| Topic: |
Science > Physics |
| User: |
"Donalbane" |
| Date: |
17 Mar 2006 11:48:08 AM |
| Object: |
Diffusion coefficient for water vapor in air |
Does anyone know a formula to approximate the temperature dependence of
the diffusion coefficient for water vapor in air? I have a couple of
values D(@T=22 C) = 2.5x10^-5 m^2/s and D(@T=8 C) = 2.39x10^-5 m^2/s,
but I need D(T) for arbitrary T.
Thanks,
Don
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| User: "tadchem" |
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| Title: Re: Diffusion coefficient for water vapor in air |
17 Mar 2006 12:26:31 PM |
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Donalbane wrote:
Does anyone know a formula to approximate the temperature dependence of
the diffusion coefficient for water vapor in air? I have a couple of
values D(@T=22 C) = 2.5x10^-5 m^2/s and D(@T=8 C) = 2.39x10^-5 m^2/s,
but I need D(T) for arbitrary T.
My old P-Chem text gives the functional form of the diffusion
coefficient D as
D = (k*T) / (6*pi*n*r)
where k is Boltzmann's constant, T is absolute temperature, n is the
viscosity of the medium (air) and r is the 'particle radius' of the
diffusion species.
This will give you a first order approximation, but be warned that the
viscostiry of air is also temperature dependent (among other things).
For a 'pure gas' kinetic molecular theory gives viscosity n as
n = (1/pi*sigma^2)*[sqrt(k*T*m/pi)]
where sigma is a distance parameter corresponding roughly to the radius
of the gas molecule and m is its mass (k and T are as before). Note
that pressure and density are *not* important to viscosity.
Good starting values for air would be sigma = 0.372 nm and m = 29
g/mole.
HTH
Tom Davidson
Richmond, VA
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