Practica

DETERMINATION OF DIFFUSION COEFFICIENTS
Academic year 2011-2012

PRACTICE 1: Free and forced heat convection
PRACTICE 2: Determination of diffusion coefficients
PRACTICE 3: The Hagen-Poiseuille law

GROUP 2
Jesús Conde Bande
Antonio Fernández Álvarez

PRACTICE 2.DETERMINATION OF DIFFUSION COEFFICIENTS

1.0BJECTIVE.
In this experiment the objective is to determine thediffusion coefficient of a gas by evaporation of the liquid surface, and to study the effect of temperature on the diffusivity.

2. THEORETICAL BACKGROUND.

The diffusivity of a solute defines the rate of transfer of the solute in a given fluid under the driving force of a concentration gradient. The mass transfer process is called diffusion. Diffusivity is classically defined as the mass ofsolute transferred per unit area per unit time under unit concentration gradient. In general, the diffusivity of a solute decreases with its molecular weight, and the molecular weight of the fluid through which it is diffusing, increases with temperature, but decreases with pressure. Diffusion is the method of by which the solutes are transported through the individual phases inchromatography systems but transport through the chromatography column results from fluid flow.
In this experiment, we use an Arnold diffusion cell to determine the diffusivity of acetone.

The constant molar flow of A can be expressed according to equation:

NAZ= JAZ+XAZNAZ+NBZ
Where:

JAZ: describes the transfer of molecular material as a result of a concentration gradient, Fick equation.JAZ= -cDAB DXAdz=-DABdCAdz
In first equation NBZ = 0, and we have that:
NAZ=-cDABDXAdz+XANAZ
NAZ1-XA=-cDABDXAdz

NAZZ1Z2dz=XA1XA2--cDAB dXA1-XA
NAZZ2-Z1=-cDABln1-XA21-XA1
(Z2-Z1) = Z = effective distance.
NAZZ=-cDABln1-XA21-XA1

We know:
XA+XB=1
P=PA+PB
PB=PXB
C=RTP
PB2=1-XA2
PB1=1-XA1
and considering the mass A that is evaporating with time, NAZcan also be expressed:
NAZ=-ρAMAAdVdt

we had a new equation:
-ρAMAAdVdtZ=-PRTDAB ln1-XA21-XA1

Integrating:
0ZZdZ=PRTMAρADABlnPB2PB10tdt
Z22Z0Z=PRTMAρADABlnPB2PB1t
-------------------------------------------------
z2-z02=2DABPMAtρaRTlnPB2PB1
Where:
slope=2DABPMAρaRTlnPB2PB1
z2-z02=axis Y
t=axis X
We know that:
PB2→P2=PA2+PB2 , but, PA2=0 because in top of capillaryit is not acetone, and for this: P2=PB2
On the other hand,
P1=PA1+PB1, where:
P1=Patm , and, PA1=PV of acetone
log10PV=A-BTºC+C
being: A= 4. 21840
B= 1197. 010
C= 218. 060
Finally, we calculate DAB equaling to the slope because we know the rest of values.

3. MATERIALS AND METHODS.
In this experiment we use:
1. Capillary.
2. Thermometer.3. Water bath.
4. Graph paper and a rule.
5. Chronometer.
6. Acetone.
7. Fan.

4. RESULTS AND DISCUSSION.

The results that we will see in this experiment are going to get as follows:
26/10/2011: Patm. = 999 mbar, 35ºC
26/10/2011: Patm. =999 mbar, 55ºC

Table 1. Shows the difference in height over time to 35ºC.
z2-zₒ2 mm | z (mm) | T(s) |
0 | 1035 | 0 |
0| 1035 | 180 |
0 | 1035 | 360 |
0 | 1035 | 540 |
0 | 1035 | 720 |
-10325 | 1030 | 900 |
-10325 | 1030 | 1080 |
-10325 | 1030 | 1260 |
-10325 | 1030 | 1440 |
-10325 | 1030 | 1620 |
-10325 | 1030 | 1800 |
-20600 | 1025 | 1980 |
-20600 | 1025 | 2160 |
-20600 | 1025 | 2340 |
-20600 | 1025 | 2520 |
-20600 | 1025 | 2700 |
-30825 | 1020 | 2880 |
-30825 | 1020 |3060 |
-30825 | 1020 | 3240 |
-30825 | 1020 | 3420 |
-30825 | 1020 | 3600 |
-41000 | 1015 | 3780 |
-41000 | 1015 | 3960 |
-41000 | 1015 | 4140 |
-41000 | 1015 | 4320 |
-41000 | 1015 | 4500 |
-41000 | 1015 | 4680 |
-51125 | 1010 | 4860 |
-51125 | 1010 | 5040 |
-51125 | 1010 | 5220 |
-51125 | 1010 | 5400 |






Graph 1. Data set the temperature to 35ºC....