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A latex balloon, wall thickness 3.091 x 10-4 m, contains helium at a concentration of 0.7 kg m-3. Under these conditions the to...Asked by kkr
A latex balloon, wall thickness 3.091 x 10-4 m, contains helium at a concentration of 0.86 kg m-3. Under these conditions the total surface area of the balloon is 0.56 m2. The diffusion coefficient of He in latex at room temperature is 4.9 x 10-9 m2s-1. Calculate the rate of helium effusion (in g/hr) from the balloon.
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Answered by
Antonio
This is a question from the second midterm of 3.091x by edx. Stop cheating, it's silly and pointless.
Answered by
Stud
Fick’s Law
M= -D•(Δρ/Δx)•A,
where
M = m/t = mass flux [kg/s],
D= 4.9•10 ⁻⁹ m ²/s is the diffusion coefficient (diffusivity) [m²/s],
Δρ/Δx is the gradient of density [kg/m⁴],
A is the area [m²].
M= - 4.9•10⁻⁹•(0 - 0.35)/3.091•10⁻⁴)•0.88 =
4.9•10⁻⁹•0.35•0.88/3.091•10⁻⁴=
=4.88•10⁻⁶ kg/s= 17.58 g/hr
M= -D•(Δρ/Δx)•A,
where
M = m/t = mass flux [kg/s],
D= 4.9•10 ⁻⁹ m ²/s is the diffusion coefficient (diffusivity) [m²/s],
Δρ/Δx is the gradient of density [kg/m⁴],
A is the area [m²].
M= - 4.9•10⁻⁹•(0 - 0.35)/3.091•10⁻⁴)•0.88 =
4.9•10⁻⁹•0.35•0.88/3.091•10⁻⁴=
=4.88•10⁻⁶ kg/s= 17.58 g/hr
Answered by
Ok
Stud your answer is incorrect, and you did not even use the surface area in the equation....ugh.
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