Correct Answer - A::B::C::D
`n=1, For monoatomic gas: C_p=(5R)/2, C_v=(3R)/2`
Cyclic process
`AtoB` rArr Isochoric process
`CtoA` rArr Isochoric compression
(a) Work done=Area of closed curve ABCA during cyclic
process. i.e. `DeltaABC`
`DeltaW=1/2xxbasexxheight=1/2V_0xx2P_0=P_0V_0`
(b) Heat rejected by the gas in the path CA during isobaric
compression process
`DeltaQ_(CA)=nC_pDeltaT=1xx(5R//2)(T_A-T_C)`
`T_C=(2P_0V_0)/(IxxR), T_A=(P_0V_0)/(IxxR)`
`DeltaQ_(CA)=(5R)/2[(P_0V_0)/R-(2P_0V_0)/R]=-5/2P_0V_0`
Heat absorbed by the gas on the path AB during
isochoric process
`DeltaQ_(AB)=nC_vDeltaT=1xx(3R//2)(T_B-T_A)`
`=(3R)/2[(3P_0V_0)/(IxxR)-(P_0V_0)/(IxxR)]=3P_0V_0`
As `DeltaU=0` in cyclic process, hence,
`DeltaQ=DeltaW`
`DeltaQ_(AB)+DeltaQ_(CA)+DeltaQ_(BC)=DeltaW`
`DeltaQ_(BC)=P_0V_0-(P_0V_0)/2=(P_0V_0)/2`
(d) Equation for Line BC is `P=-[(2P_0)/(V_0)]V+5P_0,`
`P=(RT)/V [For one mole]`
`:. RT=-(2P_0)/(V_0)V^2+5P_0V....(i)`
For maximum, `(dT)/(dV)=0,-(2P_0)/(V_0)xx2V+5P_0=0,`
`:. V=(5V_0)/4....(ii)`
Hence from equation (i) and (ii)
`RT_(max)=(-2P_0)/(V_0)xx((5V_0)/4)^2+5P_0((5V_0)/4)`
`=-2P_0V_0xx25/16+(25P_0V_0)/4=25/8P_0V_0`
`:. T_(max)=25/8 (P_0V_0)/R`