A cycle followed by an engine (made of one mole of an ideal gas in a cylinder with a piston) is shown in figure. Find heat exchanged by the engine, with the surroundings for each section of the cycle. `C_V=3/2R`

(a) A to B : constant volume
(b) B to C: constant pressure
(c) C to D: adiabatic
(d) D to A: constant pressure

Explanation

(a)For process A to B volume is constant hence work done dW=0
`rArr` From first law of thermodynamics ,
dQ=dU+dW
`therefore` dQ=dU+0
`therefore` dQ=dU
`=nC_V dT`
`=nC_V(T_B-T_A)`
`=(1)(3/2R)(T_B-T_A)`
`=3/2(RT_B-RT_A)`
`=3/2(P_B V_B -P_A V_A)`
because in equation of gas PV = RT at point A, `P_A V_A =RT_A` and at point B, `P_B V_B=RT_B`
`therefore` Heat enchanged `=3/2 (P_B V_B-P_A V_A)`
(b) For process B to C pressure is constant hence work done,
`dW=PDeltaV`
`therefore dW=P_B(V_C-V_B)`
`rArr` From first law of thermodynamics,
dQ=dU+dW
`=3/2R(T_C-T_B)+P_B(V_C+V_B)`
`therefore dQ=3/2 (RT_C -RT_B) +P_B (V_C -V_B)`
`=3/2(P_C V_C- P_B V_B) + P_B (V_C-V_B)`
`=5/2 P_B (V_C -V_B)` [ `because P_B=P_C` and `P_B=P_A` ]
`therefore` Heat exchanged `=5/2P_B (V_C-V_A)`
(d) For process C to D , it is adiabatic process so heat exchanged dQ=0
(d) For process D to A at constant pressure `P_A` the gas compressed from volume `V_D` to `V_A` The heat exchanged will be as shown in (b) `P_A` instead of `P_B`, `V_A` instead of `V_C` and `V_A` instead of `V_D` have to be taken.
`therefore` Heat exchanged `dQ=5/2 P_A(V_A-V_D)`