In the given figure, an ideal gas changes it state from `A` to state `C` by two paths `ABC` and `AC`.
(i) Find the path along which work done is the least.
(ii) The internal energy of the gas at `A` to `10J` and the amount of heat supplied to change its state to `C` through the path `AC` is `200J`. Calculate the internal energy at `C`.
(iii) The internal energy of the gas at state `B` is `30J`. Find the amount of heat supplied to the gas to go from `A to B`.

Answer

(i) along `AC`
(ii) `170J` (iii) `20J`

Explanation

(i) Along path `AC` work done is least. As areas under `P -V` diagram along `AC` is less as compare to along path `ABC`
`DeltaW_(ABC) = `Area under the `EABCDE = 15 xx 4 = 60J`.
`DeltaW_(AC) =` Area under `EACDE = 5 xx 4 +(10 xx4)/(2) = 40J`.

(ii) `DeltaW_(AC) = 40 J` (as calculated above)
`DeltaQ_(AC) = 200J`.
`:. DeltaU_(AC) = DeltaQ_(AC) - DeltaW_(AC) = 200 - 40 = 160`
`U_(AC) = U_(C) -U_(A) = 160`
`:. U_(C) = U_(A) +160 = 10 +160 = 170 J`,
(iii) `DeltaW_(AB) = 0`
Ideal gas equation `PV = nRT`
`DeltaP. V = n R DeltaT ( :' V = "const")`
`DeltaT = (DeltaP.V)/(nR)`
`DeltaU = nC_(v)DeltaT = nC_(v). (DeltaPV)/(nR) =- (C_(v))/(R ) . DeltaP.V`
`:. U_(B) = (C_(V))/(R ) P_(B). V`
`U_(A) = (C_(V))/(R)P_(A).V`
Given `U_(B) = 30`
`(C_(V))/(R) xx P_(B) xx V = (C_(V))/(R) xx 15 xx2`
`=30 xx (C_(V))/(R) = 30, (C_(V))/(R) = 1`
`:. U_(A) = (C_(V))/(R) = 30, (C_(V))/(R) =1`
`:. U_(A) = (C_(V))/(R) xx P_(A) xx V = 1 xx5 xx2 = 10 J`
`:. DeltaU_(AB) = U_(B) - U_(A) = 30 - 10 = 20J`
`:. DeltaQ_(A rarr B) = DeltaU_(A rarr B) +DeltaW_(A rarr B) = 20 +0 = 20J`.