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A latex balloon, wall thickness 3.091 x 10-4 m, contains helium at a concentration of 0.31 kg m-3. Under these conditions the t...Asked by troubled student
A latex balloon, wall thickness 3.091 x 10-4 m, contains helium at a concentration of 0.87 kg m-3. Under these conditions the total surface area of the balloon is 0.87 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.
Answers
Answered by
mango
anyone got it right?
Answered by
troubled student
Im struggling with it, at least any hints?
Answered by
chelo
What is the formula to use
Answered by
C
pls. tell
Answered by
MIT2
CHEAT!
Do NOT help this person. They are trying to cheat in a midterm exam.
Do NOT help this person. They are trying to cheat in a midterm exam.
Answered by
Ling
same here, how to insert the data into the formula?
Answered by
chelo
the formula?
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
mangozier
doesnt work for me. the answer is still wrong
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