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Homework Help: What happens to the temperature of the gas?

  1. Aug 15, 2006 #1
    An ideal gas is contained in an insulating cylinder fitted with a piston.

    The piston is suddenly moved inwards so that the volume of the gas is reduced.

    What happens to the temperature of the gas?

    Should the temperature increase?
  2. jcsd
  3. Aug 15, 2006 #2
    How does volume relate to temperature, according to the ideal gas law? Wait, that's pretty much what the question is asking. Do you know the ideal gas law equation?
  4. Aug 15, 2006 #3

    Andrew Mason

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    What you want to determine is what happens to the internal energy of the gas, U = nCvT

    U = Q - W where W = work done by the gas.

    This is an adiabatic compression (no exchange of heat with surroundings so Q = 0). So U = -W. So the question is: Is work done on the gas or is work done by the gas? That will tell you how U and T change.

  5. Aug 16, 2006 #4
    i am afraid i don't quite get you. what's C v and T?

    does the ideal gas law help? you only know the the volume decreases and the pressure increases. they are not proportional in this case since temp is yet to be determined whether is it a constant.

    thanks if you can clarify this.
  6. Aug 16, 2006 #5

    use the first law of thermodynamics: [tex] Q + W = \Delta E_{pot} + \Delta E_{kin}[/tex]

    --> the process is adiabatic, thus Q = 0. The gas is compressed. Acoording to W = p\Delta V[/tex] W < 0

    Because the increase in potential energy is neglectable, [tex] \Delta E_{kin}[/tex] > 0, because of conservation of energy.

    Because temperature rises, when the kinetic energy rises, the temperature must increase during this process.
  7. Aug 16, 2006 #6

    Andrew Mason

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    Cv is the specific (molar) heat capacity of the gas at constant volume. This is just to show you that internal energy of the gas is proportional to temperature.

    Since Q is 0, the change in internal energy of the gas will depend on whether work is done on the gas (in which case, its internal energy increases) or work is done by the gas (in which case its internal energy decreases). Since [itex]\Delta U = nC_v\Delta T = -W[/itex] where W is the work done BY the gas, all you have to do is figure out the sign of W (+ if work is done by the gas, - if done on the gas) to find the direction of the change of T.

  8. Aug 16, 2006 #7
    Those answers seem to be beyond what mengshuen has learned so far. All he needs to do is look at Charles's law.
  9. Aug 16, 2006 #8
    mengshuen, http://en.wikipedia.org/wiki/Charles%27s_Law" [Broken] can be derived from the ideal gas law, if you're more comfortable with that. But I advise you to take the following course - you'll learn more, and it always applies to these sorts of problems.

    You know the ideal gas law: [tex]PV = nRT[/tex]. We can rewrite the ideal gas law as [tex]\frac{PV}{nT} = R[/tex]. [itex]R[/itex] is a constant, so as long as the system remains ideal and closed, we have the general equation:
    [tex]\frac{PV}{nT} = \frac{P'V'}{n'T'}[/tex].
    Where the left side has the initial values, and the right side has the final values.

    Here's the fun part - cancelling stuff out. Let's say you have a problem where a heater is changing the temperature of a gas inside a closed cylinder. What is constant?

    Answer: Volume and number of moles of gas. Thus we can write that [tex]V = V'[/tex] and [tex]n = n'[/tex]. After some elementary algebra, our general equation for this particular problem becomes:
    [tex]\frac{P}{T} = \frac{P'}{T'}[/tex].

    You can apply the same methodology to your problem.

    As you can see, Charles's Law, Boyle's Law, etc. can all be derived from the ideal gas law. As long as you can understand what is held constant, the ideal gas law is all you need.
    Last edited by a moderator: May 2, 2017
  10. Aug 17, 2006 #9

    Andrew Mason

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    The problem in this question is that none of the three variables is constant: [itex]P' \ne P; T' \ne T; V' \ne V[/itex]. You have to apply the First law of Thermodynamics.

  11. Aug 17, 2006 #10
    In the case of rapid expansion/compression you could use the adiabatic laws to calculate how the Temperature changes with Volume (without considering the pressure; all you need is the initial Temperature and the Volume before and after).

    HereĀ“s a link:

    The last formula mentioned is exactly the one you need (the Temperature increases when you decrease the volume in the adiabatic case). It is derived from the formula already stated by Andrew Mason.
    Last edited: Aug 17, 2006
  12. Aug 17, 2006 #11
    none of those laws work in this instance, so i guess the first law of thermodynamics is the best thing to use now. let me try them. in the mean time, thanks for all your help.
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