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A worker in a hazardous waste recycling facility inhales and absorbes toluene at a rate of 22 m g/h. He excretes 2 L per day containg 18 mg/L toluene. If the blood volume of the individual is 5.6 L and the toluene is metabolised with a first-order rate constant of 0.08 h -1 , what is the steady-state concentration of toluene in the individual’s vascular system? Assume that storage is negligible.
A worker in a hazardous waste recycling facility inhales and absorbes toluene at a rate of 22 mg/h. He excretes 2 L per day containg 18 mg/L toluene. If the blood volume of the individual is 5.6 L and the toluene is metabolised with a first-order rate constant of 0.08 h-1, what is the steady-state concentration of toluene in the individual’s vascular system? Assume that storage is negligible.

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3 years ago

Saurabh Koranglekar
10341 Points
```							Neglecting mass transferbetween the vascular systemand the rest of the body, and either transient or time delay (lag) contributions; the toluene uptake (absorption) should equal the combined effects of excretionand metabolic depletion. The lastwas said to constitute a first-order kinetic process. Let V be the blood volume, while Cstands forthe volumetric toluene concentration in the vascular system ― treatedas homogeneous with respect to concentration ―and k is the metabolic rate constantwith units of reciprocal time. Hence we may write:-dC/dt = k·CandV·dC/dt= U - E, where U ≥ 0and E ≥ 0 stand for uptake and excretion rates, respectively; considered constant for t > 0. Integration, from initial (i) time (ti= 0) to elapsed time (t), gives: ∫ dC/C = - ∫0,tk·dt,or,Ct= Ci·exp(-k·t). Steady state(s) concentration, even if asymptotically approached, might bethought to be practically and approximately attained for Cs=- (U - E) / (k·V) ≥ 0 (k > 0), after correspondingelapsed time, ts= - [ln(Cs/Ci)] / k > 0. The initial concentration should be comprised within:E / (k·V) ≥ Ci≥ 0.Itmay be judgedreasonable to limit absorption to a relatively shorter time interval (possibly initial) compared with (final) elapsed time. In that case, a concentration maximum(Cmax) could be expected, by the end of absorption, for tmax= - [ln(Cmax/Ci)] / k ≥ 0
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9 months ago
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