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Internal energy in expansions

  1. Nov 21, 2013 #1
    I've got a few issues with the topic of thermodynamic expansion, ΔU, etc.

    With ΔU=q+w, this w is to mean work done on the system (positive contribution to ΔU) minus work done by the system (i.e. negative contribution to internal energy), so then how do we summarize this for cases of expansion? Because if we say that ΔU=q-Δ(pV) then this suggests that ΔH=q even without constant pressure because ΔH=ΔU+Δ(pV). But if we instead say that ΔU=q-pΔV which gives the right relationships ( ΔU=q at constant volume, ΔH=q at constant pressure), doesn't this assume constant pressure - how can we instead evaluate ΔU in terms of q, Δp and V for an isochoric process with changing pressure?
     
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  3. Nov 21, 2013 #2

    Andrew Mason

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    Since thermodynamics in general, and the first the law in particular, deals only with changes between states of thermodynamic equilibrium, we cannot apply the first law during the process unless the process is reversible.

    For example, you cannot say that δW = -PdV (P being the internal pressure of the gas) unless it is a reversible process. If the external pressure is 0, there is no work done during the expansion yet ∫PgasdV > 0.

    If there is no change in volume, there is no work done by or on the gas. So Q = ΔU from the first law.

    In terms of enthalpy: H = U + PV, or dH = dU + VdP + PdV. Since volume is constant:

    ΔH = ∫dH = ΔU + VΔP

    So, ΔU = ΔH - VΔP = Q

    AM
     
  4. Nov 21, 2013 #3
    Ok so you're confirming that 1) work done by the system is decreasing the internal energy of the system and work done on the system is increasing the internal energy of the system, hence the sign (-pΔV) on W, and 2) ΔU=q-pΔV is indeed correct.

    My questions still remain kind of fuzzily ... why is ΔU=q-Δ(pV) incorrect (which it must be, else ΔH=q under all conditions), as I don't see why a change in pressure should necessitate irreversibility? After all, changes in pressure occur necessarily in a container of fixed volume if a gaseous phase process/reaction occurs there, but thermodynamics readily treats these. But then what do we do, if pressure is changing but volume is not - how do we express w and ΔU? If I understood why ΔU=q-Δ(pV) was incorrect and why it must be ΔU=q-pΔV then it would be clear why, if pressure is changing but volume is not, ΔV=0 means that ΔU=q and the rest follows from enthalpy being defined as ΔH=ΔU+Δ(pV).
     
  5. Nov 21, 2013 #4
    Hi again Big Daddy. The correct (more precise) form of the first law is:
    [tex]ΔU=q-w[/tex] where
    [tex]w=\int{p_{surr}dV}[/tex]
    These equations apply both to reversible and irreversible paths. However, they assume that the initial and final states of the system are equilibrium states. So, if we are talking about ΔH, we have:
    [tex]ΔH=q-\int{p_{surr}dV}+Δ(pV)[/tex]
    where the final term refers to the system pressures and volumes in the initial to the final equilibrium states.

    If the path is reversible, then p=psurr, and the equations become:
    [tex]ΔU=q-\int{pdV}[/tex]
    [tex]ΔH=q-\int{pdV}+Δ(pV)=q+\int{Vdp}[/tex]
    Also, of course, for a reversible path,
    [tex]q=\int{TdS}[/tex]

    Chet
     
  6. Nov 21, 2013 #5

    Andrew Mason

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    Just to add to what Chet has said, the expansion of a gas only does work on the surroundings if it is able to apply a force through a distance to its surroundings. This requires an external pressure applied to the gas. If a gas freely expands against no external pressure, it does no work on its surroundings. So it is the external pressure that determines the work done in an expansion.

    So, δW = PdV (where P is the pressure of the gas and W is the work done by the gas on the surroundings) only if the gas pressure is the same as the pressure applied to the gas by the surroundings. That is the case only in a quasi-static expansion.

    AM
     
  7. Nov 21, 2013 #6
    To add to what Andrew said, in the case of irreversible (non quasi-static) expansion, the pressure of the gas within the system is typically non-uniform, so, at any time during the process, there is no single value of pressure that can be assigned to the gas. However, at the interface with the surroundings, the local pressure across the interface is continuous, and the system pressure matches the surroundings pressure at the interface. This is the pressure that is used to calculate the amount of work that the gas is doing on the surroundings. In making this calculation, it is assumed that we have enough information (such as by using pressure transducers) or by some other method (e.g., lifting a weighted piston) to know the surroundings pressure at the interface.

