An ideal gas intially has pressure `P` volume `V` and temperature `T`. Its is isothermally expanded to four times of its original volume, then it is compressed at constant pressure to attain its original volume `V`. Finally, the gas is heated at constant volume to get the original temperature `T`. (a) Draw `V-T` curve (b) Calculate the total work done by the gas in the process.(given `ln2 = 0.693)`

Explanation

(a) `V-T` curve for all process is shown in figure. The initial state is represented by the point `A.` in the first step, it is isothermally expanded to a volume `4V`. This is shown by `AB` . Then the pressure is kept constant and the gas is compressed to the initially volume `V`. From the ideal gas equation, `V//T` is constant at constant pressure `(PV = nRT)`. Hence, the process is shown by a line `BC` which passes through the origin. At point `C`, the volume is `V`. In the final step, the gas is heated at constant volume to a temperature `T`. This is shown by `CA`. The final state is the same as the initial state.
(b) Total work done by gas, `W_(Total) = W_(AB) +W_(BC) +W_(CA)`
`W_(AB) = nRT In (4V)/(V) = 2nRT In2 = 2PV In2`.
Also `P_(A) V_(A) = P_(B) V_(B)` (As `AB` is isothermal process)
or, `P_(B) = (P_(A)V_(A))/(V_(B)) = (PV)/(4V) = (P)/(4)`.
In the step `BC`, the pressure remains constant. Hence the work done is,
`W_(BC) = (P)/(4) (V- 4V) =- (3PV)/(4)`.
In the step `CA`, the volume remains constant and so the work done is zero. The net work done by the gas in the cyclic process is
`W = W_(AB) +W_(BC) +W_(CA) = 2PV In 2 - (3PV)/(4) +0`
Hence, the work done by the gas `0.636 PV`.