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I Isobaric vs. isothermal expansion

  1. Nov 23, 2016 #1
    We have a piston with ideal gas in it and a weight. The weight is placed on the piston.

    The gas is heated externally and the gas expands. Will the expansion be isobaric or isothermal?

    One argument would be: the expansion will be isobaric because the weight is providing constant pressure. The other is: "all" heat goes into mechanical work done by the gas (i.e. lifting the weight, W = mgΔh), and the temperature of the gas will not rise meaning the expansion is isothermal.

    So the question is: which one is it? What are the circumstances under which this can be determined?
  2. jcsd
  3. Nov 23, 2016 #2
    Your first statement is correct. The weight is a constraint. The system cannot help but keep a constant pressure AND it will do work on or take energy from the weight to meet that constraint. This means that the gas is not an isolated system, and rules that apply to an isolated system don't apply here.

    Now you say "all heat goes into mechanical work. Why would that be true? What conservation law requires that? Adding heat some heat goes to raising the weight and some goes into heating the gas such that the expansion preserves constant pressure.
  4. Nov 23, 2016 #3

    The energy of the weight changes by

    m g (h2 - h1) = m g (V2 - V1) / A
    = m g n R (T2 - T1) / (P A)

    P A is equal to the weight m g

    m g (h2 - h1) = n R ( T2 - T1)
    = (2/3) (3/2) n R (T2 - T1)
    = (2/3) delta U

    So 3/5 of the heat goes to raising the temp and 2/5 go to raising the weight
  5. Nov 23, 2016 #4
    For example, in this video it is described how in an isothermal expansion or compression how heat and work balance each other out in an isothermal process so that the temperature of the gas stays constant.

    So, when would this apply if the situation I described is isobaric?

    PS. Sorry for posting this on a wrong forum... I misread "high energy" as "heat energy".
  6. Nov 23, 2016 #5

    jack action

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    Ideal Gas Law: ##PV = mRT##.

    ##m## & ##R## are constant, so ##PV \propto T##

    Now, in an isobaric process ##P## is constant, so if ##V## increases, then ##T## must also increase. The heat addition corresponds to the work done + the increase in internal energy (i.e. increase in temperature).

    In an isothermal process ##T## is constant, so if ##V## increases the same way as in the isobaric process, then ##P## must have decreased (i.e. some weights were removed). The work done took its energy from the internal energy of the gas which means the gas temperature would have dropped. If it didn't (it's isothermal), then it means some heat was added to replace the «lost» internal energy. So the heat addition corresponds to the work done only.

    Note also that, even if the volume change is the same in both cases, the work done is less with the isothermal case, because the pressure is decreased during the process.

    Conclusion: If it is isothermal, you will have to remove weights.
  7. Nov 24, 2016 #6
    So, the pressure must be lower in the isothermal case. But what if we just do an experiment and see the volume growing and the weight being lifted? If we know the mass of the weight, the area of the piston and measure how much the weight is being lifted, how can we then know whether the process has been isobaric or isothermal when there are no earlier experiment results to compare to?

    In the video I linked to in my previous post, there is only a single weight on a piston that causes the pressure of F/A (A is the area of the piston, F weight). Yet, it describes an isothermal expansion/compression.
  8. Nov 24, 2016 #7
    I watched the video carefully. I see on the screen pressure is defined "P = F / A". I don't see a weight, and I didn't hear any suggestion that pressure would remain constant during the isothermal process.

    P V = n R T

    If I hold both P AND T constant I can only do work (change V) by changing n. That would be and isothermal process, but that isn't the meaning of isothermal expansion / compression and is clearly not what the video is talking about.
  9. Nov 24, 2016 #8

    jack action

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    You are kind of answering the question yourself.

    If the weight is constant and the area is constant, then the pressure is also constant. That is the definition of 'isobaric'.

    You seem to think that it is possible to raise the weight without increasing the temperature. You can't. When you want to do work with a gas, you must «loose» some portion of the energy input that will be absorbed by the gas itself.

    Imagine a single gas molecule in an enclosed space bouncing back and forth. The faster it goes (i.e. the higher the temperature), the more it will hit the walls, the better it will be able to keep the enclosed space from collapsing on itself (i.e resist pressure).

    Now, double the volume of that enclosed space (i.e. do work). The molecule has more travel to do to keep hitting the walls at the same rate, thus more chance of the enclosed space collapsing (i.e. the pressure is lower).

    What can you do? 2 things:
    • You can add a second molecule bouncing at the same speed as the first one (i.e. increase the mass ##m##);
    • You can increase the speed of the molecule (i.e. increase the temperature).
    How much energy do you need to increase the speed of the molecules? It has been found that for each joule of work you do at constant pressure, you need ##\frac{f}{2}## joules of energy added to increase the kinetic energy of the gas molecules, where ##f## is the degree of freedom of the gas molecules.

    So the total energy added (##C_P\Delta T##) is the work done (##R\Delta T##) plus the internal energy added (##C_V\Delta T = \frac{f}{2}R\Delta T##). It can be rewritten as:

    ##C_P = R + \frac{f}{2}R##
    ##\frac{C_P}{R} = 1+\frac{f}{2}##


    ##C_P = \frac{2}{f}C_V + C_V##
    ##\frac{C_P}{C_V} = 1+\frac{2}{f}##
  10. Nov 28, 2016 #9
    I shall now quote a book "Physics for Scientists and Engineers" by Douglas C. Giancoli, 4th edition (2009), page 508 (ISBN-13: 9780131495081). The text in bold is just as at appears in the original text.

