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Ideal Gas Law in "alternate" universe

  1. Oct 3, 2015 #1

    RaulTheUCSCSlug

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    1. The problem statement, all variables and given/known data
    Assume that in an alternate universe, the laws of physics are very different from ours and that "ideal" gases behave as follows:
    (i) At constant temperature, pressure is inversely proportional to the square of the volume.
    (ii) At constant pressure, the volume varies directly with the 2/3 power of the temperature.
    (iii) At 273.15K and 1.00 atm pressure, 1.00 mole of an ideal gas is found to occupy 22.4 L

    Also note that for P and T constant you can assume the volume is proportional to the number of moles of gas.

    2. Relevant equations
    PV=nRT

    P∝1/V^2 for constant T

    V∝T^2/3 for constant P

    V∝n for constant P and T

    3. The attempt at a solution

    Okay so I understand that when the ideal gas law was derived, each variable was held constant, then seen how the other variables changed and was related through a proportionality constant R. I also know that for the original ideal gas law, PV was proportional to T and PV is proportional to mT so therefore when you add the constant of proportionality R, the relation is clear that PV=nRT.

    But for the instances given I can only get that since P is proportional to 1/V^2 and V is proportional to n, then it would make since that in the final solution it is something like P/V^2=Rn^2T^2/3 in which R is constant. But that is not what the solution states.

    Solutions answer is PV^2=n^2RT^4/3.

    Why is the T^2/3 squared...? Is it because the Volume is squared? Thus both N and T have to be squared? And how does this show that P is proportional to inverse V squared. I'm just really stumped and I don't think I understand the concept of holding a variable constant or something...
     
  2. jcsd
  3. Oct 4, 2015 #2

    andrewkirk

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    First note that N is assumed to be held constant in (i) and (ii), and ideally it would say that, but (unfortunately, in my view) usual practice seems to be to take that as implied rather than stating it explicitly.

    Taking that as read, I would proceed as follows. First write the most general form of the solution, under which everything is proportional to some as-yet-unknown power of everything else:
    $$P=kV^aN^bT^c$$
    for some constant ##k## to be determined by further experiments, and unknown constants ##a, b, c##.
    Rule (i) Tells us that ##a=-2##. So our equation becomes:
    $$P=kV^{-2}N^bT^c$$
    Making V the subject, we get:
    $$V=N^\frac{b}{2}T^\frac{c}{2}\sqrt{\frac{k}{P}}$$
    Now we use (iii) to infer that ##b=2##. So our equation becomes:
    $$V=NT^\frac{c}{2}\sqrt{\frac{k}{P}}$$
    Lastly, we use (ii) to infer that ##\frac{c}{2}=\frac{2}{3}##, so ##c=\frac{4}{3}## and the final equation is:
    $$V=NT^\frac{2}{3}\sqrt{\frac{k}{P}}$$

    Squaring both sides and multiplying by ##P##, we get the result.
     
  4. Oct 4, 2015 #3

    RaulTheUCSCSlug

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    Wow, that was such a clear explanation, thank you! So is that the general process that you would take with these type of problems? and I am still a bit confused on why we would assume that N is constant other than if it was explicitly said? Is this because N can be any number, but since there will not be a change of molecules due to the terms not affecting the molecules by means of destroying or creating them in some way it is constant? Because if not, there would be some serious consequences right!?
     
  5. Oct 4, 2015 #4

    RaulTheUCSCSlug

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    Just a quick question though, why did you choose the equation to be P=k(V^a)(N^b)(T^c) instead of... oh idk...

    k=(P^a)(V^b)(N^c)(T^d)

    wouldn't that make more sense since k is a constant? Or is there a specific reason to having it the general way that you wrote it.
     
  6. Oct 4, 2015 #5

    andrewkirk

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    I think it's because they are talking about a system, and that usually means something that cannot gain or lose particles. I don't know why it means that, but that's the way it seems to be typically used in thermodynamics. I suppose because it's easier to prevent a system from gaining or losing particles than it is to isolate it from temperature or pressure changes.

    Your equation is equivalent. If we start with ##k=P^aV^bN^cT^d## then by raising each side to the power ##-\frac{1}{a}## (which we can do because the whole system is trivial if any of them are zero) and then multiply both sides by ##Pk^\frac{1}{a}## we get ##P=k'V^{a'}N^{b'}T^{c'}## where ##k'=k^\frac{1}{a},\ a'=-\frac{b}{a},\ b'=-\frac{c}{a},\ c'=\frac{d}{a}##.

    Your equation is nicer in that it is symmetrical between all the variables. But I chose the form I did because it made it most convenient to apply rule (i), which I wanted to apply first (for no other reason than that it was written first).
     
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