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Pressure inside two containers

  1. Jan 13, 2016 #1
    1. The problem statement, all variables and given/known data
    Two rigid containers contain the same type of ideal gas and are connected by a thin tube with a valve. Container two is 3 times the volume of container 1. The initial pressures inside the containers are different, as are the temperatures. When the valve is opened, the temperatures inside the containers are maintained. Find an expression for the equilibrium pressure inside the containers.

    2. Relevant equations
    PV = nRT

    3. The attempt at a solution
    P1V1=n1RT1 and P2V2=n2RT2
    But, V2=3V1
    Total volume = V1 + V2, but V2 = 3V1

    I decided to take the molar volume, as I can't calculate the number of moles, But i'm not sure if i can do this

    It's this step that really stumps me, because the temperature inside the two tanks are maintained at different values, so for this last step I took the average temperature.

    P3ΣV,m=RT3 where P3 is the equilibrium temp and T3 is the average temperature of the two tanks.

    After some rearranging i come to the answer P3=T3/((T1/P1)+(T2/3P2))

    My real questions are, was I correct in taking the molar volume, and am I incorrect in taking the average temperature of the two tanks?

    Sorry for the hard to read formatting, it's my first time posting on here.
  2. jcsd
  3. Jan 13, 2016 #2
    This equation looks incorrect to me. Please reconsider.
    If V1=V and V2 = 3V, please express the number of moles of each container in terms of V, T1, P1, T2, and P2.
    What is the total number of moles in terms of these parameters?
  4. Jan 13, 2016 #3
    Ah, I think I see.
    n1+n2=P1V1/RT1 + P2V2/RT2

    And then I can sub V2 for 3V1

    n1+n2=P1V1/RT1 + P23V1/RT2

    Bringing out the V1

    n1+n2=V1(P1/RT1 + 3P2/RT2)

    Divide left side by (P1/RT1 + 3P2/RT2) and right side by (n1+n2)

    V,m=RT1/P1 + RT2/3P2

    Which after looking up above gives the same result, so i'm not sure what i've done.
  5. Jan 13, 2016 #4
    Let P be the final pressure in both tanks. In terms of P, V1, and T1, what is the final number of moles in tank 1? In terms of P, V1, and T2, what is the final number of moles in tank 2? In terms of these parameters, what is the final sum of the number of moles in the two tanks? Has the total number of moles in the two tanks changed from the initial to the final state? What is the final pressure?
  6. Jan 13, 2016 #5
    Tank 1, n1=PV1/RT1
    Tank 2, n2=PV2/RT2

    n1+n2=PV1/RT1 + PV2/RT2

    moles of gas doesn't change so PfV1/RT1 + PfV2/RT2 = P1V1/RT1 + P2V2/RT2

    The question wants me to work out the pressure without any values of volume given, so can I equate the molar volumes?

    RT1/P1 + RT2/3P2 = RT1/Pf + RT2/3Pf

    P1/T1 + 3P2/T2 = Pf/T1 + 3Pf/T2

    Pf = ((P1/T1 + 3P2/T2)/(1/T1 + 3/T2))
    Last edited: Jan 13, 2016
  7. Jan 13, 2016 #6
    Still felt uneasy with what I worked out above. In particular the molar volume. If the molar volume is (total volume/n1+n2) then, it must be,

    So, in the case of

    n1+n2 = PV1/RT1 + P3V1/RT2

    Dividing by total volume, 4V1

    n1+n2/4V1=P1V1/4V1RT1+ P23V1/4V1RT2

    n1+n2/4V1=P1/4RT1+ 3P2/4RT2

    So, 4V1/(n1+n2)=4RT1/P1+ 4RT2/3P2

    V,m = 4RT1/P1+ 4RT2/3P2 = 4RT1/P+ 4RT2/3P

    After solving, it comes out the same as above, Pf = ((P1/T1 + 3P2/T2)/(1/T1 + 3/T2))

    I think this proof was a bit clearer, hopefully it's right.
  8. Jan 13, 2016 #7
    No. In your last equation above, you simply substitute V2=3V1. This gives you:

    Pf = ((P1/T1 + 3P2/T2)/(1/T1 + 3/T2))

    Then, to make the equation look more presentable, you multiply number and denominator by T1 T2 to obtain:

  9. Jan 13, 2016 #8
    Thank you for you help, greatly appreciated. :)
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