An ideal gas goes through a reversible cycle a→b→c→d. The V-T diagram is shown below. Processes d→a and b→c are adiabatic. The corresponding P-V diagram for the process is (all figures are schematic and not drawn to scale).

In processes a→b and c→d the volume is directly proportional to temperature of gas, hence the process is isobaric, however, the slope of curve is different in each case. using ideal gas equation PV=nRT ⇒ V=nRT/P slope m=nR/P for a high pressure the slope of the V-T curve will be less. In given diagram the slope of c→d process is lower than a→b process, hence process a→b is at higher pressure, from which option B and D can be ruled out. Now initial rate of change of volume w.r.t. temperature is very less from b→c and d→a as shown in V-T diagram. hence slope of P-V curve should tend to lim ΔV→0 ∂P/∂v = ∞ as initial change in volume is nearly zero, which can be verified in option A. Hence correct answer is option A.