a) A pressure vessel contains a large volume of CO2 gas at 10 atm pressure. A membrane composed of a poly(ether ketone) polymer with thickness 100 microns and net effective area of 100 cm2 covers a small perforated port in the container. The solubility of CO2 in the polymer at 10 atm is 6.97*10-4 moles/cm3 at 35⁰ C. The diffusivity of CO2 in the polymer is known to be 2.28*10-8 cm2/s at this temperature. How long will it take for 0.001 moles of CO2 to leak from the container at steady-state? Assume that the amount of carbon dioxide in the surroundings is insignificant.
Express your answer in seconds.
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To solve this problem, we can use Fick's Law of Diffusion, which states that the rate of diffusion is proportional to the concentration gradient and the diffusion coefficient. The equation can be written as:
J = -D * (∆C/∆x)
where J is the diffusion flux (moles per unit area per second), D is the diffusion coefficient (cm^2/s), ∆C is the difference in concentration (moles/cm^3), and ∆x is the thickness of the membrane (cm).
In this case, we have the diffusion coefficient (D) and the thickness of the membrane (∆x). We can rearrange the equation to solve for the concentration difference (∆C), which is the amount of CO2 that leaks through the membrane:
∆C = (-J * ∆x) / D
First, we need to calculate the diffusion flux (J). The question tells us that the amount of CO2 that leaks is 0.001 moles, and we want to find the time it takes for this amount to leak at steady-state. Therefore, the diffusion flux is given by:
J = (0.001 moles) / (time)
Since we want to express the answer in seconds, we'll keep the time in seconds as well.
Now we can substitute the values into the equation to find the concentration difference (∆C):
∆C = (-(0.001 moles / time) * 100 microns) / (2.28 * 10^(-8) cm^2/s)
It's important to convert all the units to the same system. In this case, we can convert microns to centimeters by dividing by 10,000.
∆C = (-(0.001 moles / time) * (100 microns / 10,000)) / (2.28 * 10^(-8) cm^2/s)
Simplifying the equation:
∆C = (-0.01 moles) / (time * 2.28 * 10^(-8) cm^2/s)
Finally, we can solve for time:
time = (-0.01 moles) / (∆C * 2.28 * 10^(-8) cm^2/s)
We know that the solubility of CO2 in the polymer at 10 atm is 6.97 * 10^(-4) moles/cm^3. We can substitute this value into the equation:
time = (-0.01 moles) / ((6.97 * 10^(-4) moles/cm^3) * 2.28 * 10^(-8) cm^2/s)
Calculating this expression will give us the time it takes for 0.001 moles of CO2 to leak from the container at steady-state.