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Effusion and setting up the calculus problem

  1. Mar 8, 2016 #1
    1. The problem statement, all variables and given/known data

    Gas escapes from a leaky cylinder so that the rate of decrease of pressure in the cylinder is proportional to P - Patm where P is the pressure in the cylinder and Patm is atmospheric pressure (1.0 atm). The initial pressure in the cylinder is 1.1 x 103 atm. If the rate of pressure decrease is 0.28 atm/min when the pressure in the cylinder is 8.5 x 102 atm, how many days will it take the pressure in the cylinder to drop to 1% of its initial value?

    2. Relevant equations
    dp/dt is proportional to P - Patm

    3. The attempt at a solution
    This appears to be a simple problem where my challenge is simply to set up the calculus problem to solve, however, I am having trouble with this. The rate of the effusion of the gas from the leak is connected to the amount of the gas still in the cylinder less 1 atm. The one data point given in the problem should allow me to determine some sort of constant of proportionality.
     
  2. jcsd
  3. Mar 9, 2016 #2

    Borek

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    Staff: Mentor

    Hard to say where your trouble is, as the problem looks quite straightforward. And yes, all your remarks about what should be done look right.
     
  4. Mar 12, 2016 #3
    I would find how much gas would have to leak for the final conditions to be met and figure out how long it takes for this to happen
     
  5. Mar 13, 2016 #4
    What is the differential equation that you wrote down to solve this problem? This seems more like a math problem than a chemistry problem.

    Chet
     
  6. Mar 13, 2016 #5

    Charles Link

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    Homework Helper

    I get a reasonably good answer for this problem. You should be able to compute the proportionality constant ## \alpha ## with the information given. (P=850, dP/dt=-.28 atm/min). Question is, what did you get for the solution to your differential equation, using this ## \alpha ## ? There is one additional constant in the homogeneous solution of the differential equation, (along with the particular solution), that is determined by the initial pressure.
     
    Last edited: Mar 13, 2016
  7. Mar 13, 2016 #6
    We're still waiting for the OP to respond. It's been almost a week since he's been seen. We shouldn't hold our breath.
     
  8. Mar 13, 2016 #7

    Charles Link

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    Homework Helper

    It was still a fun problem to calculate. The effusion rate calculation was kind of interesting.
     
  9. Mar 13, 2016 #8
    Why don't we wait a couple of more days (say until Tues), and, if we still haven't heard back from the OP, feel free to present your solution. As a Mentor, I give you a dispensation to do so.

    Chet
     
  10. Mar 15, 2016 #9

    Charles Link

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    Homework Helper

    Per Chestermiller's permission above, I will present the solution. It really is a very good application of first order differential equation principles. It makes for a good review of the concepts, even for those who have a solid mathematics background.
    The differential equation for the system is ## dP/dt= - \alpha (P-P_o) ## where ## P_o ##=1 atm
    Given that ## dP/dt ##=-.28 atm/min when ## P=## 850 atm, we find ## \alpha=.28/849 ##
    The first order (inhomogeneous) differential equation ## dP/dt+\alpha P=\alpha P_o ## has a simple particular solution ## P=P_o ##
    We need to include the solution to the homogeneous equation ## dP/dt + \alpha P=0 ## which is ## P=Aexp(-\alpha t) ## where A is a constant (comes from ## dP/P=-\alpha dt ## so that ## \ln P=-\alpha t +C ## where C is a constant. This gives ## P=Aexp(-\alpha t) ## where A is a constant ) The full solution is ## P=Aexp(-\alpha t)+P_o ## and at t=0, ## P## =1100 atm. This gives us ## A ## =1099 atm.
    The problem asks us to find the time t when the pressure is 1 % of its initial value, so we need to find t for ## P## =11 atm. 11=1099*exp(-\alpha t)+1 This gives ## t= \ln(1099/(11-1))*849/.28 ## =1.40 E+4 minutes/((60*24)minutes/day) =9.7 days. Hopefully my arithmetic is correct. I welcome your feedback.
     
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