CHEMICAL ENGINEERING

218 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

As the vapour pressure of tolueneD54 kN/m^2 , the partial pressure of toluene from
Raoult’s lawD 1  0. 59
ð 54 D 22 .14 kN/m^2 DPT 1 and:

PT 2 D 1  0. 70

 ð 101. 3 D 30 .39 kN/m^2

For toluene: NTD 0. 051 ð 10 ^4
 30. 39  22. 14
/ 8. 314 ð 365 ð 0. 0002

D 6. 93 ð 10 ^5 kmol/m^2 s

For benzene:PB 1 D 101. 3  22. 14 D 79 .16 kN/m^2

PB 2 D 101. 3  30. 39 D 70 .91 kN/m^2

Hence, for benzene:NBD 0. 051 ð 10 ^4
 70. 91  79. 16
/ 8. 314 ð 365 ð 0. 0002

D 6. 93 ð 10 ^5 kmol/m^2 s

Thus the rate of interchange of benzene and toluene is equal but opposite in direction.

PROBLEM 10.3

By what percentage would the rate of absorption be increased or decreased by increasing
the total pressure from 100 to 200 kN/m^2 in the following cases?

(a) The absorption of ammonia from a mixture of ammonia and air containing 10% of

ammonia by volume, using pure water as solvent. Assume that all the resistance

to mass transfer lies within the gas phase.

(b) The same conditions as (a) but the absorbing solution exerts a partial vapour pres-

sure of ammonia of 5 kN/m^2.

The diffusivity can be assumed to be inversely proportional to the absolute pressure.

Solution

(a) The rates of diffusion for the two pressures are given by:

NADD/RTL

P/PBM

PA 2 PA 1

 (equation 10.34)

where subscripts 1 and 2 refer to water and air side of the layer respectively and subscripts
AandBrefer to ammonia and air.

Thus: PA 2 D 0. 10 ð 100
D10 kN/m^2 andPA 1 D0kN/m^2

PB 2 D 100  10

 D90 kN/m^2 andPB 1 D100 kN/m^2

PBMD 100  90 

/ln 100 / 90

 D 94 .91 kN/m^2

∴ P/PBMD 100 / 94. 91
D 1. 054

Hence: NADD/RTL

1. 054  10  0
   D 10. 54 D/RTL