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PV=nRT – Why isn’t ‘T’ inversely proportional to ‘V’?

  1. May 31, 2012 #1
    What am I missing? According to the ideal gas law, as volume increases, temperature increases. However I’m not sure I understand why. I certainly understand the motion of the molecules equals the kinetic energy of the molecules. The faster the molecules bounce about, the more thermal energy the system contains. Further, the larger the number of bouncing molecules contained within the system, the more thermal energy the system contains. But if you reduce the volume without changing the number of moles why would that change the internal energy.

    In fact, why doesn’t the gas get colder as you increase volume?
    PV=nRT tells me if we increase the volume, the temperature increases – which means the internal energy increase. But what if we increased the volume of my system to the size of our universe – wouldn’t it be awfully cold in my system? Aren’t I describing the space of our universe – near absolutely zero in temperature with just a relative handful of molecules?

    Further, if I reduce the volume why would that necessarily alter the temperature? Sure, it would increase the number of bounces per unit time, but since each bounce transfers its energy to its dancing partner, why would more bounce mean either more energy or less energy?

    I know I’m missing something obvious but I just don’t see it. This is classical physics, not quantum, so the answer should be apparent to me. It’s not.
     
  2. jcsd
  3. May 31, 2012 #2
    Actually it's the other way round!

    We usually write the independent variable on the right and the dependent variable on the left.

    eg y=3x

    read as 'y is three times x'

    So if you heat something (even a gas) up it expands (volume increases).

    However the universal gas law is telling you something else.

    It is telling you that it isn't only the volume which increases with temperature it is also the pressure and that the end result is an interplay between increase in both. That is a given increase in temperature can result in a bigger an increase in pressure and a smaller (or even none at all) increase in volume. It is even possible to think of a negative volume increase ( a decrease) with sufficient increase in pressure.
     
    Last edited: May 31, 2012
  4. May 31, 2012 #3

    AlephZero

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    That's right. The ideal gas equation says that if you increase the volume and keep the pressure constant, the temperature increases.

    To keep the pressure constant when the volume increases, you have to add energy to the gas. if you don't add energy (or if you remove energy, as in a heat engine whcih does mechanical work as the gas expands) the temperature will decrease.
     
  5. May 31, 2012 #4

    Ken G

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    Your question comes up a lot in the ideal gas law, and it's actually quite an important question because it gets to the heart of what equations mean in physics. An equation is just something that will be true, it doesn't say why it's true or come with some logical explanation-- you have to supply that yourself, and often it helps to rewrite the equation in a different form. For example, take the equation F=ma. This equation can also be written a = F/m, and it's the same equation mathematically, but note that it is "saying" something completely different in that form. Written F=ma, the equation seems to tell us that if you have a mass m, and you see an acceleration a, then there must be a net force F on the object. But if you write it a = F/m, it says that if you have a mass m, and impose a force F, you should expect an acceleration a = F/m. Note this second form is much more like a "cause and effect" explanation of the equation-- we don't say that having acceleration a "causes" force F to appear magically, we say that having a force F causes an acceleration a.

    Now consider the ideal gas law. It is customary to write it PV=nRT, as you did, but this is not actually a very logically "cause and effect" kind of way to write it in most of the situations where it gets used. That's because n and T do not determine P and V in most situations. One common situation is when you have a container of fixed V with fixed n, and we control T and let P vary (say by keeping it in equilibrium with an environment at some T, this is called "isothermal"), then P = nRT/V. Another common situation is that we control P and let T and V vary (say with a piston, and insulate the container from any outside heat, this is called "adiabatic"), such that V/T= nR/P. This seems to be the case you are imagining, but note that is no problem with V decreasing when T increases, because this is all accomplished by an even larger increase in P. Here you need a second constraint, coming from conservation of energy, to determine how V and T behave-- the ideal gas law only tells you how V/T behaves. So the "cause and effect" is that P increases cause V/T to decrease, but conservation of energy is what causes V and T to go in opposite directions.

    Incidentally, a place where a similar question emerges is in the Earth's atmosphere. Students notice that the T drops as you go up, but the density n/V also drops, and this doesn't make sense to them because the ideal gas law seems to say n/V is inversely proportional to T. But that assumes P stays fixed, and the real "cause and effect" logic of the ideal gas law in the atmosphere emerges when we write it n/V = P/RT. Both P and T are set by other processes in the atmosphere, and the ideal gas law is just an equation for the density n/V, which responds to P and T in such a way that the small decrease in T is overwhelmed by the large decrease in P, and n/V drops as you go up.
     
    Last edited: May 31, 2012
  6. May 31, 2012 #5
    I hope you are not implying V is not equal to nRT/P for an ideal gas?

    The problem with writing equations in fractional form is always what happens when the quantitity on the bottom is zero.
     
  7. May 31, 2012 #6
    OK - that might be what I was missing. (And I was glad to hear I'm not the only one confused about this.)

    I can see I still need to think about this and also play with some problems.
     
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