Question 10.35: For the diffusion of carbon dioxide at atmospheric pressure ...

$\frac{∂C_{A}}{∂t} =D\frac{∂^{2}C_{A}}{∂y^{2}}$

where $C_{A}$is concentration of solvent undergoing mass transfer.
The boundary conditions are:

Taking Laplace transforms then:

$\frac{\overline{∂C_{A}} }{∂t} =\int_{0}^{\infty }\left(\frac{∂C_{A}}{∂t} \right) e^{-pt}dt$

$=[C_{A}e^{-pt}]_{0}^{\infty }-\int_{0}^{\infty }(-pe^{pt})C_{A}dt=0+p\overline{C} _{A}$

$\frac{\overline{∂^{2}C_{A}} }{∂y^{2}} =\frac{∂^{2}\overline{C}_{A} }{∂y^{2}}$

Thus:   $p\overline{C} _{A}=D\frac{∂^{2}\overline{C}_{A} }{∂y^{2}}$

$\frac{∂^{2}\overline{C}_{A} }{∂y^{2}}-\frac{p}{D} \overline{C} _{A}=0$

$\overline{C}_{A}=Ae^{\sqrt{p/D}y }+Be^{-\sqrt{p/D}y }$

For t > 0;

Thus:  $\frac{C_{As}}{p} =B.1$

and:  $\overline{C} _{A}=\frac{C_{As}}{p} e^{-\sqrt{p/D} y}$

Inverting:  $\frac{C_{A}}{C_{As}} =\text{erfc}\frac{y}{2\sqrt{Dt} }$  (See Table in Volume 1, Appendix)

Differentiating with respect to y:

$\frac{1}{C_{As}} \frac{∂C_{A}}{∂y} =\frac{∂}{∂y} \left\{\frac{2}{\sqrt{\pi } }\int_{y/2\sqrt{D}t }^{\infty }e^{-y^{2}/4Dt}d\left(\frac{y}{2\sqrt{Dt} } \right) \right.$

$=\frac{2}{\sqrt{\pi } } .\frac{1}{2\sqrt{D}t } (e^{-y^{2}/4Dt})=-\frac{1}{\sqrt{\pi Dt} } e^{-y^{2}/4Dt}$

The mass transfer rate at t, y, N = $-D\left\{\frac{-1}{\sqrt{\pi Dt} } e^{-y^{2}/4Dt}\right\} C_{As}$

when: t > 0, then: $\left(-D\frac{∂C_{A}}{∂y}\right) _{y=y}=C_{As}\sqrt{\frac{D}{\pi Dt} } e^{-y^{2}/4Dt}$    (i)

At t > 0  and  y = 0, then: $N_{0}=C_{As}\sqrt{\frac{D}{\pi t} }$     (ii)

For a concentrated 1% of surface value at y = 1 mm, $C_{A}/C_{As}$ = 0.01 and:

$0.01 = \text{erfc}\left\{\frac{10^{-3}}{2\sqrt{1.5\times 10^{-9}t} } \right\}$

Writing     erf x = 1 – erfc x, then:
0.99 = erf(12.91$t^{-1/2})$

From tables     1.82 = 12.91$t^{-1/2}$
t = 50.3 s

The mass transfer rate at the interface at t = 50.3 s is given by equation (ii) as:

$N_{0}=C_{As}\sqrt{\frac{D}{\pi t} } =C_{As}\sqrt{\frac{1.5\times 10^{-9}}{\pi \times 50.3} }$

= 3.08 × $10^{-6} C_{As}$ kmol/m² s

The mass transfer rate at y = 1 mm. and t = 50.3 s is given by equation (i) as:

$N=N_{0}e^{-y^{2}/4Dt}=3.08\times 10^{-6}e^{-10^{-6}/(4\times 1.5\times 10^{-9}\times 50.3)}C_{As}$

= 1.121 × $10^{-7} C_{As}$ where $C_{As}$ is in kmol/m³.

Henry’s law constant, K = 1.08 × 10⁶, where: K = P/X.
Also: P = 760 mm Hg
X = Mol fraction in liquid
X = 760/(1.08 × 10⁶) = 7.037 × $10^{-4}$ kmol CO₂ kmol solution (≈ per kmol water)
Water molar density = (1000/18) kmol/m³

and:  $C_{As}=7.037\times 10^{-4}\left(\frac{1000}{18} \right) =0.0391$ kmol/m³

When y = 0, then: $N_{A0}$ = 1.204 × $10^{-7}$ kmol/m² s.
When y = 1 mm, then: $N_{A}$ = 4.38 × $10^{-9}$ kmol/m² s.