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Helmholtz potential for mixture of simple ideal gases

  1. Feb 12, 2016 #1
    1. The problem statement, all variables and given/known data
    Two subsystems within a 20 l cylinder are separated by an internal piston. Each of them is initially composed of 1 mole of component 1 and one mole of component 2, both of which will be treated as a monatomic ideal gas. The cylinder has diathermal walls and is in contact with a thermal reservoir (meaning its temperature should be constant and equal to [itex]T_r = 373 K[/itex] . The piston is permeable to component 1 but impermeable to component 2. The piston is in the center. We're to prove that the Helmholtz potential for this system is:
    [itex]F=N\frac{T}{T_o} - \frac{3}{2}NRTln\frac{T}{T_0} - N_1RTln(\frac{VN_o}{V_0N_1}) - N_2^1RTln(\frac{V^1N_0}{V_0N_2^1}) - N_2^2RTln(\frac{V^2N_0}{V_0N_2^2})[/itex]
    Where [itex] T_o, f_o, V_o , N_0 [/itex] are attributes of a standard state. N is the total mole number and superscripts represent the subsystem, while subscripts refer to components ([itex] N_2^1 [/itex] is the mole number of component2 at subsystem 1 and so on).
    Afterwards, we must calculate the work required to push the piston to a position such that [itex]V^1=5 [/itex] and [itex]V^2=15 [/itex] in two ways : first by direct integration (the relation dW= PdV is valid, for the process is assumed to be carried out quasi-statically)
    The answer is W=893J
    2. Relevant equations
    We know that the Helmholtz potential of a mixture of simple ideal gases is the is the sum of the individual potentials. The fundamental equation of a monatomic gas in the Helmholtz potential is:
    [itex]F=NRT[\frac{F_0}{N_0RT_0} - ln(\frac{T}{T_0}^{3/2} \frac{V}{V_0} (\frac{N}{N_0})^{-1})][/itex]

    3. The attempt at a solution
    The Helmhholtz potential is additive, so I wrote both F as a sum of the potentials of each subsystem. Namely:
    [itex] F^1=N_1^1RT[\frac{F_{011}}{RN_{01}T_0} - \frac{3}{2}ln\frac{T}{T_0} -ln(\frac{V^1}{V_{01}}\frac{N_{011}}{N_1^1})] + N_2^1RT[\frac{F_{012}}{RN_{021}T_0} - \frac{3}{2}ln\frac{T}{T_0} - ln(\frac{V^1}{V_{01}}\frac{N_{012}}{N_2^1})] [/itex]
    And
    [itex] F^2=N_1^2RT[\frac{F_{021}}{RN_{02}T_0} - \frac{3}{2}ln\frac{T}{T_0} -ln(\frac{V^2}{V_{01}}\frac{N_{021}}{N_1^2})] + N_2^2RT[\frac{F_{022}}{RN_{022}T_0} - \frac{3}{2}ln\frac{T}{T_0} - ln(\frac{V^2}{V_{01}}\frac{N_{02}}{N_2^2})] [/itex]
    Summing then, I find :
    [itex] F-Nfo\frac{T}{T_0} - \frac{3}{2}RTln\frac{T}{T_0} - N_1^1RTln(\frac{V^1N_0}{V_0N_1^1}) - N_1^1RTln(\frac{V^2N_0}{V_0N_1^2}) - N_1^2RTln(\frac{V^1N_0}{V_0N_2^1}) - N_1^2RTln(\frac{V^2N_0}{V_0N_2^2}) [/itex]
    Where I've grupped all constants together. That's still different from the result I'm expected to prove, and the only different part seems to regard component one, which oddly enough is the component the piston is permeable to. Can't that be a mistake from the book? Even if it is, my expression wouldn't agree with the one I need, so I'd still need help.
    As for the second part, the variation of F is vanishing identically for me. If there's a mistake and the membrane is actually permeable to 1 instead of 2, I get an expression which makes sense, but comes in terms of [itex] N_2^1 and N_2^2 [/itex] . I wasn't able to eliminate both of them to get a number for the work
     
    Last edited: Feb 12, 2016
  2. jcsd
  3. Feb 12, 2016 #2
    Shouldn't ##N_1^1/V^1=N_1^2/V^2=N_1/V##?
     
  4. Feb 12, 2016 #3
    That would allow me to group the those two terms!!!! But why? Is it because of Dalton's partial pressure law? If i write the ideal gas law for [itex]N_1^1[/itex] and [itex]N_1^2[/itex]I guess the RT's will cancel out...
     
  5. Feb 12, 2016 #4
    Yes. If there is a (semi-permeable) membrane that allows passage of one component, but not the other, the chemical potentials (partial pressures) of that component on both sides of the membrane must be the same.
     
  6. Feb 13, 2016 #5
    What about the integration? I figure I must use [itex]dw= P^1dV^1 + P^2dV^2 [/itex], but if I write [itex]W= \int_{10}^5 N^1RT dV^1 + \int_{10}^{15} N^2RT dV^2[/itex] , where both N's are treated as constants, it gives the wrong answer. I can't think of another relation that would enable the integration.
     
  7. Feb 13, 2016 #6
    No. N1 will change between the two compartments as you move the piston (so that the partial pressures of this species in the two compartments are always equilibrated).
     
  8. Feb 13, 2016 #7
    . Does that mean the work from [itex]N_1^1[/itex] and [itex]N_1^2[/itex] will cancel each other? So the net work is [itex] \int_{10}^5 \frac{N_2^1RT}{V^1} dV^1 + \int_{10}^{15}\frac{N_2^2RT}{V^2} dV^2[/itex]? but that gives RTln(¾), so I'm still missing something.
     
    Last edited: Feb 13, 2016
  9. Feb 13, 2016 #8
    When I substitute into this equation, I get the given answer.
     
  10. Feb 13, 2016 #9
    I'm dumb, RTln(3/4) is just the answer with changed signs, which makes sense because I calculated the work delivered by the work source when I should calculate the work delivered to it o0)
     
  11. Feb 13, 2016 #10
    I don't think you're dumb. The only help I really gave you was how to address the semi-permeable membrane.
     
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