Find the maximum attainable temperature of ideal gas in each of the following process :
(a) `p = p_0 - alpha V^2` ,
(b) `p = p_0 e^(- beta v)`,
where `p_0, alpha` and `beta` are positive constants, and `V` is the volume of one mole of gas.

Explanation

(a) `p = p_0 - alpha V^2 = p_0 - alpha ((RT)/(p))^2`
(as, `V = RT//p` for one mole of gas)
Thus, `T = (1)/(R sqrt alpha) p sqrt(p_0 - p) = (1)/(R sqrt alpha) sqrt(p_0 p^2 - p^3)`…(1)
For `T_(max), (d)/(dp)(p_0 p^2 - p^3)` must be zero.
which yields, `p = (2)/(3) p_0` ...(2)
Hence, `T_(max) = (1)/(R sqrt(alpha)).(2)/(3) p_0 sqrt(p_0 - (2)/(3) p_0) = (2)/(3) ((p_0)/(R)) sqrt((p_0)/(3 alpha))`
(b) `p = p_0 e^ - beta V= p_0 e^(- beta RT//p)`
so `(beta RT)/(p) = 1 n(p_o)/(p)`, and `T = (p)/(beta R) 1 n(p_0)/(p)`.....(1)
For `T_(max)` the condition is `(dT)/(dp) = 0`, which yields
`p = (p_0)/(e)`
Hence using this value of `p` in Eq. (1), we get
`T_(max) = (p_0)/(e beta R)`.