    "Let us assume that the gas is enclosed in a container fitted with a movable piston, Fig. 19-7, and that the gas is in contact with a heat reservoir (a body whose mass is so large that, ideally, its temperature does not change significantly when heat is exchanged with our system). We also assume that the process of compression (volume decreases) or expansion (volume increases) is done quasistatically ("almost statically"), by which we mean extremely slowly, so that all of the gas moves between a series of equilibrium states each of which are at the same temperature. If an amount of heat Q is added to the system and temperature is to remain constant, the gas will expand and do an amount of work W on the environment (it exerts a force on the piston and moves it through a distance). The temperature and mass are kept constant, so from Eq. 19-1, the internal energy does not change: ΔEint = 2/3nRΔT=0. Hence, by the first law of thermodynamics, Eq. 19-4, ΔEint =Q-W=0, so W=Q: the work done by the gas in an isothermal process equals the heat added to the gas."

    Now, Fig 19-7 is nothing but a cylindrical piston with text "ideal gas" in the container and the text "movable piston". There isn't any additional weight on the piston, but, of course, the piston has a mass of its own and therefore a weight, a constant weight. The text clearly says that work is done on the piston to move it , so it wouldn't make sense to assume that the piston is weightless. Even if that was assumed, there would still be a constant force due to atmospheric pressure (F = P0Apiston)! If another constant weight was added on top of the piston, that would just equal a heavier piston, right? Therefore, the weight is constant in this process, i.e. the external pressure is constant, not the pressure of the gas (just like in the video that I posted earlier!). Changing the external pressure (due weight) is not discussed at any point (also true for the video).

    Indeed, the pressure of the gas (assuming n = constant) cannot be constant in an isothermal process if the volume changes. We know this from the ideal gas law, PV = nRT. The piston should stop moving once the pressure of the gas and the external pressure are equal, Pin=Pout ⇒Fin=Fout. The internal pressure has to drop to meet the external pressure. That's why isotherms for expasions are decreasing graphs in PV-diagrams.

    In an isobaric process, both the external pressure (due weight plus atmospheric pressure) and the pressure of the gas are equal from start to finish, so there is no net force on the piston. It means that the piston can either stay still or move at constant velocity (Newton I).

    Could the key word to this whole thing be quasistatic? That is to say, if the process is not done slowly enough, the process will be isobaric. If the gas is allowed to go through equilibrium states, the process will be isothermal.
  11. Nov 30, 2016 #10
    This is the very same thing we have been telling you about the video. The text did not say constant pressure, it did not mean constant pressure, and I assure you it cannot be constant pressure. You are imposing all this reasoning about gravity, the weight of the piston, or atmospheric pressure because of the picture of a piston. All of that is coming from you, not from the video you cite or the text you quote. They do not describe the piston, they do not talk about weight or atmospheric pressure or the spring or the magnet or the rocket engine that may well be holding the thing in. They do not talk about the relationship between pressure and position of the piston in any way. The issue is not some magic because it is quasi-static. The issue is your reading of constant pressure into examples that make no such claim, which is weird because your whole reason for starting the thread is that you yourself thought that must be impossible!

    Your instinct in starting the thread was correct. Of this you can be sure: for an ideal gas if the temperature does not change, and the pressure does not change, and the number of atoms does not change then the volume does not change and no work is done.


    Always. Every time. Without exception because when we postulate this fictional stuff called "ideal gas" that is what we mean. It is assured by premise.
  12. Dec 1, 2016 #11
    But there is constant external pressure due to at least the weight of the piston and atmospheric pressure, though? That is, unless we're doing the experiment in space or something.

    This pressure is external, not necessarily the same as the pressure of the gas when it is heated.

    I never said that constant pressure would be impossible. Where do you get this?

    I didn't have any instinct in the beginning.

    I said there are two arguments about the nature of the process and I was wondering what are the circumstances needed for the process to be isobaric (ΔP=0) and what are the circumstances needed for the process to be isothermal (ΔT=0). I didn't claim that the process must be either isobaric or isothermal. I knew from the beginning that they are both possible.

    You are correct.

    Here is a graph of an isothermal process (in red). You can see that the pressure and volume both do change. This makes it possible for the temperature to stay constant. Like I told you before, it's a descending graph.

    Source: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/imgheat/isoth.gif isotermi.gif

    For monoatomic ideal gas, the formula of internal energy is ΔU=3/2*nRΔT. If the temperature is constant, i.e. ΔT=0, it means that ΔU=0.

    From this it follows that ΔU = Q-W ⇔ 0 = Q-W ⇔ Q= W. So the heat brought into the gas must equal the work done by the gas. No energy goes into heating the gas. All Q goes into doing work and the temperature of the gas stays constant.

    Now, in the isobaric process, the temperature of the gas does change. In terms of internal energy of the gas, this would mean that the heat brought into the gas must be more than work done by the gas: ΔU = Q-W ≠ 0 ⇒ Q ≠ W, meaning in this case Q > W.

    This was the core of the problem, and I think I've gotten my answer to my initial question. It's all quite clear to me now, so I hope I have been able to make sense to you as well.
    Last edited: Dec 1, 2016
  13. Dec 1, 2016 #12
    Just to avoid further confusion: of course they can't be possible simultaneously. So, if the process is isothermal, it can't be isobaric - and vice versa.

    However, I didn't initially claim that only one of the processes is possible. They are both possible depeding on the heat that flows into the gas and the energy done by the gas.
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