    Chet
     
  8. Nov 21, 2013 #7
    Thanks very much, these are extremely helpful posts.

    I have a couple more related issues. Firstly, I have seen the ideal gas approximation applied to say that ∫Vdp=ΔngasRT, under the condition of constant volume (so ∫Vdp=VΔp). Is this correct (to the ideal gas approximation), and if so, why is it specifically that ∫Vdp=ΔngasRT rather than ∫pdV or Δ(pV)?
     
  9. Nov 21, 2013 #8
    It's very hard to comment confidently on this without knowing more about the specific context. Can you give a specific example, please?
    Chet
     
  10. Nov 22, 2013 #9
    Ok, so for one example, I can give a problem from my chemistry textbook:

    "Gaseous NH3 is burned with O2 in a container of fixed volume according to the equation given below:

    4 NH3 (g) + 3 O2 (g) -> 2 N2 (g) + 6 H2O

    The initial and final states are at 298K. After combustion with 14.40 g of oxygen, some of the NH3 remains unreacted. Calculate the heat given out during the process."

    Now, it's not that I can't solve this problem. All you need is that ΔU=q=ΔH-Δngas*R*T where Δngas is the change in number of moles of gas per mole of reaction; then you would multiply this q by the number of moles of reaction that occur, to find the total heat taken in (and -q(total) is the heat given out).

    The question is how to reach this ΔU=q=ΔH-Δngas*R*T. As you have demonstrated above, it is certainly fair to say that at constant volume ΔU=q=ΔH-VΔp. But why is VΔp=Δngas*R*T, as required in the substitution for the solution ... whereas presumably Δ(pV)≠Δngas*R*T and pΔV≠Δngas*R*T (because if they were that would mean Δ(pV)=pΔV=VΔp etc., which are not merely wrong but in fact inconsistent with each other).
     
  11. Nov 22, 2013 #10
    Hi Big Daddy. Thanks for providing this specific problem. This is very helpful. I think I understand what the problem is about, and how to answer your questions. However, please bear with me. I want to take a little more time to think about it before I give you my answer.

    Chet
     
  12. Nov 23, 2013 #11
    The question is really this: If, under ideal gas conditions, you know the heat of reaction (which is the amount of heat that must be removed at constant temperature and pressure when the reactants are converted into the products), how do you get the amount of heat that must be removed if the reaction is carried out at constant volume? This is a bit of a tricky question, and it is understandable that you struggled with it.

    If there is no change in the number of moles as a result of the reaction, there's no problem because, if the reaction is carried out at constant pressure, the volume doesn't change. So under these circumstances, the amount of heat that must be removed at constant volume must equal the amount of heat that must be removed at constant pressure.

    Now, let's consider the more general case in which there is a change in the number of moles. To get the amount of heat that is removed at constant volume, you need to determine the change in the internal energy of the constant volume system (since no work is done). But, before we do that, let's first consider the change in the internal energy if the reaction is carried out at constant pressure (and, of course, constant temperature). Under these circumstances, we have:
    [tex]ΔU=ΔH-Δ(PV)=ΔH-Δn(RT)[/tex]
    So this is the change in internal energy if the reaction takes place at constant pressure and temperature. But, we need to know the change in internal energy if the reaction takes place at constant volume and temperature.

    Now, another thing we know is that the internal energy of an ideal gas is independent of pressure. Therefore, the internal energy of the reaction products at the final pressure in the constant volume container (which differs from the initial pressure, because the number of moles has changed) is the same as they would be if the final gas were at the initial pressure. Therefore, ΔU for carrying out the reaction at constant volume is the same as ΔU for carrying out the reaction at constant pressure. The same goes for ΔH. These are general results for ideal gas reactions.
     
  13. Nov 24, 2013 #12
    So because Δ(pV)=Δngas*R*T and as before Δ(pV)=∫pdV+∫Vdp the results we need drop out. Δ(pV)=Δngas*R*T itself is a simple statement of the ideal gas equation, Δ(pV)=p2V2-p1V1=n2RT-n1RT=(Δn)RT.

    Thanks very much for all the help.
     
