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Suppose a cell is suspended in a solution containing a solute of constant concentration \(C_{x}\). Suppose further that the cell has constant volume \(V\) and that the area of its permeable membrane is the constant \(A\). By Fick's law the rate of change of its mass \(m\) is directly proportional to the area \(A\) and the difference \(C,-C(t)\), where \(C(t)\) is the concentration of the solute inside the cell at any time \(t\). Find \(C(t)\) if \(m=V C(t)\) and \(C(0)=C_{11}\). See Figure \(3.25\).

Short Answer

Expert verified
The concentration inside the cell is \( C(t) = C_x + (C_{11} - C_x) e^{-\frac{kA}{V}t} \).

Step by step solution

01

Understand Fick's Law

Fick's Law states that the rate of change of mass through a membrane is directly proportional to the membrane area and the concentration difference across the membrane. This can be expressed as \( \frac{dm}{dt}=kA(C_x - C(t)) \), where \( k \) is a proportionality constant, \( A \) is the area, \( C_x \) is the external concentration, and \( C(t) \) is the internal concentration.
02

Relate Mass and Concentration

The mass \( m \) of the solute in the cell is proportional to the concentration \( C(t) \) by the relationship \( m=V C(t) \). We substitute this into Fick's Law equation, leading to \( \frac{d(V C(t))}{dt} = kA(C_x - C(t)) \).
03

Simplify the Differential Equation

Since \( V \) is a constant, the equation simplifies to \( V \frac{dC(t)}{dt} = kA(C_x - C(t)) \). Dividing through by \( V \), we get \( \frac{dC(t)}{dt} = \frac{kA}{V}(C_x - C(t)) \).
04

Solve the Differential Equation

This is a first-order linear differential equation. Rearrange to \( \frac{dC(t)}{dt} + \frac{kA}{V}C(t) =\frac{kA}{V}C_x \). The integrating factor \( \mu(t) \) is \( e^{\int \frac{kA}{V} dt} = e^{\frac{kA}{V}t} \). Multiply through by the integrating factor: \( e^{\frac{kA}{V}t} \frac{dC}{dt} + \frac{kA}{V} e^{\frac{kA}{V}t} C = \frac{kA}{V} C_x e^{\frac{kA}{V}t} \).
05

Integrate to Find Solution

The left-hand side becomes \( \frac{d}{dt}(e^{\frac{kA}{V}t} C(t)) = \frac{kA}{V} C_x e^{\frac{kA}{V}t} \). Integrate both sides: \( e^{\frac{kA}{V}t} C(t) = C_x e^{\frac{kA}{V}t} + C_1 \). Solve for \( C(t) \): \( C(t) = C_x + C_1 e^{-\frac{kA}{V}t} \).
06

Apply Initial Condition

Use the initial condition \( C(0) = C_{11} \) to find \( C_1 \). Substitute into the equation: \( C_{11} = C_x + C_1 \), giving \( C_1 = C_{11} - C_x \). Substitute \( C_1 \) back into the equation: \( C(t) = C_x + (C_{11} - C_x) e^{-\frac{kA}{V}t} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First-order Linear Differential Equation
A first-order linear differential equation consists of a derivative of a function, the function itself, and perhaps some extra constant terms. It often takes the form: \( \frac{dy}{dt} + P(t)y = Q(t) \). In this equation, \( y \) is the function of time \( t \), \( P(t) \) is a coefficient of \( y \), and \( Q(t) \) is a separate function or constant. These equations are called "first-order" because they involve the first derivative of the function.In our exercise involving Fick's Law, we encounter a first-order linear differential equation in the form of \( \frac{dC(t)}{dt} = \frac{kA}{V}(C_x - C(t)) \). By rearranging, we write it in a form similar to our general linear equation: \( \frac{dC(t)}{dt} + \frac{kA}{V}C(t) = \frac{kA}{V}C_x \), setting us up to apply further solving techniques, like using an integrating factor. Understanding this form is crucial for solving differential equations using systematic methods.
Initial Conditions
Initial conditions are the values provided for solving a differential equation at a specific point, often at the start of an observation, typically time \( t=0 \). These are essential because they allow us to find a specific solution that fits our particular scenario, rather than just a general solution.In our context, we have the initial condition \( C(0) = C_{11} \). This means that at time \( t=0 \), the concentration inside the cell is \( C_{11} \). By substituting this into our general solution, we can find the constant of integration, \( C_1 \), which helps us pin down the exact form that \( C(t) \) needs to take. So, without initial conditions, we would be left with potentially infinite possibilities of solutions.
Concentration Gradient
The concentration gradient represents the difference in solute concentration across a boundary, like a cell membrane. It is this difference that drives the diffusion process according to Fick's Law.In the exercise, the concentration gradient is the difference between the external concentration \( C_x \) and the internal concentration \( C(t) \). This gradient is fundamental in determining the rate and direction of solute movement. Essentially, solutes move from areas of higher concentration to areas of lower concentration, which helps achieve equilibrium over time. The role of the concentration gradient is made evident in the rate equation \( \frac{dm}{dt} = kA(C_x - C(t)) \), where the gradient directly influences the rate of mass change.
Integrating Factor
The integrating factor is a clever mathematical tool used to solve first-order linear differential equations. It is a multiplier that simplifies the differential equation into a form that can be easily integrated.The formula for the integrating factor \( \mu(t) \) is generally given by \( e^{\int P(t) \ dt} \), where \( P(t) \) is the coefficient of the function in the equation. In our exercise, the integrating factor was found to be \( e^{\frac{kA}{V}t} \). By multiplying the entire differential equation by this factor, it transforms into a form where the left-hand side becomes the derivative of a product: \( \frac{d}{dt}(e^{\frac{kA}{V}t} C(t)) \). This simplification allows straightforward integration to solve for \( C(t) \), showcasing the power and utility of integrating factors in solving these types of equations.

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Most popular questions from this chapter

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