Water evaporating from a pond does so as if it were diffusing across an air film 0.15 cm thick. The diffusion coefficient of water in 20 !C air is about 0.25 cm2/sec. If the air out of the film is fifty percent saturated, how fast will the water level drop in a day?

[tex]P_{sat}[/tex] is the saturated water pressure, which is look up from the saturated water property at 20°C and 0.5 saturation given as 2.34 Pa

[tex]P_a[/tex] is the air pressure which is given as 0.5 times of [tex]P_{sat}[/tex]

R is the universal gas constant as [tex]8.314 kJ/kmol-K[/tex]

T is the temperature in Kelvin scale which is [tex]20+273= 293K[/tex]

By substituting values in the equation

[tex]\bigtriangleup c= \frac{P_{sat}-P_a}{RT} \\ \bigtriangleup c= \frac{P_{sat}-0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5 \times 2.34}{8.314 \times 293} \\\bigtriangleup c= 0.48 mol/m^3[/tex]

Converting [tex]\bigtriangleup c[/tex] into [tex]cm^3/cm^3[/tex]

As 1 mole of water 18 [tex]cm^3[/tex] so

[tex]\bigtriangleup c= 0.48 mol/m^3 \\ \bigtriangleup c= 0.48 \times 18 \times 10^{-6} cm^3/cm^3 \\ \bigtriangleup c= 8.64 \times 10^{-6} cm^3/cm^3[/tex]

Putting this in the equation of mass flux equation gives

[tex]j_{H_2O}=\frac{D}{l} \bigtriangleup c \\ j_{H_2O}=\frac{0.25}{0.15} \times 8.64 \times 10^{-6} \\ j_{H_2O}=14.4 \times 10^{-6} cm/s[/tex]

For calculation of water level drop in a day, converting mass flux as

[tex]j_{H_2O}=14.4 \times 10^{-6} \times 24 \times 3600 cm/day\\ j_{H_2O}=1.24 cm/day[/tex]

So the water level will drop by about 1.24 cm in 1 day.