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.

Find the pressure of air in a vessel being evacuated as a function of

1.	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» Correct Answer - `p = p_(0)e^(-Ct//V)` `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))`

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