1. Water evaporation
To find the rate of evaporation, we need to find the flux of water across the air film:
DH
j1 = Δc1
l
Since the film is made up of air, H = 1. We are given D and ℓ, so we need to find Δc1. We first
calculate the difference in the partial pressure of water across the film. From a steam table we
find that at 20 °C, psat = 2.3388 kPa. Since the air immediately above the water is saturated we
have:
Δp1 = psat − 0.5 psat = 0.5 psat = 0.5(2.3388 kPa ) = 1.1694 kPa = 1169.4 Pa
Assuming an ideal gas, we can find the concentration difference:
Δp 1169.4 Pa mol mol
Δc1 = 1 = = 0.48 3 = 4.8 ⋅10 −7
RT ⎛ J ⎞ m cm3
⎜ 8.3145 ⎟(293 K )
⎝ K ⋅ mol ⎠
We can now calculate the flux:
cm 2
0.25
j1 = s ⎛⎜ 4.8 ⋅10 −7 mol ⎞⎟ = 8.0 ⋅10 −7 mol
0.15 cm ⎝ cm3 ⎠ cm 2 s
To find the height change, we divide by the molar density of water:
mol
8.0 ⋅10 −7
cm 2 s = 1.44 ⋅10 −5 cm = 1.25 cm
g 1 mol s day
1 3⋅
cm 18.015 g
2. Diffusion across a monolayer
Recalling that the resistance is the inverse of the permeance, we have:
1 s cm
=2 ⇒ P = 0.5
P cm s
We know that the permeance is given by:
DH Pl ⎝
⎛
⎜ 0.5
cm ⎞
( −7
⎟ 2.5 ⋅10 cm )
− 6 cm
2
s ⎠
P= ⇒D= = = 7 ⋅10
l H 0.018 s
3. Diffusion coefficient of NO2 in water
We treat the water as a semiinfinite slab. From Eq. (2.3-18) we know that the flux at the surface
is given by:
, Chapter 2 Diffusion in dilute solutions page 2-2
D
j1 z =0 = (c10 − c1∞ )
πt
The concentration at the surface is given by Henry’s Law:
p 0.93 atm mol
c10 = = = 2.5 ⋅10 −5 3
H cm atm
3
cm
37000
mol
Since we have a semiinfinite slab we assume c1∞ = 0. To find the flux over the total time t, we
integrate:
[ t]
t t t
D
(c10 − 0)dt = Ac10 D ∫ dt = 2 Ac10 D Dt
t
N1 = ∫ A j1 z =0 dt = A∫ = 2 Ac10
πt π 0 t π π
0
0 0
Assuming an ideal gas, the flux N1 is given by:
N1 =
PV
=
( )
(0.93 atm ) 0.82 cm3 = 3.2 ⋅10 −5 mol
RT ⎛ cm3atm ⎞
⎜⎜ 82.06 ⎟(289 K )
⎝ K ⋅ mol ⎟⎠
Solving for D, we have:
2
⎡ ⎤
⎛ N1 ⎞ π
2
⎢ −5
3.2 ⋅10 mol ⎥ π − 6 cm
2
⎜
D=⎜ ⎟
⎟ ⋅ = ⎢ ⎥ ⋅ = 5.4 ⋅10
⎢⎣
( )
⎝ 2 Ac10 ⎠ t ⎢ 36.3 cm 2 ⎛⎜ 2.5 ⋅10 −5 mol ⎞⎟ ⎥ 180 s
3 ⎥
cm ⎠ ⎦
s
⎝
4. Permeability of water across a polymer film
The rate of water loss can be found by linear regression or by simply using the first and last data
points:
14.0153 g − 13.5256 g g
N1 = = 0.031
16 days day
The molar flux is given by:
g 1 mol 1 day 1 hr
0.031 ⋅ ⋅ ⋅
N1 day 18.015 g 24 hr 3600 s mol
j1 = = = 2.3 ⋅10 −10
A 85.6 cm 2
cm 2 s
From Eq. (2-2.10) we have:
DH
j1 = Δc1
l
We need to find Δc1. We first calculate the difference in the partial pressure of water across the
film. From a steam table we find that at 35 °C, psat = 0.0555 atm. Since the air inside the bag is
saturated, we have:
Δp1 = psat − 0.75 psat = 0.25 psat = 0.25(0.0555 atm ) = 0.0139 atm
Assuming an ideal gas, the concentration difference is:
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