A pressure vessel is lined with a membrane to reduce the loss of helium. The membrane is made of a material through which the diffusivity of helium at 25 C is 3.091 x 10-10 m2/s. The concentration at the inner surface is maintained constant at 4.00 x 10-6 kg/m3. The concentration at the outer surface of the membrane is maintained constant at 2.00 x 10-6 kg/m3. The membrane thickness is 3.091 x 10-6 m and its area is 2.50 m2. How much helium escapes per hour through the membrane? Express your answer in kg.
1.8e-6
To calculate the amount of helium that escapes through the membrane per hour, we need to use Fick's Law of Diffusion. Fick's Law states that the rate of diffusion is proportional to the surface area and the concentration difference of the substance being diffused.
The formula for diffusion, according to Fick's Law, is:
J = -D * A * dC/dx
Where:
J = Diffusion rate (kg/s)
D = Diffusivity of the gas (m2/s)
A = Surface area of the membrane (m2)
dC/dx = Concentration gradient (kg/m4)
We are given:
D = 3.091 x 10-10 m2/s
A = 2.50 m2
dC/dx = (4.00 x 10-6 kg/m3 - 2.00 x 10-6 kg/m3) / 3.091 x 10-6 m
Substituting the values into the formula, we get:
J = - (3.091 x 10-10 m2/s) * (2.50 m2) * [(4.00 x 10-6 kg/m3 - 2.00 x 10-6 kg/m3) / 3.091 x 10-6 m]
Simplifying the expression, we find:
J = - (3.091 x 10-10 m2/s) * (2.50 m2) * (1.00 x 10-6 kg/m3 / 3.091 x 10-6 m)
J = - 2.50 x 1.00 x 10-6 kg/s
The negative sign indicates that the helium is escaping, so we'll take the absolute value:
J = 2.50 x 1.00 x 10-6 kg/s
To convert the rate per second to per hour, we need to multiply by the number of seconds in an hour (3600 seconds):
J = 2.50 x 1.00 x 10-6 kg/s * 3600 s/h = 0.009 kg/h
Therefore, the rate at which helium escapes through the membrane is 0.009 kg/h.