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Pressure, Volume, Temperature, etc.

  1. Jan 7, 2005 #1
    Ok so I am doing this problem and it is making me insane. The problem states that you have a container, the container has a partition and in compartment A you have an ideal gas at 5 atm and 400K. In compartment B you have another ideal gas at 8 atm and 400K. The partition is removed and the gasses are allowed to mix together. The molar fraction for gas A is found to be 25/43. The total volume of both of the compartments is 29 liters. Find the original volumes of compartments A and B.

    So here is how I approached it.

    the molar fraction is [N
    k/Ntotal] = [Pk/Ptotal].

    Using this formula the total number of moles is 43 and using PV=nRT you get the total pressure equal to 48.669 atm. Plugging this number back into the molar fraction formula gives you a partial pressure for gas A of 28.30 atm.

    Now it may just be me but from the get go this does not make a lick of sense as you originally had pressure for gas A of 5 atm and pressure for gas B of 8 atm. If the container is not flexible and the total volume of the container is 29 liters I personally would not expect the combined pressure of the two gasses to be more than 3.5 times greater than the sum of their individual pressures. I have gone on to try and figure out what has gone wrong several different ways but am at a total loss. The only thing I can think of is that the container is flexible, but we have not started studying that and it does not mention it anywhere in the text. I am sure it is something glaringly obvious and if anyone would like to point it out to me I would be eternally grateful. My homework is not due until next friday but this is making me insane.
  2. jcsd
  3. Jan 7, 2005 #2
    First of all, you made a mistake finding the number 48.669 using PV=nRT.You don't have n. 25/43 is a ratio ONLY and tells you nothing about how many moles of gas particles in the box.
    Hope this answer you question
  4. Jan 7, 2005 #3
    Further Questions Pressure, Temp Volume etc

    Thanks for you quick response,
    Ok so I looked back in my text and it specifically defines Dalton's Law as Ptotal=[ntotalRT/V] where ntotal is the total number of moles of gas present. the molar ratio is defined as Xk=nk/ntotal=Pk/Ptotal. I would really love it if the total number of moles present was not 43 but nowhere in the question does it give the identity of the gasses. Can you nudge me onto the next step or how you would figure out the total pressure?
  5. Jan 7, 2005 #4


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    Though i haven't slept over the last 17hrs,this problems is far too simple for me to let the fatigue alter my judgements.

    Initial state for compartment A:
    [tex] p_{1}V_{1}=\nu_{1}RT [/tex](1)
    Initial state for compartment B:
    [tex]p_{2}V_{2}=\nu_{2}RT [/tex](2)

    Divide (1) through (2) and u'll get
    [tex] \frac{p_{1}V_{1}}{p_{2}V_{2}}=\frac{\nu_{1}}{\nu_{2}} [/tex] (3)
    Plug in the numbers and find
    [tex] \frac{5}{8}\frac{V_{1}}{V_{2}}=\frac{25}{18} [/tex](4)
    ,from where
    [tex]V_{1}=\frac{20}{9} V_{2} [/tex](5)

    But we know that
    [tex] V_{1}+V_{2}=29l [/tex] (6)
    [tex] V_{1}=20l;V_{2}=9l [/tex] (7)



    PS.[tex] \frac{\nu_{1}+\nu_{2}}{\nu_{1}}=\frac{43}{25}\Rightarrow \frac{\nu_{1}}{\nu_{2}}=\frac{25}{18} [/tex] =>RHS from (4).
  6. Jan 7, 2005 #5

    Andrew Mason

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    Correct. (BTW, it is easier to write with subscripts: Nk/Ntotal = Pk/Ptotal)

    It just gives the mole fraction as 25/43. It doesn't say the total no. of moles is 43.

    I assume the system does not exchange heat with the environment. So the process is adiabatic ([itex]\Delta Q = 0[/itex]). Since there is no work done by the gas either ([itex]\Delta W = 0[/itex]), this process results in no change to the internal energy of the system (So: [itex]\Delta U = 0[/itex])

    This means:
    [tex]P_AV_A + P_BV_B = P_fV_f[/tex]


    [tex]n_ART_i + n_BRT_i = (n_A + n_B)RT_f[/tex]

    which means that

    [tex]T_i = T_f[/tex] so lets just call it T.

    [tex]n_A = P_AV_A/RT = (25/18)n_B[/tex] and [tex]n_B = P_BV_B/RT[/tex]


    [tex]n_A = (25/18)P_BV_B/RT = P_AV_A/RT [/tex]

    [tex]25P_BV_B = 18P_AV_A [/tex]

    [tex]25*8*V_B = 18*5*V_A [/tex] where [itex]V_A=29-V_B[/itex]

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