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.

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a) 1.74

b=0.551

You have 333 mL of alkaline solution at pH = 11.1. You wish to neutralize this by reacting it with 222 mL of acid. What must be the value of the pH of the acid?

To find out how long it will take for 0.001 moles of CO2 to leak from the container at steady-state, we can use Fick's first law of diffusion.

Fick's first law of diffusion states that the rate of diffusion (J) is equal to the product of the diffusion coefficient (D), the concentration gradient (∆C/∆x), and the area (A) through which the diffusion occurs.

J = -D * (∆C/∆x) * A

Since we are assuming steady-state conditions, the concentration gradient is constant, and we can rearrange the equation to solve for time (t):

t = (V * ∆C) / (J * A)

Where:
V = volume of CO2 to be leaked (0.001 moles)
∆C = change in concentration (∆C = solubility at 10 atm - solubility in surroundings)
J = rate of diffusion
A = effective area of the membrane (100 cm^2)

First, let's calculate ∆C:
∆C = (6.97 x 10^-4 moles/cm^3) - 0
∆C = 6.97 x 10^-4 moles/cm^3

Next, let's calculate the rate of diffusion:
J = -D * (∆C/∆x) * A

Since we are assuming steady-state conditions, the concentration gradient (∆C/∆x) is constant (-∆C/∆x = ∆C/L), where L is the thickness of the polymer membrane.

J = -D * (∆C/∆x) * A
J = -D * (∆C/L) * A
J = -D * (∆C * A/L)

Using the values given:
D = 2.29 x 10^-8 cm^2/s
∆C = 6.97 x 10^-4 moles/cm^3
A = 100 cm^2
L = 100 microns (converted to cm by dividing by 10^4)

J = - (2.29 x 10^-8 cm^2/s) * (6.97 x 10^-4 moles/cm^3 * 100 cm^2) / (100 microns / 10^4)
J = - (2.29 x 10^-8 * 6.97 x 10^-2) moles/s

Now, we can substitute the values into the equation to find the time (t):
t = (V * ∆C) / (J * A)
t = (0.001 moles * 6.97 x 10^-4 moles/cm^3) / ((2.29 x 10^-8 * 6.97 x 10^-2) moles/s * 100 cm^2)

Calculating this will give you the time (t) in seconds.

To convert the time from seconds to hours, divide it by 3600 (since there are 3600 seconds in an hour).

For part (b), the diffusivity is observed to double when the temperature is increased by 10 degrees C. This suggests that the diffusivity (D) and the temperature (T) are related by the equation:

D2 = D1 * exp((Ea/R) * ((1/T2) - (1/T1)))

Where:
D2 = diffusivity at temperature T2
D1 = diffusivity at temperature T1
Ea = activation energy
R = gas constant (8.314 J/(mol*K))
T2 = final temperature (35 + 10 = 45 degrees C, or 318 K)
T1 = initial temperature (35 degrees C, or 308 K)

By rearranging the equation, we can solve for the activation energy (Ea):

(Ea/R) = ln(D2/D1) / ((1/T2) - (1/T1))

Then, substitute the values:

(Ea/8.314) = ln(2) / ((1/318) - (1/308))

Calculating this will give you the activation energy (Ea) in joules. To convert it to electron volts (eV), divide it by 1.602 x 10^-19 J/eV.