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.

Answer

(a) `T_(max) = (2)/(3)(p_(0)//R) sqrt(p_(0)//3 alpha)]`
(b) `T_(max) = p_(0)//e betaR`.

Explanation

(a) `pV = RT`
`V = (RT)/(p)`
`p = p_(0) - alphaV^(2)`
`p = p_(0) - alpha ((RT)/(p))^(2)`
`T = (1)/(Rsqrt(alpha)) sqrt(p_(0)p^(2)-p^(3))`
for maximum value `T, sqrt(p_(0)p^(2)-p^(3))` should be maximum
`(d)/(dP) (p_(0)p^(2)-p^(3)) = 0`
`p = (2p_(0))/(3)`
`T = (1)/(Rsqrt(alpha)) sqrt(p_(0)p^(2)-p^(3))`
`p =(2p_(0))/(3)`
`T_(max)= (2)/(3) (p_(0))/(R ) sqrt((p_(0))/(3alpha))`
(b) `p = p_(0)e^(-betaV)`
or `p = p_(0) e^(-(betaRT)/(p))`
for maximum value of `T`
`(dT)/(dp) = 0`
`p = p_(0)e^(-(betaRT)/(p))`
`ln (p) = ln p_(0) -(betaRT)/(p)`
`ln ((p)/(p_(0))) =- beta (RT)/(p)`
`T =- (p)/(betaR) ln ((p)/(p_(0)))`
for `T_(max)`
After solving `p = (p_(0))/(e) rArr T_(max) = (p_(0))/(e betaR)`.