Two moles of helium gas undergo a cyclic process as shown in Fig. Assuming the gas to be ideal, calculate the following quantities in this process

(a) The net change in the heat energy
(b) The net work done
(c) The net change in internal energy

Explanation

The process from A to B is isobaric (at constant P)
`:. W_(AB)=PdV= nRdT= 2 R (400-300)= 200R`
The process from B to C is isothermal (at constant temperature)
`:. W_(AB)=int_(V_(B))^(V_(C)) pdV= int_(P_(B))^(P_(C))-VdP`
`= int_(P_(B))^(P_(C))-nRT((dP)/P)= nRT "log"_(e)(P_(C))/(P_(B))= nRT"log"_(e)(P_(B))/(P_(C))=2Rxx400 log_(e)2`
The process from C to D is isobaric (at constant P)
`:. W_(CD)=nR DeltaT= 2R(300-400)= -200R`
The process from D to A is isothermal (at constantT)
`:. W_(DA)= -nRT"log"(P_(A))/(P_(D))= -2Rxx300"log"_(e)2`
Total work done in the whole cycle
`W=W_(AB)+W_(BC)+W_(CD)+W_(DA)= 200R+800R "log"_(e)2-200R-600R "log"_(e)2`
`W=200R"log"_(e)2= 200xx8.32xx2.303xx0.3010= 1153.5J`
(ii) As the system returns finally to its initial state, there is no change in internal energy, i.e., `DeltaU=0`
(iii) From first law of thermodynamics, `DeltaQ= DeltaU+ DeltaW=0+1153.5= 1153.5J`