One mole of an ideal monatomic gas is taken round the cyclic process ABCA as shown in figure. Calculate

(a) the work done by the gas.
(b) the heat rejected by the gas in the path CA and the heat absorbed by the gas in the path AB,
(c) the net heat absorbed by the gas in the path BC,
(d) the maximum temperature attained by the gas during the cycle.

Answer

A, B, C, D

Explanation

`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`