8/18/2019 Diffusion problem

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1. Calculate the rate of diusion of sugar through a stagnant lm of coee

0.10 cm thick when the concentrations at 15% and 5%, respectively, on

either side of the lm. ssume the diusivity of sugar through coee

under the given conditions to !e 0."0 # 10$5 cm&s and the density of a

10% solution is 1.01'( g&cc.

)iven*

+v 0."0 # 10$5 cm&s

-10% wt solution 1.'( g&cc

/euired*

2olution* 3nicomponent

+iusion

  xA 1=

15

342

15

342  +

85

18

xA 1=0.00920

  xA 2=

15

342

15

342  +

95

18

xA 2=0.00276

8/18/2019 Diffusion problem

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  x10%=

10

342

10

342  +

90

18

x10%

=0.00581

 M 10=0.00581 (342g /mol)+(1−0.00581)(18 g /mol )

 M 10=¿ 1(.44 g&mol

C  Ave=  ρ

10

 M 10=

  1.0139g /cc19.8824 g/mol  

C  Ave=0.0510mol /cc

 N  A=0.70 x 10

−5 cm2

s (0.0510molcc )0.10cm

  ln

(1−0.00276

1−0.00920 ) 

 N  A=2.3129 x10−8  mol

cm2s≈2.3129 x10

−7 kmol

m2s

. n rnold cell is used to measure the diusivity of acetone in air at 0C

and 100 k6a pressure. t time 0, the liuid acetone surface is 1.10 cm

from the top of the tu!e and after 4 hours of operation, the liuid surface

drops to .05 cm. if the concentration of acetone in air that 7ows over the

top of the tu!e is 8ero, what is the diusivity of acetone in air9 t 0C,

the vapor pressure of acetone is k6a and density is "(0 kg&m'.

)iven*

8/18/2019 Diffusion problem

3/4

6acetone k6a

- acetone "(0 kg&m'

/euired* + acetone$air : 0C,

100 k6a

2olution*

 P(¿¿T − P A) L P A1− P A2

¿¿

2 t f  ρm MW  ¿

 Dv=( z

2

2− z1

2)( ρacetone)¿

 ρm=

 PT  RT  =

100 kPa(   1atm101.325kPa )0.08205

 m3atm

kmolK (20+273 ) K 

 ρm=0.0411kmol /m3

 P A1

= P ° A x A=24kPa (1 )=24 kPa

 P A2

=0kPa

8/18/2019 Diffusion problem

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 P

 P P

(¿¿T − P A2)−

 (¿ ¿T − P A1)

ln

( PT − P A

1

 PT − P A2 )

=(100−0)−(100−24)

ln

(  100−0

100−24 )

  =87.4518 kPa

¿(¿¿T − P A) L=¿

¿

 Dv=

(0.02052−0.01102 )m2(790 kgm3 )2 (8h )(0.0411 kmolm3 )(58

  kg

mol )(   24−0

87.4518 )( 3600 s1h   )

 Dv=6.2737 x 10−6m

2

s