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.

Explanation

From the ideal gas equation `p = (m)/(M)(R T)/(V)`
`(dp)/(dt) = (RT)/(MV) (dm)/(dt)` ….(1)
In each stroke, volume `v` of the gas is ejected, where `v` is given by
`v = (V)/(m_N) [m_(N - 1) - m_N]`
In case of continous ejection, if `(m_(N - 1))` corresponds to mass of gas in the vessel at time `t`, then `m_N` is the mass at time `t + Delta t`, where `Delta t`, is the time in which volume `v` of the gas has come out. The rate of evacuation is therefore `(v)/(Delta t)` i.e.,
`C = (v)/(Delta t) = - (V)/(m(t + Delta t)). (m(t + Delta t) - m(t))/(Delta t)`
In the limit `Delta t - 0`, we get
`C = (V)/(m) (dm)/(dt)` ...(2)
From (1) and (2)
`(dp)/(dt) = - (C)/(V) (m RT)/(MV) = -(C)/(V) p` or `(dp)/(p) = - (C)/(V) dt`
Integrating `int_p^(p_0) (dp)/(p) = - (C)/(V) int_t^0 dt` or `1 n (p)/(p_0) = -(C)/(V) t`
Thus `p = p_0 ^(-C t//V)`.