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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 567
�e process is very slow; even a�er 1 hour the concentration at a radius of 10
cm is zero to within the precision of the calculation.
P16C.6 (a) �e generalised di�usion equation is given by [16C.9–710], where c is
concentration, t is time, D is the di�usion coe�cient, x is displacement
and υ is the velocity of convective �ow.
∂c
∂t
= D ∂
2c
∂x2
− υ
∂c
∂x
An expression for c(x , t) is a solution of the di�usion equation if substi-
tution of the expression for c(x , t) into each side of the di�usion equation
gives the same result.�e proposed solution is
c(x , t) = c0
(4πDt)1/2
e−(x−x0−υt)2/4Dt
LHS
∂c
∂t
= c(x , t) [− 1
2t
+ (x − x0)2
4Dt2
− υ2
4D
]
RHS
D
∂2c
∂x2
− υ
∂c
∂x
= D ∂
∂x
[−(x − x0 − υt)
2Dt
c(x , t)] − υ [−(x − x0 − υt)
2Dt
c(x , t)]
= D [(x − x0 − υt)2 − 2Dt
(2Dt)2
c(x , t)] + υ [(x − x0 − υt)
2Dt
c(x , t)]
= c(x , t) [(x − x0 − υt)2
4Dt2
− 1
2t
+ υ
2Dt
(x − x0 − υt)]
= c(x , t) [− 1
2t
+ (x − x0)2
4Dt2
− υ2
4D
]
�e LHS is equal to the RHS, and so the proposed function is indeed a
solution to the generalised di�usion equation.
As t → 0 the exponential term e−x
2/4Dt falls o� more and more rapidly,
implying that in the limit t = 0 all thematerial is at x = 0.�e exponential
function dominates the term t1/2 in the denominator.
(b) Using the values D = 5×10−10 m2 s−1 and υ = 10−6 ms−1, Fig. 16.3 shows
the concentration pro�le against time at di�erent distances (x − x0) =
0.5mm, 1.0mm, 5.0mm. As expected, the initial concentration at a given
distance is zero. As time passes, the concentration begins to increase to
a maximum value when (x − x0 = υt) and then, as di�usion continues
in the x-direction, the concentration begins to decrease toward zero at
longer times.�e value of c(x , t)/c0 reaches its maximum value at longer
times as the value of (x−x0) is increased since it takes longer for the solute
to reach distance (x − x0).