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Molar Ratio of Two Ideal Gases

  1. Apr 1, 2014 #1

    enokoner

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    Gold Member

    1. The problem statement, all variables and given/known data

    A mixture of two gases, A and B, exists at pressure p1, volume V, and temperature T1. Gas A is subsequently removed from the mixture in a constant-volume process. The remaining gas B is found to have a pressure p2, volume V, and temperature T2. Express the ratio of the number of moles of gas B to the number of moles of gas A in the terms of p1, p2, T1 and T2.

    a. [itex] \frac{p_2 T_1}{p_1 T_2 - p_2 T_1}[/itex]

    b. [itex] \frac{p_2 T_1^{2}}{T_2(p_1 T_2 - p_2T_1)}[/itex]


    *Options c and d were not written because they contained specific gas constants which do not pertain to molar equations.




    2. Relevant equations


    Ideal Gas: [itex]pV = N \overline{R}T[/itex]

    Dalton's Law: [itex] p = \sum p_i [/itex]


    3. The attempt at a solution

    Universal gas constant crosses out. Volume stays constant and also crosses out.

    ∴ [itex] \frac{N_B}{N_A}= \frac{p_B T_A}{T_B p_A} [/itex]


    Relating pA and pB to p1

    [itex] p_1= p_A + p_B [/itex]


    and to p2

    [itex] p_2 = p_B [/itex]


    Also, [itex] T_A = T_1 \ T_B = T_2 [/itex]


    ∴ [itex] \frac{N_B}{N_A}= \frac{p_2 T_1}{T_2(p_2 - p_1)}[/itex]


    This is not an option. I have a feeling its because I assumed [itex] p_2 = p_B [/itex]. That assumption doesn't feel right. I don't know how else to relate these two. Thank you for considering this.
     
  2. jcsd
  3. Apr 1, 2014 #2

    Borek

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    Staff: Mentor

    I am not sure I get it.

    I have a feeling you can be a victim of your own choice of confusing indices.
     
  4. Apr 1, 2014 #3

    enokoner

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    I agree that the indices are out of control. But I double checked. Sorry for the step-skipping.

    [itex]

    N = \frac{pV}{\overline{R}T} \\

    ∴ \frac{N_B}{N_A} = \frac{\frac{p_BV}{\overline{R}T_B}}{\frac{p_A V}{\overline{R}T_A}} \\

    =\frac{p_B T_A}{p_A T_B}


    [/itex]
     
  5. Apr 1, 2014 #4

    Borek

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    Staff: Mentor

    Aren't pA and pB measured at different temperatures?

    I got 'a', starting from

    [tex]p_1V=(N_A+N_B)RT_1[/tex]

    [tex]p_2V=N_BRT_2[/tex]
     
  6. Apr 1, 2014 #5

    enokoner

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    OP Second Attempt by solving in reverse.

    I got the answer but I don't know how. The book states that the answer is B. I got it by solving the problem in reverse. I used a common equation that relates pressure and temperature wrt Ideal gases. The problem is I don't understand why I'm supposed to used that or why I should set certain values to each other.


    2.1 Relevant equations


    Ideal Gas: [itex]pV = N \overline{R}T[/itex]

    Dalton's Law: [itex] p = \sum p_i [/itex]

    Assuming System is isentropic: [itex] \frac{T_2}{T_1} = (\frac{p_2}{p_1})^{\frac{k-1}{k}} [/itex]


    3. Second attempt at a solution

    Universal gas constant crosses out. Volume stays constant and also crosses out.

    ∴ [itex] \frac{N_B}{N_A}= \frac{p_B T_A}{T_B p_A} [/itex]


    Relating pA and pB to p1:

    [itex] p_1= p_A + p_B [/itex]


    and to p2:

    Assuming (I HAVE NO BASIS FOR THIS ),

    [itex] p_B = p_2 ; \ p_2 = p_1 ; \ k = 0.5 [/itex]


    Therefore,

    [itex] \frac{T_2}{T_1} = (\frac{p_B}{p_2})^{\frac{0.5-1}{0.5}} \\

    = \frac{P_2}{P_B} \\

    ∴ p_B = p_2 \frac{T_1}{T_2} \\

    ∴ p_A = p_1 - p_B = p_1 - p_2(\frac{T_1}{T_2}) [/itex]

    Also, [itex] T_A = T_1 ; \ T_B = T_2 [/itex]


    Substituting,

    [itex] \frac{N_B}{N_A}=\frac{p_2(\frac{T_1}{T_2})T_1}{T_2(p_1-p_2(\frac{T_1}{T_2})} \\


    = \frac{p_2 T_1^{2}}{T_2(p_1 T_2 - p_2 T_1)} [/itex]


    I know setting those values arbitrarily is nonsense. But I thought perhaps it could give me some insight. It has not. I'm grasping at straws. Can anyone help?
     
    Last edited: Apr 1, 2014
  7. Apr 1, 2014 #6

    enokoner

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    Gold Member

    Thank you so much.

    I got your same answer using your very clear setup. (You rock man). The solution was too good. It was elegant and made sense the whole way through...

    Setting both eqs. equal to V/R:

    [itex]\frac{N_A}{N_B}+ \frac{N_B}{N_B} =\frac{T_2 p_1}{p_2 T_1} \\

    \frac{N_B}{N_A} = \frac{1}{\frac{T_2 p_1}{p_2 T_1} -1} \\


    = \frac{p_2 T_1}{T_2 p_1 -p_2 T_1}



    [/itex]

    Thats how it should be. So I went hunting for the online errata. I can't believe I didn't do this sooner!

    errata.png



    Amazing Borek! I gotta go over all the errata. Thanks a million.
     
    Last edited: Apr 1, 2014
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