is a semi-circle and CDA is half of an ellipse. Then

heat flows out of the gas during the path B→C→D

work done during the path A→B→C is zero

the process during the path A→B is isothermal

positive work is done by the gas in the cycle ABCDA

Solution:

ΔQ=ΔU+W
For process B→C→D ΔU is negative and W is also negative, so ΔQ is also negative, hence heat flows out during this process.
A to B, work is +ve. B to C work is −ve. But +ve work is more, so net work is not zero in process A→B→C.