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 `AtoB` is isothermal
B. work done during the path `BtoCtoD`
C. heat flows out of the gas during the path `BtoCtoD`
D. the process during the path `AtoB` is isothermal

Answer

B, D

Explanation

(b,d)
In case of an isothermal process we get a rectangular
hyperbola in a P-V diagram. Therefore option (a) is wrong.
`T_DltT_B.` Therefore in process `BtoCtoD, DeltaU` is negative.
PV decreases and volume also decrease, therefore W is
negative. From first law os thermodynamic, Q is negative
i.e. There is a heat loss option (b) is correct.
`W_(AB)gtW_(BC).`
Therefore work done during path `AtoBtoC` is positive,
option (c) is wrong
Work done is clockwise cycle in a PV diagram is positive.
Option (d) is correct.