Question

Suppose a cell is suspended in a solution containing a solute of constant concentration${\mathit{c}}_{\mathbf{S}}$. Suppose further that the cell has constant volume Vand that the area of its permeable membrane is the constant A. By Fick’s law the rate of change of its mass mis directly proportional to the area Aand the difference ${\mathit{C}}_{\mathbf{s}}\mathbf{-}\mathit{C}\mathbf{\left(}\mathit{t}\mathbf{\right)}$, where $\mathit{C}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{}$is the concentration of the solute inside the cell at time t. Find $\mathit{C}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{}$if $\mathit{m}\mathbf{=}\mathit{V}$.$\mathit{C}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{}$and$\mathit{C}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathit{C}}_{\mathbf{0}}\mathbf{.}$See Figure 3.R.2.

Short answer

The solution is $C\left(t\right)=C_s-(C_s-C_0)/e^{(}kA/Vt).$

Step-by-step solution

State Fick’s Law

Fick’s first law states that the flux is directly proportional to the concentration gradient. Mathematically the Fick’s law, is written as $dm/dt=k\cdot A\cdot (C_s-C).$Also it is given that $m=V\cdot C\left(t\right).$.

Substitute in the fick’s law

Take the derivative on both sides of the equation $m=V\cdot C\left(t\right).$

$\frac{dm}{dt}=V\cdot \frac{dC}{dt}$

Substitute$V\cdot \frac{dC}{dt}$for $\frac{dm}{dt}$into the fick’s law,

$V\cdot \frac{dC}{dt}=k\cdot A\cdot (C_s-C)$

Solve the differential equation using the initial condition $C\left(0\right)=C_0$provided that $k,A,C_s$and Vare constants.

$$\begin{gathered} V\cdot \frac{dC}{dt}=k\cdot A\cdot (C_s-C) \\ \frac{dC}{dt}=\frac{kA}{V}\cdot (C_s-C) \\ \int \frac{dC}{C_s-C}=\int \frac{kA}{V}(t+C_1) \\ ln(C_s-C)=\frac{kA}{V}(t+C_1) \end{gathered}$$

Substitute 0 for t and $C_0$for$C\left(0\right)$, into $ln(C_s-C)=kA/V(t+C_1)$to find the value of the constant

$-ln(C_s-C_0)=\frac{kA}{V}(0+C_1)$

$-ln(C_s-C_0)\frac{V}{kA}=C_1$

Find the solution

Substitute $-ln(C_s-C_0)\frac{V}{kA}$for $C_1$into $ln(C_s-C)=\frac{kA}{V}(t+C_1)$find $C\left(t\right)$

$$\begin{gathered} -ln(C_s-C)=\frac{kA}{V}(t-ln(C_s-C_0\left)\frac{V}{kA}\right) \\ -ln(C_s-C)=\frac{kA}{V}t-ln(C_s-C_0) \\ ln(C_s-C_0)-ln(C_s-C)=\frac{kA}{V}t \\ ln\frac{C_s-C_0}{C_s-C}=\frac{kA}{V}t \end{gathered}$$

Simplify further as

$$\begin{gathered} \frac{C_s-C_0}{C_s-C}=e^{\frac{kA}{V}t} \\ \frac{C_s-C_0}{e^{\frac{kA}{V}t}}=C_s-C \\ {C}_s-\frac{C_s-C_0}{e^{\frac{kA}{V}t}}=C \end{gathered}$$

Hence the solution is $C\left(t\right)=C_s-\frac{C_s-C_0}{e^{\frac{kA}{V}t}}$.