COURSE NUMBER ChE 302
TITLE STUDY OF DIFFUSION
EXPERIMENT NO 02
DATE OF PERFORMANCE 15/01/2014
DATE OF SUBMISSION 22/01/2014
SUBMITTED TO
DR.MD.EASIR ARAFAT KHAN
ASSISTANT PROFESSOR
DEPARTMENT OF CHEMICAL ENGINEERING
BUET,DHAKA-1000,BANGLADESH
SUBMITTED BY
KAMONASHIS SARKAR (TENDUL)
STUDENT ID 1002008(L-3/T-1)
GROUP NUMBER 02 (A1)
PARTNERS STUDENT ID 1002006,1002007,1002009,1002010
CHEMICAL ENGINEERING DEPARTMENT
BANGLADESH UNIVERSITY OF ENGINEERING AND TECHNOLOGY
Summary
The main objective of this experiment is to determine the liquid diffusion co-efficient of NaCl solution in
distilled water by using the concept of mass transfer.Mass transfer is the net movement of mass from one
location, usually meaning a stream, phase, fraction or component, to another. Mass transfer occurs in
many processes, such as absorption, evaporation, adsorption, drying, precipitation, membrane filtration,
and distillation.The phrase is commonly used in engineering for physical processes that involve diffusive
and convective transport of chemical species within physical systems. Mass transfer takes place in either a
gas phase or a liquid phase or in both simultaneously. Mass transfer is a result of diffusion. Diffusion is
the phenomenon of material transport by atomic motion. The driving force of diffusion process is the
concentration gradient of the component. Diffusivity is one of the most important fluid properties.
Diffusion is of great importance in Fluid flow and mass transfer operations such as chemical process plant
design, catalyst design in chemical industry, formation of alloys etc.
Experimental setup
Figure1 Schematic diagram for liquid diffusivity measurement
Diffuser
Conductivity
meter
Capillaries
Conductivity
Probe
Magnetic
stirrer
Stirrer bar
Observed data
Volume of water, V = 1.1 Liter
Length of capillaries, x = 0.5 cm
Diameter of capillaries, d = 0.1 cm
Numbers of capillaries, N = 97
Table1 Observed data for determining diffusivity of NaCl solution.
Observation
number
Time (seconds)
1M NaCl solution 4M NaCl solution
Conductivity, k (µS) Conductivity, k (µS)
01 0 0.70 11.29
02 300 83.42 641.60
03 600 100.60 758.90
04 900 114.20 845.60
05 1200 124.20 892.50
06 1500 133.20 940.50
07 1800 141.20 980.30
08 2100 148.60 1013.00
09 2400 155.60 1045.00
Calculated data
Observation
number
Concentration
of NaCl
solution
,M(mol/L)
Rate of conductivity change
over
time(slope),dk/dt(µs/sec)
Electrical conductivity
change per unit molar
concentration
change,Cm(µs/mol/L)
Diffusivity
,D(cm2/sec)
01 1 0.0331 85543.96 2.79×10-4
02 4 0.1804 85543.96 3.81×10-4
Graphs
y = 0.0331x + 80.381 for 1M NaCl solution
0
20
40
60
80
100
120
140
160
180
0 500 1000 1500 2000 2500 3000
Con
du
ctiv
ity,µ
s
Time,sec
Graph-1(Conductivity vs time)
y = 0.1804x + 646.12 for 4M NaCl solution
0
200
400
600
800
1000
1200
0 500 1000 1500 2000 2500 3000
Con
du
ctiv
ity,µ
s
Time,sec
Graph2(Conductivity vs time)
Graph3 Determination of conductivity change over time
y = 0.0331x + 80.381
for 1M solution
y = 0.1804x + 646.12 for 4M solution
0
200
400
600
800
1000
1200
0 500 1000 1500 2000 2500 3000
Co
nd
uct
ivit
y,µ
s
Time,sec
Graph3(Conductivity vs time)
Sample calculation
Determination of electrical conductivity change per unit molar concntration change
Water was 500ml and 2ml 1M NaCl solution was added,then we can write
V1S1=V2S2 V1=Volume of NaCl solution=2ml
S2=(V1S1/V2) S1=Concentration of NaCl solution
=1M
S2=(2×1)/502 V2=Volume of solution=502ml
S2=3.98×10-3M S2=Concentration of solution=?
Water was 500ml and 4ml 1M NaCl solution was added,then we can write
V1S1=V3S3 V1=Volume of NaCl solution=2ml
S3=(V1S1/V3) S1=Concentration of NaCl solution
=4M
S3=(4×1)/504 V3=Volume of solution=502ml
S3=7.94×10-3M S3=Concentration of solution=?
