(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 at 10 atm is 6.97 x 10-4 moles/cm3 at 35 C. The diffusivity of CO2 in the polymer is known to be 2.29 x 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 hours.
(b) If the diffusivity is observed to double when the temperature is increased by 10 degrees C, what is the activation energy for diffusion? Express your answer in units of eV
a) 1.74
b=0.551
Wrong
(a) To calculate the time it takes for 0.001 moles of CO2 to leak from the container at steady-state, we can use Fick's law of diffusion. Fick's law states that the rate of diffusion is proportional to the concentration gradient and the diffusion coefficient. In this case, we know the diffusion coefficient (D) of CO2 in the polymer and the concentration gradient can be determined using the solubility of CO2 at 10 atm.
The concentration gradient (ΔC) can be calculated as the difference between the solubility of CO2 at 10 atm in moles/cm^3 and the solubility of CO2 in the surroundings (which we assume to be insignificant). From the given data, the solubility of CO2 at 10 atm is 6.97 x 10^(-4) moles/cm^3.
ΔC = 6.97 x 10^(-4) - 0 = 6.97 x 10^(-4) moles/cm^3
Now, we can use Fick's law to calculate the rate of diffusion (J) in moles/second:
J = -D * ΔC/A,
where D is the diffusion coefficient, ΔC is the concentration gradient, and A is the effective area of the port covered by the membrane.
Plugging in the given values, we have:
J = - (2.29 x 10^(-8) cm^2/s) * (6.97 x 10^(-4) moles/cm^3) / (100 cm^2)
Now, we can calculate the time it takes for 0.001 moles of CO2 to leak (t) using the formula:
t = 0.001 moles / J
Substituting the value of J, we have:
t = 0.001 moles / [(2.29 x 10^(-8) cm^2/s) * (6.97 x 10^(-4) moles/cm^3) / (100 cm^2)]
Simplifying this expression will give us the time in seconds. To convert it to hours, we divide by 3600 (as there are 3600 seconds in an hour).
(b) To calculate the activation energy for diffusion, we use the Arrhenius equation:
D2 / D1 = exp(-Q/RT),
where D1 and D2 are the diffusivities at temperatures T1 and T2 respectively, Q is the activation energy, R is the gas constant (8.314 J/(mol K)), and T is the absolute temperature in Kelvin.
From the given information, we know that doubling the temperature results in doubling the diffusivity (D2 = 2D1). By substituting these values into the equation, we can solve for the activation energy Q.
2D1 / D1 = exp(-Q / (R * (T1 + 10))),
2 = exp(-Q / (8.314 * (T1 + 10))).
By taking the natural logarithm of both sides:
ln(2) = -Q / (8.314 * (T1 + 10)).
Rearranging this equation, we can solve for the activation energy Q:
Q = - ln(2) * 8.314 * (T1 + 10).
Substituting the value of T1, and solving the equation will give us the activation energy in eV.