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CHE391: Unit Operation Laboratory

Experiment No. 10

VAPOUR IN AIR DIFFUSION

I/C

Prof. Siddharth Panda Prof. R.K. Gupta

Department of Chemical Engineering Indian Institute of Technology, Kanpur

Date of Experiment:

7-03-2013

Date of Report Submission:

14-03-2013

Any revision sought: None

Regular

Regular

Group No. 2

Group Member:--

Nitin Kumar Singh 10460

Palash Agarwal 10473

Pankaj Kumar 10476

Name of T.A.: K.Anitha

Marks: _______________________

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Contents

1. Objective 3

2. Introduction 3

3. Theory 3

4. Apparatus Description 5

5. Procedure 6

6. Observations and Calculation 6

7. Result and discussion 10

8. Conclusion 10

9. Nomenclature 11

10. Precautions 11

11. Sources of error 11

12. References 12

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Objective

1. To determine the diffusion coefficient of organic vapor, carbon tetrachloride (CCl4) in air

2. To study the effect of temperature on the diffusion coefficient.

Introduction

Diffusion is the process by which molecules of a given substance move from an area of relatively high concentration to an area of lower concentration. When the molecules have diffused so that they are in a uniform concentration, this state is called equilibrium. The driving force behind this movement of molecules is their thermal energy. Increase in temperature means an increase in molecules' speed (kinetic energy). So the molecules move faster and there will be more spontaneous spreading of the material which means that diffusion occurs quicker. So diffusivity is a function of temperature too.

Theory

Steady state of diffusion is achieved when two gases are undergoing continuous diffusion with a steady supply of fresh gas. After equilibrium is attained, diffusion occurs at a steady rate and this is known as steady state diffusion. In this experiment, CCl4 vapors are steadily diffusing in an environment of stagnant air at a total pressure of 1 atm.

Let A be CCl4, and B be air. Dalton’s law gives, C = CA + CB dCA/ dx = -dCB/ dx (as total concentration C remains constant) Let 1 and 2 be the boundaries of diffusion with 1 being the boundary near the liquid and 2 being the boundary at the top of the T-tube. The integrated form of Fick’s law gives, NA=DAB (PA1-PA2)/RTx

Also, P = PA1 + PB1 = PA2 + PB2 Therefore, NA=DAB (PB2-PB1)/ RTx

Total flux = (NAPA/P) + (NBPB/P) (If diffusion is within both component)

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We know, NA = NPA/P + JA NA = NAPA/P + JA (Since, air here is stagnant, NB = 0) NA = (NAPA/P)-(DAB/RT) dPA/dx (From Fick’s law) NA=NA [1-(PB/RT)] + (DAB/RT) dPB/dx

NA X

dx0

= (DABP/RT)B

PB

PB

B PdP /

2

1

Integrating, NA = (DABP/RTx) ln(PB2/PB1) The rate of diffusion would correspond to the fall in the liquid level in the tube. NA = (DABP/RTx)ln(PB2/PB1)=(ρl/M)(dx/dt) Integrating, (DABP/RT)ln(PB2/PB1)(Mt/ρl) = (x2

2-x12)/2

In terms of concentration, (x2 - x0

2) = (2tMADABCAC)/ρlCBM Where CBM = (CB1-CB2)/ln (CB1/CB2)

Rewriting,

The above equation implies that the plot between t/(x-x0) against (x-x0) should yield a straight line with slope S given by:

At any temperature T, C is given by:

t

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C = P/RT k mole/m3

CA = [(P.V)A/P]C k mole/m3 CB1=C k mole/m3 CB2= [(P-(V.P)A)/P] k mole/m3 Vapor pressure relation for CCl4: VP = (0.019)T2 – (0.2519)T + 7.5239 (T is temperature in K)

DAB=const. T1.5 / P

Apparatus Description

CCl4 is held in a glass T-tube (inner diameter = 3mm) maintained at a constant temperature water bath. Temperature of the bath is controlled by the DTC. An air pump is used to blow air through the T-tube while the change in level is observed by a sliding microscope. The image of the meniscus appears inverted.

Utilities required: Electric supply: Single phase, 220 V AC, 50 Hz, 5-15 Amp and socket with earth

connection Floor area required: 1m*1m CCl4 and laboratory glassware

Fig#1 Vapour in air diffusion apparatus

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Procedure

1. Clear apparatus & make it free from dust. 2. Fill 3/4th water bath with water. 3. Set the water bath temperature at the desired level upto 700 C wait till the bath attains

the set temperature. Note the steady temperature of the bath. 5

4. Fill the t- tube with CCl4 upto within 2 cm of the top of capillary liquid . note down the initial diffusion height of liq. In the capillary.

