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 total surface area of the balloon is 0.11 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.
21 answers
9.31
can someone give me a formula?
formula, please?
The answer 9.31 is not right can any give a formula.
Same here, can anyone show how to insert the data into the formulas. Thank you
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 total surface area of the balloon is 0.52 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.
Anyone with the formula and how to insert the information from the question above would be great. Thank you
This is a question from the second midterm of 3.091x by edx. Stop cheating, it's silly and pointless.
can someone give me a formula? and what data to use in the formula? thank you
Rate of effusion of gases are inversely propotional to the square root of their molar masses.
Rate1 / Rate2 = square root of (Mass2 / Mass 1)
Rate1 / Rate2 = square root of (Mass2 / Mass 1)
How to input the data into the formula?
Could any 1 help us plz .. i just need to know the formula
42.8 is wrong! could you input the data above into the formula? Thank you
maybe by Shakti:
0.52*(4.9*10^-9)*(concentration in mole/0.52)
then multiply solution (molar flow rate) with molar mass of helium.
solution should be effusion rate.
its just a suggestion, all other interested are invited to post their opinion/formula
0.52*(4.9*10^-9)*(concentration in mole/0.52)
then multiply solution (molar flow rate) with molar mass of helium.
solution should be effusion rate.
its just a suggestion, all other interested are invited to post their opinion/formula
Where does the data goes in formula?
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
How did you convert kg/s to g/hr
kg/s -> g/hr
M(kg/s)*1000*60*60= M(g/hr)
M(kg/s)*1000*60*60= M(g/hr)
wat is the value of change in density if area ia 0.55 m^2
formula - it
D C/(1000 x) S 3600
D=4.9 x 10-9
C= 0.31
x=3.091 x 10-4
S=0.11
3600 is time
D C/(1000 x) S 3600
D=4.9 x 10-9
C= 0.31
x=3.091 x 10-4
S=0.11
3600 is time
Thank you all.