Assessment 11: Fick’s laws
1. Which statement is not true about the concept of steady state diffusion as it applies to
diffusion. (5 pts)
A. Steady-state diffusion is the situation wherein the rate of diffusion into a given system
is just equal to the rate of diffusion out
B. There is no net accumulation or depletion of diffusing species
C. the diffusion flux is independent of time
D. None of the above
E. All of the above
2. What is the driving force for steady-state diffusion? (5 pts)
A. Temperature
B. Concentration gradient
C. Atomic size differences
3. The purification of hydrogen gas by diffusion through a palladium sheet is discussed
in Section 6.3. Compute the number of kilograms of hydrogen that pass per hour
through a 5-mm-thick sheet of palladium having an area of 0.20 m2 at 500C.
Assume a diffusion coefficient of 1.0  10-8 m2/s, that the respective concentrations at
the high- and low-pressure sides of the plate are 2.4 and 0.6 kg of hydrogen per cubic
meter of palladium, and that steady-state conditions have been attained. (10 pts)
A. 2.6 x 10^-3 kg/h
B. 5.9 x 10^-3 kg/h
C. 6.7 x 10^-2 kg/h
D. 4.9 x 10^-4 kg/h
Solution
This problem calls for the mass of hydrogen, per hour, that diffuses through a Pd sheet. It first
becomes necessary to employ both Equations 6.1a and 6.3. Combining these expressions and
solving for the mass yields
M = JAt =  DAt

C
x
0.6  2.4 kg / m3 
=  (1.0  10-8 m2 /s)(0.20 m2 ) (3600 s/h)

 5  103 m 
= 2.6  10-3 kg/h

4. A sheet of steel 1.5 mm thick has nitrogen atmospheres on both sides at 1200C and
is permitted to achieve a steady-state diffusion condition. The diffusion coefficient
for nitrogen in steel at this temperature is 6  10-11 m2/s, and the diffusion flux is
found to be 1.2  10-7 kg/m2-s. Also, it is known that the concentration of nitrogen in
the steel at the high-pressure surface is 4 kg/m3. How far into the sheet from this
high-pressure side will the concentration be 2.0 kg/m3?
concentration profile. (10 pts)
Assume a linear
A. 3.5 mm
B. 1 mm
C. 0.12 mm
Solution
This problem is solved by using Equation 6.3 in the form
C  CB
J =  D A
xA  xB
If we take CA to be the point at which the concentration of nitrogen is 4 kg/m3, then it becomes

necessary to solve for xB, as
C  C 
B
xB = xA + D  A

J


Assume xA is zero at the surface, in which case

4 kg / m3  2 kg / m3 
xB = 0 + (6  10-11 m2 /s) 

1.2  107 kg / m2 - s 

= 1  10-3 m = 1 mm
5. Determine the carburizing time necessary to achieve a carbon concentration of 0.45
wt% at a position 2 mm into an iron–carbon alloy that initially contains 0.20 wt% C.
The surface concentration is to be maintained at 1.30 wt% C, and the treatment is to
be conducted at 1000C. Use the diffusion data for -Fe in Table 6.2. (15 pts)
A. 6.7 hours
B. 98 hours
C. 37 hours
D. 19.7 hrs
Solution
In order to solve this problem it is first necessary to use Equation 6.5:
C x  C0
 x 
= 1  erf 

2 Dt 
Cs  C0
wherein, Cx = 0.45, C0 = 0.20, Cs = 1.30, and x = 2 mm = 2  10-3 m. Thus,

 x 
Cx  C0
0.45  0.20
=
= 0.2273 = 1  erf 

2 Dt 
Cs  C0
1.30  0.20
or

 x 
erf 
= 1  0.2273 = 0.7727
2 Dt 
By linear interpolation using data from Table 6.1

z
0.85
z
erf(z)
0.7707
0.7727
0.90
0.7970
z  0.850
0.7727  0.7707
=
0.900  0.850 0.7970  0.7707

From which
z = 0.854 =
x
2 Dt
Now, from Table 6.2, at 1000C (1273 K)



148,000 J/mol
D = (2.3  10-5 m2 /s) exp 

 (8.31 J/mol- K)(1273 K) 
= 1.93  10-11 m2/s

Thus,
0.854 =
Solving for t yields
2  103 m
(2) (1.93  1011 m2 /s) (t)

t = 7.1  104 s = 19.7 h
6.
The outer surface of a steel gear is to be hardened by increasing its carbon content.
The carbon is to be supplied from an external carbon-rich atmosphere that is
maintained at an elevated temperature. A diffusion heat treatment at 850C (1123 K)
for 10 min increases the carbon concentration to 0.90 wt% at a position 1.0 mm
below the surface. Estimate the diffusion time required at 650C (923 K) to achieve
this same concentration also at a 1.0-mm position. Assume that the surface carbon
content is the same for both heat treatments, which is maintained constant. Use the
diffusion data in Table 6.2 for C diffusion in -Fe. (10 pts)
A. 63.9 min
B. 104 min
C. 45 min
D. 201 min
Solution
0.90 wt%
at a position 1.0 mm below the surface we must employ Equation 6.6b with position (x) constant;
that is
Dt = constant
Or
D850t850 = D650t650
In addition, it is necessary to compute values for both D850 and D650 using Equation 6.8. From

Table 6.2, for the diffusion of C in -Fe, Qd = 80,000 J/mol and D0 = 6.2  10-7 m2/s.
Therefore,


80,000 J/mol
D850 = (6.2  10-7 m2 /s) exp 

 (8.31 J/mol- K)(850  273 K) 
= 1.17  10-10 m2/s



80,000 J/mol
D650 = (6.2  10-7 m2 /s) exp 

 (8.31 J/mol- K)(650  273 K) 
= 1.83  10-11 m2/s

Now, solving the original equation for t650 gives
t650 =
 =

(1.17
D850 t850
D650
 1010 m2 /s) (10 min)
1.83  1011 m2 /s
= 63.9 min