5. Make the connection with the air or vaccum pump & allow a gentle current of air to flow over the capillary.

6. Record the height of liquid in the capillary after every 15 min. Repeat step 1 through 5 for different water bath temperature

8. Use different organic fluids like ethanol, toluene , acetone , hexane etc. & tabulate the results & discuss .

Observations and Calculations

P = 101.325 kN/m2 ρ = 1.59 x 103 Kg/m3 R = 8.314 J/mol-K MA = 153.82 gm/mol

Case 1 : For T = 333.15 K po = 0.019 T2 + 0.2519 T + 7.5239

= 2032.393 N/m2 CT = P/RT = 36.582 mol/m3 CA = (po /P)CT = 0.734 mol/m3 CB1 = CT = 36.582 mol/m3 CB2 = (1-po/P)CT = 35.848 mol/m3 CBM = (CB1 – CB2)/ln(CB1/CB2) = 36.214 mol/m3

Initial Height = Xo =7.443 cm

Table#1 For T = 333.15 K

S.No. Time

ϴ(sec)

X0(cm) X(cm) X0-X(cm) ϴ/(X0-X)

(sec/cm)

1 1140 7.443 7.359 0.084 13571.43

2 2100 7.443 7.300 0.143 14685.31

3 3000 7.443 7.260 0.183 16393.44

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From Graph #1, we get;

Slope = S = 27780 sec/cm2

DAB = ρCBM/2MACACTS = 0.251 cm2/sec

Case 2: For T = 323.15 K po = 0.019 T2 + .2519 T + 7.5239

= 1910.215 N/m2 CT = P/RT = 37.714 mol/m3 CA = (po/P)CT = .711 mol/m3 CB1 = CT = 37.714 mol/m3 CB2 = (1-po/P)CT = 37.003 mol/m3 CBM = (CB1 – CB2)/ln(CB1/CB2) = 37.357 mol/m3 Initial Height = Xo = 7.09 cm

Table#2 For T = 323.15 K

S.No. Time

ϴ(sec) X0(cm) X(cm) X0-X(cm)

ϴ/(X0-X)

(sec/cm)

1 900 7.09 7.056 0.034 26470.59

2 1800 7.09 7.03 0.060 30000

3 2700 7.09 7.00 0.090 30000

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From graph #2, we get;

Slope = S = 61420 sec/cm2 DAB = ρCBM/2MACACTS = 0 .117 cm2/sec

Case 3: For T = 313.15 K

po = 0.019 T2 + .2519 T + 7.5239

= 1791.837 N/m2

CT = P/RT = 38.918 mol/m3

CA = (po/P)CT = .688 mol/m3

CB1 = CT = 38.918 mol/m3

CB2 = (1-po/P)CT = 38.230 mol/m3

CBM = (CB1 – CB2)/ln (CB1/CB2) = 38.573 mol/m3 Initial Height = Xo = 6.823 cm

Table#3 For T = 313.15 K

S.No. Time

ϴ(sec)

X0(cm) X(cm) X0-X(cm) ϴ/(X0-X)

(sec/cm)

1 900 6.823 6.811 0.012 75000

2 1800 6.823 6.802 0.021 85714.29

3 2700 6.823 6.795 0.028 96428.57

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From graph #3, we get; Slope = S = 549450 sec/cm2 DAB = ρCBM/2MACACTS = 0.014 cm2/sec To determine the variation of diffusivity coefficient with temperature we plot the following graph

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According to the theory DAB is proportional to T1.5 so we plot a graph between them with linear regression

Results and Discussion

Temperature(K) T1.5(K1.5) DAB(cm2/s)

333.15 6080.78 .251

323.15 5809.07 .117

313.15 5541.51 .014

From the Graph#4, we can note that as the temperature of the bath increases, the diffusion

coefficient also increases.

From Graph #5 we see that the diffusion coefficient varies linearly with T1.5 which

confirms the theory.

Slight error which is observed in graphs may be due to human error and apparatus error.

Conclusion

We can conclude that diffusivity of a vapor in air is dependent on temperature in such a way that a rise in temperature causes a rise in the diffusivity. Diffusivity is approximately proportional to T1.5 (T in Kelvins).

DAB α T1.5

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Nomenclature

CA: Molar concentration of A (CCl4), mol /m3

CB: Molar concentration of B (air), mol /m

3

CT: Total molar concentration, mol/m

3

CBM: Log mean concentration of CCl4, mol/m

3

DAB: Diffusion coefficient, cm

2/s

MA: Molecular weight of CCl4, g/mol P: Total pressure, N/m

2

R: Universal Gas constant, J/mol-K (V.P.)A: Vapor pressure of A,

N/m2 x: Final height, cm

xo: Initial height, cm xo - x = Drop in liquid level in time, cm

ϴ = Time interval for evaporation, min ρ = Density of CCl4, kg/m

3

S = Slope of graph

Precautions

1. The readings of meniscus should not be effected by water surface so water should be

filled accordingly

2. Carbon tetrachloride should be colorless.

3. The heater should not be switched on before filling the water in the bath.

4. Microscope focus should be clear.

5. Readings on vernier should be measured accurately.

Sources of Error

1. Imperfect reading due to water surface which can affect the view while watching from

microscope.

2. Imperfect mixing may lead to uneven temperature inside the water bath.

3. Manual error in noting the reading from the microscope.

4. External air current may affect the diffusion rate.

5. Velocity of external air can change the pressure exerted.

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References

1. Laboratory manual, ChE 391, chemical engineering laboratory

2. Robert E. Treybal, “ Mass-Transfer Operations”, 3rd

edition, McGraw-Hill Book

Company, 1981, Page 22-28

3. McCabe, Smith, “Unit Operations in Chemical Engineering”, 7th

edition, McGraw-Hill,

NY, 2005, Page 528-532.

4. Binay K Datta. Principles of Mass Transfer and Separation Processes. Prentice Hall of

India Pvt Ltd. ND. 2007. pp 14-42.

5. Robert E Treybal. Mass Transfer Operations. McGraw Hill. 3rd

edition. pp 26-35.

6. http://lotus.com.my/images/LS-32000(done)/32136.png

7. http://www.wisegeek.com/what-is-diffusion.htm