A  human being  (75  kg ) consumes about  6000 Kj   of food    per day. Assume  that the food   is all glucose   and  that  overall reaction   is- C6H12O6  +6O2 ------à 6CO2 +6H20  , Δ H= -2816 KJ

Find  man's   metabolic   rate ( the rate   of living ,loving  and laughing )  in terms   of  moles   of oxygen  used   per   m3 of person per second.

Solution   -

Ro2= -ro2  = - 1/V*(Dno2/dt )  =  mol O2 used/m3  of person -s

Let us  evaluate  the two  terms   in this equation. First   of all,  from  our  life  experience , we estimate   the density  of man  to be-

Ρ=  1000 kg/m3

So,  for the   person   in the question-

Vperson =  75/1000 =0.075  m3

Next  , noting  that   each mole  of glucose  consumed  uses  6  moles of   oxygen  and releases   2816 kj  of energy,

DNO2/dt=  dno2/dt  =6000/2816 *1/6 =12.8  mol  O2/day -ro2 '' = 1/0.075*12.8  mol o2 used/day  *1/24*3600  =0.002  mol -O2/M3-S