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Pressure equalization between two tanks

  1. Jun 10, 2015 #1
    1. The problem statement, all variables and given/known data
    Given 2 tanks A and B.
    Tank A has a volume of 2m3 and contains air at pressure 1 bar absolute
    Tank B has a volume of 1m3 and contains air.
    Tank A and B and interconnect with a pipe and has a stopcock and is currently closed.
    What is the pressure or air that would be needed in Tank B such that after the stopcock is opened and pressures in both tanks have equalized the pressure in both tanks now is 0.8 bar.

    2. Relevant equations
    Can the equation
    Code (Text):
    P1V1 = P2V2
    be used for solving this problem.
    I'm a bit wary about this since boyles law applies to a
    since this problem involves 2 cyclinders can we consider the gas to be
    3. The attempt at a solution
    P1 = ?
    V1 = 1m3
    P2 = 0.8bar
    V2 = 2 + 1 = 3m3

    P1 = (0.8 * 3) / 1
    P1 = 2.4bar
    Which seems to be incorrect, intuitively I would have said that the pressure in the smaller tank (tank B) should have been below 1 bar to get the equalization pressure to 0.8 bar.
    Where am I going wrong?
  2. jcsd
  3. Jun 10, 2015 #2
    Sing ## P_1,P_2,V_1,V_2 ## before and ##P,V## after mixing and do your calculation again.
  4. Jun 10, 2015 #3


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    Gold Member

    Maybe you will solve the problem easier if you imagine that there is a very thin plastic sheet between the two volumes of gas such that they do not mix after you open the valve. So the two quantities of gas will change their volumes until the pressure equalizes. At that point they will each have the same 0.8 bar of pressure.

    Or, to put it another way, you have two fixed amounts of gas, each of which obeys the equation you mention.

    p1 v1 = p2 v2
    p3 v3 = p4 v4

    You know the before volumes for each, the before pressure for one, and the after pressure for each.
  5. Jun 10, 2015 #4

    For the first amount of gas in TankA
    V1 = 2m3
    P2 = 0.8bar
    V2 = 3m3 ??
    For the second amount of gas in TankB
    P3 = ?
    V3 = 1m3
    P4 = 0.8bar
    V4 = 3m3 ??
    Now that's where I get stuck. :confused:
  6. Jun 10, 2015 #5


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    Education Advisor
    Gold Member

    ##3 m^3## is the volume for both quantities. What is the volume for the original quantity that was at 1 bar and now is at 0.8 bar?

    Again, ##3 m^3## is the volume for both. The other quantity got calculated above. So what is V4? And so then what is P3?
  7. Jun 10, 2015 #6
    Apparently, you are are supposed to assume that the temperature does not change (as would be the case for an ideal gases with no heat entering or leaving and now work being done on the surroundings). Let the temperature be T. In terms of T and the gas constant R, how many moles of gas are there in tank A to start with? In terms of T, R, and the initial pressure P of tank B, how many moles of gas are there in tank B to start with? In terms of these parameters, how many moles of gas are in both tanks to start with? In terms of T and R, how many moles of gas are present in 3 cubic meters of gas at 0.8 bars? Set the initial total number of moles equal to the final number of moles, and you will have your equation for determining P.

  8. Jun 10, 2015 #7
    For Tank A The tank volume would be constant at 2m3 but air volume would increase since pressure has dropped by 20% the volume would increase by 20% and that would be 2.4 m3.

    For tank B we know tank volume 1m3 and final Pressure 0.8bar if there was 1 bar abs in the beginning again at 0.8 bar pressure the volume would go up by 20% and it should be 1.2m3. :confused:
    But can I assume this 1bar as initial pressure of tank B ?
  9. Jun 10, 2015 #8
    Use ##PV=nRT## by mass conservation.
  10. Jun 10, 2015 #9
    PV = nRT
    P = 1bar
    V = 2 m3
    R = 8.3144621 * 10-5 m3.bar/K. mol
    T = 27 deg. C = 273 + 27 = 300 deg K
    n = PV/RT = 2/(8.3 * 10-5) * 300
    n = 200000/2490
    n = 80.3 moles

    I do not know the initial pressure of tank B

    Again initial pressure of tank B is unknown.

    n = 96.22 moles

  11. Jun 10, 2015 #10
    You know that there are 80.3 moles in tank A to start, and 96.22 moles in tanks A and B combined. So how many moles are in tank B to start? So you know the number of moles in tank B to start, and you know the temperature and volume of tank B. So, what is the initial pressure in tank B?

  12. Jun 11, 2015 #11
    Is it possible to find moles without knowing the initial pressure??:oldsurprised:
    If The total moles of gas inside the 2 cylinders will be constant through out the experiment...
    Then the moles in tank B would be 96.22 - 80.3 = 15.92
    but this seems to be quite low compared to tank A which was 80.3 ??:confused:
  13. Jun 11, 2015 #12
    Would it be right to use this
    P1 * V1 + P2 * V2 = Ptot * Vtot
  14. Jun 11, 2015 #13
    Not really. The pressure is going to be lower and the volume is only half.

    So, now that that is settled, what is the original pressure in tank B?

  15. Jun 11, 2015 #14
    PV = nRT
    P = ?
    V = 1 m3
    R = 8.3144621 * 10-5 m3.bar/K. mol
    T = 27 deg. C = 273 + 27 = 300 deg K
    n = 15.92 moles
    P = nRT / V
    P = 0.4 bar

    So 0.4 bar of pressure would be needed in Tank B to get the final equalization pressure of 0.8 bar in both tanks.
  16. Jun 11, 2015 #15
    Having said that why couldn't we just use

    P1 * V1 + P2 * V2 = Ptot * Vtot
  17. Jun 11, 2015 #16
    Yes. Do you know why?
  18. Jun 11, 2015 #17
    For Tank A and Tank B, P*V = constant
    so when we connect Tank A and Tank B that becomes say a large Tank C so after equalization the PV of this tank will also be constant
    so we can say
    P1 * V1 + P2 * V2 = Ptot * Vtot
  19. Jun 11, 2015 #18
    Well, I start with n = PV/(RT) and n(initial) = n(final). Then, when I write the conservation of moles equation, RT cancels on both sides. In other words, PV is constant because n (and hence nRT) is constant. There are different approaches that work; just so you're comfortable with yours.
  20. Jun 11, 2015 #19
    This is OK, if you understand fundamentally what it is saying.

    If you divide this equation by RT, then you end up with the mole balance that that I was discussing in my posts, and that Insightful was referring to. Both of us felt that using the mole balance was an easier way of understanding what's happening. In any event, it is always very worthwhile for a student to examine two different approaches to the same problem (if two approaches exist).

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