Water was 500ml and 6ml 1M NaCl solution was added,then we can write
V1S1=V4S4 V1=Volume of NaCl solution=2ml
S4=(V1S1/V4) S1=Concentration of NaCl solution
=1M
S4=(6×1)/506 V4=Volume of solution=502ml
S4=11.86×10-3M S4=Concentration of solution=?
From conductivity meter we have measured conductivity(K).
when S2=3.98×10-3
M then K2=400.9 µs
when S3=7.94×10-3
M then K3=741 µs
when S4=11.86×10-3
M then K4=1075 µs
Now electrical conductivity change per unit molar conductivity change(C)
C1=(K3-K2)/(S3-S2) =(741-400.9)/((7.94-3.98)×10-3) =85883.84 (µs/mol/L)
C2=(K4-K3)/(S4-S3) =(1075-741)/((11.86-7.94)×10-3) =85204.08 (µs/mol/L)
Cm=(C1+C2)/2 =(85883.84+85204.08)/2 =85543.96 (µs/mol/L)
Here
Diffusivity, D=
Wher V=Volume of water in diffusion vessel,(mol/L).
x=Length of capillaries,(cm).
d=Capillaries diameter,(cm).
N=Number of capillaries.
M=Molar concentration of NaCl solution,(mol/L).
CM=The electrical conductivity change per unit molar concentration
change,(µs/mol/L).
=Rate of conductivity change over time,(µs/sec).
For 1M NaCl solution
D=
D=2.79×10-4
cm2/s
For 4M NaCl solution
D=
D=3.81×10-4
cm2/s
Results and Discussions
In this experiment for different values of concentration we have obtained the following results:
For 1M NaCl solution the value of diffusivity is 2.79×10-4
cm2/s
For 4M NaCl solution the value of diffusivity is 3.81×10-4
cm2/s
Diffusion coefficient is inversely proportional to concentration and directly proportional to the
slope of conductivity vs. time graph. As we use the more concentrated solution, the slope of the
graph also increases Now here we have two plots (concentration vs time).Both the plots are
y=mx+c types state line.Their slopes are
For 1M NaCl solution the value of slope is 0.0331 (µs/sec)
For 4M NaCl solution the value of slope is 0.1804 (µs/sec)
Actually diffusivity needs to be constant for both case and the statelines needs to be parallel for
satisfying this condition.But in reality are not parallel and even they intersect each other at a
certain point on the graph paper.Thats why there are some ups and downs in the results we
have.Behind this result there are some following reasons which are listed below:
a) Trapped air bubbles in the capillaries which can hinder mass transfer.
b) Incomplete filling of the cell could create a trapped air pocket.
c) Overfilling of the cell, resulting in NaCl spilling out into the water when the cell is placed in
the diffusion vessel.
d) As the 4M NaCl solution was not freshly prepared, its concentration may not be precisely
equal to 4M.
e) In order to keep the external concentration of salt as low as possible, and in order to get
conductivity readings that reflect the total amount of salt in the tank, we have to keep fluids
in the tank well-mixed. To keep the fluids well-mixed a magnetic stirrer were used.The rate
of stirring should be kept low enough to prevent the fluid flow out of the capillaries.
f) Conductivity vs. time graph for 1M and 4M solutions should be parallel to each other. But
from our experiment it is seen that the lines for 1M and 4M solutions intersect each other. We
were supposed to use dematerialized water. Initial conductivity reading for 1M solution were
found higher than it should be for completely dematerialized water which means the 1M
solution we have used were contaminated. May be this contamination were responsible for
the change in the relationship graph between conductivity and time.
g) Fick’s first law holds good when concentration is constant(steady state) and when it varies
with time then fick’s second law appears to satisfy the condition.The second law is
But when steady state condition applies to fick’s second law it becomes
fick’s first law that is
where c=concentration, D=diffusivity , x=X-direction, j=diffusion flux.
These two law can describe the reason of deviation from ideal condition.
List of References
Mc Cab, Smith , Harriott Unit Operation of Chemical Engineering, Sixth edition
Cussler E.L., Diffusion-Mass transfer in fluid systems, third edition
Holman J.P., Heat Transfer, Ninth edition, McGraw-Hill
http://en.wikipedia.org/wiki/Mass_transfer
http://wiki.answers.com/Q/What_are_the_factors_affecting_the_rate_of_diffusion
http://classof1.com/homework_answers/chemical_engineering/ficks_law/