Answer:
1.25 cm/day
Explanation:
An air thickness , (l) = 0.15 cm
Air Temperature =
[tex](T_a)=20^0C = (20+273)K\\(T_a)=293K[/tex]
Mass Diffusion coefficient (D) = [tex]0.25cm^2/sec[/tex]
If the air pressure [tex](P_a) = 0.5 P_{sat}[/tex]
We are to determine how fast will the water [tex](H_2O)[/tex] level drop in a day.
From the property of air at T = 20° C
[tex]P_{sat} = 2.34[/tex] from saturated water properties.
The mass flow of [tex](H_2O)[/tex] can be calculated as:
[tex]H_2O = \frac{D}{\phi} \delta C[/tex]
where:
[tex]\delta C = \frac{P_{sat}*P_a}{RT }[/tex]
R(constant) = 8.314 kJ/mol.K
[tex]\delta C = \frac{2.34*0.5}{8.314*293 }[/tex]
[tex]\delta C = 4.803*10^{-4}\\ \delta C =0.48*10^{-3} mol/m^3\\ \delta C = 0.48*10^{-6} mol/cm^3[/tex]
Since 1 mole = 18 cm ³ of water
[tex]0.48*10^{-6}mol/cm^3[/tex] will be: [tex](0.48*10^{-6}mol/cm^3 *18)cm^3/cm^3[/tex]
[tex]\delta C = 8.64 * 10^{-6}[/tex]
Again:
[tex]H_2O = \frac{D}{\phi} \delta C[/tex]
[tex]= \frac{0.25}{0.15}*8.69*10^{-6} \frac{cm^2/sec}{cm}[/tex]
[tex]=1.4481*10^{-5} \frac{cm^2/sec}{cm}[/tex]
Converting the above value to cm/day: we have:
[tex]1.448*10^{-5}*3600*24\frac{cm}{s}*\frac{s}{yr}*\frac{yr}{day}[/tex]
= 1.25 cm/day
∴ the rate at which the water level drop in a day = 1.25 cm/day
