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

Step-by-step solution

Is an adiabatic process T V &#947; - 1 = c o n s t &#8658; V T 1 &#947; - 1 = c o n s t &#8658; as T increase V decreases at non-uniform rate In process a &#8594; b P = constant as V &#8733; T In process c &#8594; d P &#8242; = constant s V &#8733; T But since slope of V - T graph &#8733; 1 P since slope of ab < slope of cd &#8658; P a b &#62; P c d Also in adiabatic process d &#8594; a as T is increasing V in decreasing &#8658; P is increasing, so P - V diagram is as below