One mole of an ideal monoatomic gas is taken along the path ABCA as shown in the PV diagram. The maximum temperature attained by the gas along the path BC is given by:

Temperature at A =3P0V0/nR
Temperature at b =P0×2V0/nR
Maximum temperature can be between B and C
P-V equation for process BC
P-P0 = P0(2V0 - V0)/(V0 - V0) (V - V0)
P = P0V/(V0) + 2P0
Multiply by V
PV = P0V^2/(V0) + 2P0V
RT = P0V^2/(V0) + 2P0V
Make dT/dV = 0
This gives T = 25P0V0/8R