(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.

conversion for seconds to hours is 60*60 or multiply 3600.

could you please explain the formula

i tried everything
thanks for your time

18

18

Thanx

So what is a) and what is b)?

18 is not right

someone please help regarding answer of a and b. 18 is not showing green check

Same here. Received a red X on the answer.

formula for formula in #8

4.5

WONG ANSWER

what would be answer of b. would it be 45

B) is not 45 and A) is not 18

Impossible for someone to explain

a) 1.74

thanks

Thanks

Thanks for a) any thoughts on b)?

somebody please tell the answer of b. I tried with all equation.

D=D0e^-Ea/RT
Created Eq.1 for 35+273K
and Eq.2 for 45+273k

i got Ea 1.55*10^4J/mol
even after conversion with j/mol to EV not getting green check
1KJ/mol=1.03*10^-2

if some body got green check please suggest me.

Suggestion: do the work yourself rather than cheating. If you're desparate for a green tick, draw one in a graphics program

MIT the best university in the world. why are you cheating? whay are you not reading and understand the leacture?

Bad student >>>

how to convert j/mol to EV?

1KJ/mol=1.03*10^-2EV

please somebody tell about b

return to lecture and you get b

this is Honor. Please stop cheating

this is not useful for you

b=0.551

Thank you