A chamber of volume `V = 87 1` is evacuated by a pump whose evacuation rate (see Note the foregoing problem) equals `C = 10 1//s`. How soon will the pressure in the chamber decrease by `eta = 1000` times ?

A chamber of volume `V = 87 1` is evacuated by a pump whose evacuation

1.	A chamber of volume `V = 87 1` is evacuated by a pump whose evacuation rate (see Note the foregoing problem) equals `C = 10 1//s`. How soon will the pressure in the chamber decrease by `eta = 1000` times ?
Answer» Let `rho` be the instantaneous density, then instantaneous mass `= V_p`. In a short interval `dt` the volume is increased by `Cdt`. So, `V_rho = (V + Cdt) (rho + d rho)` (because mass remains constant in a short interval `dt`) so `(d rho)/(rho) = - (C)/(V) dt` Since pressure `alpha` density `(dp)/(p) = - (C)/(V) dt` or `int_(p_1)^(p_2) - (dp)/(p) = (C)/(V) t`, or `t = (V)/(C) 1n (p_1)/(p_2) = (V)/(C) 1n (1)/(eta) 1.0 min`.

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A chamber of volume `V = 87 1` is evacuated by a pump whose evacuation rate (see Note the foregoing problem) equals `C = 10 1//s`. How soon will the pressure in the chamber decrease by `eta = 1000` times ?

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