A Carnot engine cycle is shown in the Fig. (2). The cycle runs between temperatures `T_(H) = alpha T_(L) = T_(0) (alpha gt T)`. Minimum and maximum volume at state `1` and states `3` are `V_(0)` and `nV_(0)` respectively. The cycle uses one mole of an ideal gas with `C_(P)//C_(V) = gamma`. Here `C_(P)` and `C_(V)` are the specific heats at constant pressure and volume respectively. You must express all answers in terms of the given parameters `{alpha, n, T_(0), V_(0) ?}` and universal gas constant `R`.
(a) Find `P, V, T` for all the states
(b) Calculate the work done by the engine in each process: `W_(12), W_(23),W_(34),W_(41)`.
(c) Calculate `Q`, the heat absorbed in the cycle.

Explanation

`P_(1) =(R alpha T_(0))/(V_(0)), P_(2) = (alpha^((gamma)/(gamma-1))RT_(0))/(nV_(0))`
`P_(3) = (RT_(0))/(nV_(0)), P_(4) = (RT_(0))/(alpha^((1)/(gamma-1))V_(0))`
`V_(1) = V_(0), V_(2) = (nV_(0))/(alpha^((1)/(gamma-1)))`
`V_(3) = nV_(0), V_(4) = alpha^((1)/(gamma-1))V_(0)`
`T_(1) = alpha T_(0), T_(2) = alpha T_(0)`
`T_(3) = T_(0), T_(4) = T_(0)`
(b) `W_(12) =- (R )/(gamma-1) (T_(0)-alphaT_(0))`
`W_(34) = RT_(0) (alpha^((1)/(gamma-1))/n)`
`W_(41) = - (R)/(gamma-1) (alphaT_(0)-T_(0))`
(c ) `Q = RT_(0) (alpha-1) ln ((n)/(alpha^(((1)/(gamma-1)))))`