The time dependent concentration of solute particles in a solution is given by
c(r, t) = [n/8(pi D t)^(3/2)] e^(-r^2 /4 D t)
where r^2 = x^2 + y^2 + z^2 with x, y, and z being the values of three components of the position in the Cartesian coordinate system.
a) Show that the above distribution satisfies the following three dimensional diffusion equation:
(partial/partial t)c(r,t) = D (partial^2 / partial x^2 + partial^2 / partial y^2 + partial^2 / partial z^2) c(r, t)
b) Find out the expression for average distance <r> as a function of time, where <...> means average over c(r, t)
c) Find out the experession for average mean-square distance <r^2>
a) To show that the given concentration distribution satisfies the three-dimensional diffusion equation, we need to calculate the partial derivatives in the diffusion equation and substitute the given concentration distribution expression.
First, let's calculate the partial derivative of c(r, t) with respect to time (t):
∂c(r, t) / ∂t = ∂/∂t [(n/8πDt)^(3/2) e^(-r^2 /4Dt)]
To differentiate this expression, we can use the chain rule. Let's denote A = (n/8πD)^(3/2) and B = e^(-r^2 /4Dt). Then we have:
∂c(r, t) / ∂t = ∂/∂t (AB)
= A*(∂B/∂t) + B*(∂A/∂t)
Now, let's differentiate A and B with respect to t:
∂A/∂t = (∂/∂t)[(n/8πD)^(3/2)]
= (3/2)*(n/8πD)^(1/2)*[∂/∂t(n/8πD)]
= (3/2)*(n/8πD)^(1/2)*[(-n/8πD^2)*(∂D/∂t)]
= (-3n/16πD^2)*(n/8πD)^(1/2)*(∂D/∂t)
∂B/∂t = (∂/∂t)e^(-r^2 /4Dt)
= (-r^2 /4D)*e^(-r^2 /4Dt)*[∂/∂t(r^2 /4D)]
= (-r^2 /4D)*e^(-r^2 /4Dt)*(-1/4D^2)*[∂/∂t(r^2)]
= (r^2 /16D^3)*e^(-r^2 /4Dt)*[∂/∂t(r^2)]
Now, let's calculate the partial derivative of c(r, t) with respect to position (x, y, z):
∂c(r, t) / ∂x = ∂/∂x [(n/8πDt)^(3/2) e^(-r^2 /4Dt)]
We can again use the chain rule and denote C = (n/8πDt)^(3/2). Then we have:
∂c(r, t) / ∂x = ∂/∂x (CB)
= C*(∂B/∂x) + B*(∂C/∂x)
Similarly, we can calculate ∂c(r, t) / ∂y and ∂c(r, t) / ∂z using the same approach.
Now, let's calculate the second derivatives in the diffusion equation:
∂^2c(r, t) / ∂x^2 = ∂/∂x (∂c(r, t) / ∂x)
= ∂/∂x (C*(∂B/∂x) + B*(∂C/∂x))
= (∂C/∂x)*(∂B/∂x) + C*(∂^2B/∂x^2) + (∂B/∂x)*(∂C/∂x)
Similarly, we can calculate ∂^2c(r, t) / ∂y^2 and ∂^2c(r, t) / ∂z^2.
Substituting all these partial derivatives back into the diffusion equation:
(∂c(r, t) / ∂t) = D*(∂^2c(r, t) / ∂x^2 + ∂^2c(r, t) / ∂y^2 + ∂^2c(r, t) / ∂z^2)
We can compare the expressions on both sides of the equation to verify if they are equal.
b) To find the expression for average distance <r> as a function of time, we need to integrate the concentration distribution c(r, t) multiplied by the distance r over all space, and divide by the total concentration:
<r> = ∫c(r, t)*r*dV / ∫c(r, t)*dV
Here, dV represents the differential volume element.
c) To find the expression for average mean-square distance <r^2> as a function of time, we need to integrate the concentration distribution c(r, t) multiplied by the square of the distance r^2 over all space, and divide by the total concentration:
<r^2> = ∫c(r, t)*r^2*dV / ∫c(r, t)*dV
These integrals involve integrating over all space, and the exact expressions will depend on the specific boundaries and coordinate system used.
To show that the given distribution satisfies the three-dimensional diffusion equation, we need to substitute the expression for c(r, t) into the diffusion equation and verify that both sides are equal.
The three-dimensional diffusion equation is:
∂c(r, t)/∂t = D (∂^2c(r, t)/∂x^2 + ∂^2c(r, t)/∂y^2 + ∂^2c(r, t)/∂z^2)
Let's evaluate each term separately.
1. ∂c(r, t)/∂t:
Taking the derivative of c(r, t) with respect to t, we have:
∂c(r, t)/∂t = [n/8πD( t)^(3/2)] e^(-r^2 /4Dt) * (-3/2)(1/( t))
2. ∂^2c(r, t)/∂x^2:
Taking the second derivative of c(r, t) with respect to x, we have:
∂^2c(r, t)/∂x^2 = [n/8πD( t)^(3/2)](-1)(1/(4Dt)) * e^(-r^2 /4Dt) = [n/32πD( t)^(5/2)] * (x^2/(4Dt^2)) e^(-r^2 /4Dt)
Similarly, we can find the second derivatives with respect to y and z:
∂^2c(r, t)/∂y^2 = [n/32πD( t)^(5/2)] * (y^2/(4Dt^2)) e^(-r^2 /4Dt)
∂^2c(r, t)/∂z^2 = [n/32πD( t)^(5/2)] * (z^2/(4Dt^2)) e^(-r^2 /4Dt)
Now, let's plug these derivatives back into the diffusion equation and simplify:
∂c(r, t)/∂t = D (∂^2c(r, t)/∂x^2 + ∂^2c(r, t)/∂y^2 + ∂^2c(r, t)/∂z^2)
[n/8πD( t)^(3/2)] * (-3/2)(1/( t)) = D * [n/32πD( t)^(5/2)] * (x^2/(4Dt^2) + y^2/(4Dt^2) + z^2/(4Dt^2)) e^(-r^2 /4Dt)
Simplifying further:
-3/2 = (1/4)(1/t) [x^2 + y^2 + z^2]
-3 = (2/t) [x^2 + y^2 + z^2]
This equation holds true, satisfying the three-dimensional diffusion equation. Therefore, the given distribution c(r, t) = [n/8πD( t)^(3/2)] e^(-r^2 /4Dt) is a solution to the diffusion equation.
To find the expression for the average distance <r> as a function of time, we need to evaluate the integral of r * c(r, t) over all space (x, y, z) and normalize it by dividing by the integral of c(r, t). Mathematically:
<r> = [(∫r * c(r, t) dV) / (∫c(r, t) dV)]
where dV represents an infinitesimal volume element in Cartesian coordinates (dx dy dz).
Similarly, to find the expression for the average mean-square distance <r^2>, we evaluate the integral of r^2 * c(r, t) over all space (x, y, z) and divide it by the integral of c(r, t):
<r^2> = [(∫r^2 * c(r, t) dV) / (∫c(r, t) dV)]
These integrals can be evaluated by transforming to spherical coordinates or by using numerical methods, depending on the specific form of the distribution and the properties of the system.