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Thermodyn.: ideal gases in 2 two chambers separated by a sliding barrier

  1. Oct 19, 2012 #1
    1. The problem statement, all variables and given/known data
    Two different ideal gases are separated by a sliding barrier that can move vertically. The gas in the upper chamber 1 has n moles of material, whilst the gas in the lower chamber 2 has 3n moles of material. At T0 the weight of the barrier is such that the volumes of the two chambers are equal. What is the ratio of V2/V1at T1.

    2. Relevant equations

    P = F/A
    W = mg
    PV = nRT

    3. The attempt at a solution

    Equilibrium at T0: F2 = F1 + W
    where the Fi are the forces from the i gas on the barrier.

    therefore W = F2 - F1 = A(P2 - P1)
    [itex] \frac{3AnRT_{0}}{V_{2}} - \frac{AnRT_{0}}{V_{1}} = \frac{2AnRT_{0}}{V_{0}} [/itex], because [itex]V_{0} = V_{1}(T_{0}) = V_{2}(T_{0})[/itex]
    further [itex] W = \frac{4AnRT_{0}}{V} [/itex], because [itex]V_{0} = \frac{V}{2}[/itex] where V is the total volume of the two chambers!

    Now I want to check equilibrium forces at T1 but using F2 = F1 + W for T1 and substituting for constant weight will not allow me to calculate the ratio V2/V1!
     
  2. jcsd
  3. Oct 19, 2012 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Hi, sunrah.

    Can you use the fact that Vtotal = V1 + V2 to eliminate V2, say, from your equation F2=F1+W at T1? Then maybe you can solve for V1 in terms of T1, T0, and Vtotal. Once you have V1 you should be able to get an expression for V2 and then construct the ratio of the final volumes.
     
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