Figure. Shows an ideal gas changing its state `A` to state `C` by two different path `ABC` and `AC`.
a. Find the path along which the work done is the least.
b. The internal energy of the gas at `A` is `10 J` and the amount of heat supplied to change its state to `C` through the path `AC` is `200 J`. Find the internal energy at `C`.
c. The internal energy of the gas at state `B` is `20 J`. Find the amount of heat supplied to the gas to go from state `A` to state `B`.

Explanation

(i) for path ABC,
`W= W_(AB)+W_(BC)=P(V_(B)-V_(A))+P_(B)(V_(C)-V_(B))`
`= zero + 15 (6-2)= 60J`
For path AC,
`W_(AC)= "area" ACFEA= "area of" DeltaACD+ area of rectangular ADFE`
`=1/2ADxxDC+ADxxFD= 1/2xx4xx10+4xx5= 40J`
`:.` work done is least along the path AC.
(ii) `U_(A)= 10J, DeltaQ_(AC)=200J`
According to first law of thermodynamics, `DeltaQ_(AC)=DeltaU_(AC)+DeltaW_(AC)=(U_(C)-U_(A))+ DeltaW_(AC)`
`U_(C)= DeltaQ_(AC)+U_(A)-DeltaW_(AC)= 200+10-40= 170J`
(iii) `U_(B)= 20J, DeltaQ_(AB)=?`
According to first law of thermodynamics , `DelatQ_(AB)= DeltaU_(AB)+DeltaW_(AB)=(U_(AB)-U_(A))+DeltaW_(AB)`
`DelatQ_(AB)= (20-10)+Zero= 10J`