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

a. We know work done is given by the area below `PV` curve thus by observing two `PV` curves, `AC` and `ABC`, we can say that work done in path `AC` is less than that in path `ABC`,
b. Work done in path `AC` by the gas is
`W_(AC) =` area of `ACFEDA`
= area of `ACF +` area of `AFED`
`= (1)/(2) xx (15 - 5) xx (6 - 2) xx 5`
`= 20 + 20 = 40 J`
It is given that heat supplied in process `AC` is
`Q_(AC) = 200 T`
Thus change in intenal energy of gas in path `AC` is, from the first law of thermodynamics, given as
`Q_(AC) = W_(AC) + Delta U_(AC)`
or `Delta U_(AC) = Q_(AC) - W_(AC) = 200 - 40 = 160 J`
As it is given that at state `A`, internal energy of gas is `10 J` thus at state `C`, internal energy is
`Delta U_(AC) = U_(C ) - U_(A)`
or `U_(C ) = Delta U_(AC) + U_(A) = 16 + 10 = 170 J`
c. As in process `AB` no volume change takes place thus no work is done by or on the during path `AB`. Thus according to the first law of thermodynamics, we have
`Q_(AB) = Delta U_(AB) + W_(AB)`
`Q_(AB) = U_(B) + 0`
or `Q_(AB) = 20 - 10 = 10 J`