(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.

1.74

a) 1.74

thanks but b>

yes i fond it ,, 0.551

(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. According to Fick's Law, the rate of diffusion (J) through a membrane is given by:

J = (D * A * ΔC) / Δx

Where:
- J is the diffusion rate (moles/s)
- D is the diffusivity of CO2 in the polymer (cm2/s)
- A is the net effective area of the membrane (cm2)
- ΔC is the concentration difference across the membrane (moles/cm3)
- Δx is the thickness of the membrane (cm)

First, let's calculate the concentration difference across the membrane (ΔC). We are given the solubility of CO2 at 10 atm, which is 6.97 x 10-4 moles/cm3. Since the pressure inside the container is also 10 atm, the concentration difference ΔC would be 10 atm * 6.97 x 10-4 moles/cm3 = 6.97 x 10-3 moles/cm3.

Now, we can substitute the known values into Fick's Law and solve for the diffusion rate (J):

J = (2.29 x 10-8 cm2/s * 100 cm2 * 6.97 x 10-3 moles/cm3) / 100 microns

Convert the thickness of the membrane to cm:
Δx = 100 microns * 1 cm/10,000 microns = 0.01 cm

J = (2.29 x 10-8 cm2/s * 100 cm2 * 6.97 x 10-3 moles/cm3) / 0.01 cm

Simplifying the equation:
J = 1.6 x 10-12 moles/s

Now, we can calculate the time it takes for 0.001 moles to leak. We can use the formula:

time = (moles to be leaked) / (diffusion rate)

time = 0.001 moles / (1.6 x 10-12 moles/s)

Let's calculate the time:
time = 625,000 s

Finally, we need to express the time in hours:
time = 625,000 s * 1 min/60 s * 1 hour/60 min = 347.22 hours

So, it will take approximately 347.22 hours for 0.001 moles of CO2 to leak from the container at steady-state.

(b) To determine the activation energy for diffusion, we can use the Arrhenius equation:

D = D0 * exp(-Q/RT)

Where:
- D is the diffusivity
- D0 is a pre-exponential factor
- Q is the activation energy
- R is the gas constant (8.314 J/(mol*K))
- T is the absolute temperature (in Kelvin)

We are given that the diffusivity (D) is observed to double when the temperature (T) is increased by 10 degrees Celsius. Let's assume the original temperature is T1 and the increased temperature is T2.

Then, we have:
D2 = 2 * D1

Substituting the expressions for D:
D0 * exp(-Q/T2) = 2 * D0 * exp(-Q/T1)

Dividing both sides by D0 and simplifying:
exp(-Q/T2) = 2 * exp(-Q/T1)

Taking the natural logarithm of both sides:
-Q/T2 = ln(2) - Q/T1

Rearranging the equation to isolate Q:
Q/T2 - Q/T1 = ln(2)

Multiplying both sides by T1 * T2:
Q(T1 - T2) = T1 * T2 * ln(2)

Dividing both sides by (T1 - T2):
Q = T1 * T2 * ln(2) / (T1 - T2)

Now we can substitute the known values:
T1 = 35°C + 273.15 = 308.15 K (temperature before increase)
T2 = (35°C + 10°C) + 273.15 = 318.15 K (temperature after increase)

Q = 308.15 K * 318.15 K * ln(2) / (308.15 K - 318.15 K)

Calculating Q:
Q ≈ 0.623 eV

Therefore, the activation energy for diffusion is approximately 0.623 eV.