Molar flux through membrane


> 
> Consider a liquid of pure $\ce{A}$ covered by a composite membrane through which
>  the vapor of $\ce{A}$ is diffusing. The membrane consists of two layers of thickness $l\_1$ and $l\_2$. The diffusion coefficient of $\ce{A}$ in the two layers are $D\_{\ce{A},1}$ and $D\_{\ce{A},2}$ respectively. On the lower side of the membrane, the concentration of $\ce{A}$ ($c^s\_\ce{A})$ is controlled by the vapor pressure of $\ce{A}$. On the upper side the concentration of $\ce{A}$ is zero. The solubulity of $\ce{A}$ in the two membrane layers is the same.
> 
> 
> (a) Derive the expression for the flux $J\_\ce{A}$ through the composite membrane
> 
> 
> [![enter image description here](https://i.stack.imgur.com/9G1yw.png)](https://i.stack.imgur.com/9G1yw.png)
> 
> 
> 


The flux through a membrane is given by the equation:


$$J\_\ce{A} = \frac{D\_{AB}}{L}\left(c^{(m)}\_{\ce{A},F} - c^{(m)}\_{\ce{A},P}\right)$$


where, $c^{(m)}\_{\ce{A},F}$ and $c^{(m)}\_{\ce{A},P}$ are the concentrations at the edges inside the membrane.


So I would assume that $c^{(m)}\_{\ce{A},F}$ = $c^s\_\ce{A}$ and $c^{(m)}\_{\ce{A},P}$ = $c\_\ce{A}$


and because there are two membranes I would have to combine the diffusion coefficients as well as the lengths, which I do by addition to get the equation:


$$J\_\ce{A} = \frac{D\_{\ce{A},1} +D\_{\ce{A},2}}{l\_1 +l\_2} (c^s\_\ce{A} - c\_\ce{A})$$


I am unsure if this is the correct way of deriving the expression for the flux when there are two membranes. Am I thinking correct?


**Edit**


I named the third concentration in the membrane $c\_m$. Then I made two equations for the molar flux.


$$J\_\ce{A} = \frac{D\_{A,1}}{l\_1}(c^s\_A - c\_m)$$


$$J\_\ce{A} = \frac{D\_{A,2}}{l\_2}(c\_m - c\_A)$$


When rearranging and solving for $J\_A$, I get:


$$J\_\ce{A} = \frac{(c^s\_A - c\_A)}{\frac{l\_1}{D\_{A,1}}+\frac{l\_2}{D\_{A,2}}}$$


But that is the same answer as I got before


