Respuesta :
Solution :
From Fick's law:
[tex]$\frac{D_{AB}}{\Delta z } \times (C_{A1}-C_{A2})=N_a$[/tex]
Mass balance: Exits = Accumulation
[tex]-N_A A = \frac{dm}{dt}[/tex]
[tex]-N_A A = \frac{dVp}{dt}[/tex]
[tex]-N_A A = \frac{dV}{dt}p[/tex]
[tex]-N_A A = \frac{dhA}{dt}[/tex]
[tex]-N_A A = \frac{dh}{dt} \times Ap[/tex]
From the last step, area cancels out and thus leaves :
[tex]-N_A = \frac{dh}{dt} \times p[/tex]
So now we can substitute the [tex]$N_A$[/tex] by the Fick's law
[tex]$\frac{D_{AB}}{\Delta z } \times (C_{A1}-C_{A2})=\frac{dh}{dt} p$[/tex]
Substituting the values we get
[tex]$=\frac{-4 \times 10^{-13}}{0.0001} \times (3 \times 10^{26} - C_{A2}) = \frac{dh}{dt} \times 2.46$[/tex]
[tex]$=-4 \times 10^{-9} \times( 3 \times 10^{26} - C_{A2}) = 0.0001 \times 2.46$[/tex]
[tex]$= -7800 \times 4 \times 10^{-9} \times (3 \times 10^{26}-C_{A2})=0.000246$[/tex][tex]$=-(3 \times 10^{26}-C_{A2}) = 7.8846 \times 100 \times \frac{1}{69.62} \times 6.022 \times 10^{23}$[/tex]
[tex]$C_{A2} = 6.85 \times 10^{28} \ \text{ boron atoms} /m^3$[/tex]
