An ideal gas whose adiabatic exponent equals `gamma` is expanded so that the amount of heat transferred to the gas is equal to the decrease of its internal energy. Find :
(a) the molar heat capacity of the gas in the process ,
The equation of the process in the variables `t, V` ,
( c) the work performed by one mole of the gas when its volume increases `eta` times if the initial temperature of the gas is `T_0`.

Explanation

(a) `Delta U = (vR)/(gamma - 1) Delta T` and `Q = vC_n Delta T`
where `C_n` is the molar heat capacity in the process. It is given that `Q = - Delta U`
So, `C_n Delta T = (R)/(gamma - 1) Delta T`, or `C_n = -(R)/(gamma - 1)`
(b) By the first law of thermodynamics, `dQ = dU + dA`,
or, `2 dQ = dA (as dQ = -dU)`
`2 vC_n dT = pdV`, or `(2 Rv)/(gamma - 1) dT + pdV = 0`
So, `(2 RV)/(gamma -1) dT + (vRT)/(V)dV = 0`, or `(2 dT)/((gamma - 1)T) + (dV)/(V) = 0`
or, `(dT)/(T) + (gamma -1)/(2)(dV)/(V) = 0`, or, `TV^(gamma-1//2) = constant`
( c) We know `C_n = ((n - gamma)R)/((n -1)(gamma - 1))`
But from part (a), we have `C_n = -(R)/(gamma - 1)`
Thus `-(R)/(gamma - 1) = ((n - gamma)R)/((n - 1)(gamma - 1))` which yields
`n = (1 + gamma)/(2)`
From part (b) , we know `TV^(gamma - 1)//2) = constant`
So, `(T_o)/(T) = ((V)/(V_0))^((gamma - 1)//2) = eta^((gamma - 1)//2)` (where `T` is the final temperature)
Work done by the gas for one mole is given by
`A = R((T_0 - T))/(n - 1) = (2 RT_0[1 - eta^((1 - gamma)//2)])/(gamma - 1)`.