The 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,

Options

A. The process during the path `A to B` is isothermal
B. Heat flows out of the gas during the path `B to C to D`
C. Work done during the path `A to B to C` is zero
D. Positive work is done by the gas in the cycle ABCDA

Answer

B, D

Explanation

`Delta Q = Delta U+W`
Fro process ` B to C to D`
`Delta U` is is negative and W is also negative, so 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 to B to C`.