find the minimum attainable pressure of an ideal gs in the process `T = t_0 + prop V^2`, where `T_(0)n` and `alpha` are positive constants and (V) is the volume of one mole of gas.

Explanation

Given,
`T = T_(0) + alpha V^(2)` (i)
For 1 mol of a gas, `PV = RT`
or `V = (RT)/(P)`
Substituting this value in Eq. (i) we get
`T = T_(0) + alpha ((RT)/(P))^(2) = T_(0) + alpha (R^(2) T^(2))/(P^(2))`
or `TP^(2) = T_(0)P^(2) + alpha R^(2) T^(2)`
or `P = sqrtalpha RT (T - T_(0))^(-1//2)`
After differenating, we get
`(dP)/(dT) = sqrt alpha R [(T - T_(0))^(-1//2) (1)/(2) T (T - T_(0))^(-3//2)]`
For minimum pressure,
`(dP)/(dT) = 0`
`:. 0 = sqrt alpha R [(T - T_(0))^(-1//2) (1)/(2) T (T - T_(0))^(-3//2)]`
After solving, `T = 2T_(0)`
Form Eq. (ii),
`P_(min) = sqrt alpha R 2T_(0) (2 T_(0) - T_(0))^(-3//2)`
`= 2 R sqrt(alpha T_(0))`