Find the pressure of air in a vessel being evacuated as a function of evacuation time `t`. The vessel volume is `V`, the initial pressure is `p_0`. The process is assumed to be isothermal, and the evacuation rate equal to `C` and independent of pressure.
The evatuation rate is the gas volume being evacuated per unit time, with that volume being measured under the gas pressure attained by that moment.

Answer

`p = p_(0)e^(-Ct//V)`

Explanation

`m = V_(rho)`
in small interval dt the increase in volume `dV = C dt`
`m = V_(rho) = (V +C dt) (rho +d rho)`
`V_(rho) = V_(rho) +Vd_(rho) +rhoC dt`
`rho C dt =- V d rho ………..(i)`
but `P alpha rho`
`P = k rho`
`(dP)/(P) = (d rho)/(rho).........(2)`
`C dt = - (V d rho)/(rho) =- (VdP)/(P)` [from (1) and (2)]
`C int_(0)^(t) dt =- V int_(P_(0))^(P) (dP)/(P)`
`ln ((P)/(P_(0))) = (-C)/(V)t`
`(P)/(P_(0)) = e^(-(Ct)/(V))`
`P = P_(0) e^(-(Ct)/(V))`