The correct options are
B the work done by the gas in the process A to B exceeds the work that would be done by it if the system were taken from A to B along the isotherm
C in the T-V diagram, the path AB become part of parabola
D in going from A to B, the temperature T gas of the gas first increases to a maximum values then decreases
Work done = Area under P−V graph
As A1>A2
∴Wgiven process>Wisothermal process
Thus (A) is correct.
In the given process P−V equation will be a straight line with negative slope and positive intercept, i.e.,
P=−aV+b
Here, a and b are positive constants.
P=−aV+b
⇒PV=−aV2+bV
⇒nRT=−aV2+bV[∵PV=nRT(From ideal gas equation)]
⇒T=1nR(−aV2+bV)
The above equation is of parabola.
Thus (B) is correct.
T=1nR(−aV2+bV)
dTdV=0
⇒1nR(−2aV+b)=0
⇒−2aV+b=0
⇒V=b2a
Now,
d2TdV2=−2a=−ve
∴T has some maximum value.
Now, from ideal gas equation,
T∝PV
∵(PV)A=(PV)B
∴TA=TB
We conclude that temperatures are same at A and B but in between the temperature has a maximum value.
Hence in going from A to B, the temperture will first increase to a maximum value and then decrease.
Thus (D) is correct.