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.
This is a question from the second midterm of 3.091x by edx. Stop cheating, it's silly and pointless.
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
Stud your answer is incorrect, and you did not even use the surface area in the equation....ugh.
To calculate the rate of helium effusion from the balloon, we need to use Fick's Law of diffusion. The formula is as follows:
Rate of Effusion = (D * A * ΔC) / Δx
Where:
D is the diffusion coefficient of helium in latex (given as 4.9 x 10^-9 m^2/s),
A is the total surface area of the balloon (given as 0.56 m^2),
ΔC is the change in concentration of helium inside and outside the balloon,
and Δx is the thickness of the balloon wall (given as 3.091 x 10^-4 m).
In this case, the concentration difference ΔC is the difference between the concentration of helium inside the balloon (0.86 kg/m^3) and the concentration of helium in the surrounding air (which we assume to be negligible).
Let's plug in the values and calculate the rate of helium effusion:
Rate of Effusion = (4.9 x 10^-9 m^2/s * 0.56 m^2 * (0.86 kg/m^3 - 0 kg/m^3)) / 3.091 x 10^-4 m
Now, let's simplify the equation:
Rate of Effusion = (4.9 x 10^-9 m^2/s * 0.56 m^2 * 0.86 kg/m^3) / 3.091 x 10^-4 m
Rate of Effusion = 0.024096 kg/s
Finally, we need to convert the rate of effusion from kg/s to g/hr:
Rate of Effusion = 0.024096 kg/s * 1000 g/kg * 3600 s/hr
Rate of Effusion ≈ 87.14 g/hr
Therefore, the rate of helium effusion from the balloon is approximately 87.14 g/hr.