  14. Dec 31, 2013 #13
    Hi, I was revisiting this and found something that seems to complicate matters for me. The original form of the first law, ΔU=q+w, should incorporate all forms of work done on (+ve) and by (-ve) the system, but for a reaction this should include chemical work as well (this maximum value of this chemical work being ΔG) - yet in the derivations above, for equations often applied in my experience to chemical reactions, we've included only pV-type work in ΔU. If we add in chemical work to ΔU, I can't see an obvious way of deriving the equations we came to above. How do we account for this?
     
  15. Dec 31, 2013 #14
    Please provide a precise definition of the term "chemical work."
     
  16. Jan 2, 2014 #15
    We could define chemical work as the non-PV work that can be done by a system wherein a reaction is going towards chemical equilibrium, such that the maximum value of this chemical work is ΔG.

    I'm not sure how to specify it, mostly because I'm unclear on the concept. Any work involved, other than expansion-type work, in the progress of a chemical reaction (which is what I've seen is equal to ΔG) is still unclear to me how it fits in with ΔU, the derivations above, ΔH=q at constant pressure etc.
     
  17. Jan 3, 2014 #16
    I think you're thinking of the equation:

    ΔG≤-wE

    where wE represents the work done by the system on the surroundings over and above p-V work. This equation applies to a closed system that has experienced a process in which the system temperature is equal to a fixed (constant) boundary temperature at the beginning and end of the process, the system pressure is equal to a fixed (constant) boundary pressure at the beginning and end of the process, and, of course, the system is at equilibrium in the initial and final states. But what kind of work could wE represent? An example cited by Denbigh (Principles of Chemical Equilibrium) is electrical work, if the system in question were a galvanic cell. As might be expected, in many practical situations, wE would be zero. In such cases, ΔG≤0.

    So the question really is, "how do you analyze systems in which chemical reactions are occurring, and does this change any of the previous results?" The answer is that it doesn't change any of the previous results at all, but, to analyze such systems, you need to be familiar with how to determine the internal energy U or the Gibbs free energy G when you have a mixture (solution) of chemical species present. To simplify things, let's focus on the specific example of an ideal gas mixture of chemical species. Do you know how to determine the internal energy of an ideal gas mixture of chemical species from knowledge of the internal energy of each of the pure chemical species at the same temperature and pressure of the mixture? Do you know how to determine the Gibbs free energy of an ideal gas mixture of chemical species from knowledge of the Gibbs free energy of each of the pure chemical species at the same temperature and pressure? Same question for the enthalpy of a mixture of ideal gas species.

    Let's focus on a specific case. In the initial state of the closed system, you have barriers separating the pure chemical species that can react with each other. What is the Gibbs free energy of this system? At time zero, you remove the barriers, and the chemical species then react until they reach chemical equilibrium (while in contact with a constant temperature reservoir at the initial temperature and a constant pressure surroundings held at the initial pressure). What is the final Gibbs free energy of the system in the new equilibrium state? What is the change in free energy?

    Chet
     
  18. Jan 5, 2014 #17
    Ah. Perhaps before looking at work, then, I have to look at Gibbs' free energy more thoroughly. I always thought ΔG pertained to a chemical reaction or individual process rather than to the system as a whole (in the sense that, if reversible reaction A => B is in progress and ΔG<0, A -> B would be the overall forward direction whereas if ΔG>0 then B -> A would dominate, until the rates come to be equal, at equilibrium, so ΔG=0).

    You did latch on to the definition of work I was thinking of. But as the chemical reactants mix in your thought experiment and head towards equilibrium, isn't ΔG inevitably heading, from -infinity (since there are no products at the beginning, only reactants), through all negative values, to 0 at equilibrium? So, according to your equation, the (maximum) work done would be positive, but its value somehow decreasing as the reaction goes on ... which doesn't seem to make sense?
     
  19. Jan 5, 2014 #18
    No. If the reaction is heading in the forward direction, G is decreasing. If the reaction is heading in the reverse direction, G is decreasing also. At equilibrium, no matter which direction it is approached from, G is at a minimum. Pick any combination of concentrations for the reactants and products, and calculate the Gibbs free energy (for fixed temperature and pressure). The value you get will always be greater than if the set of concentrations were equilibrium values.
    No. The free energy of a mixture is the sum of the number of moles of each species times the chemical potential of that species. If there are no products present, their number of moles in the mixture is zero, and they make no contribution to the free energy of the mixture. Sure, the chemical potential has a term in it which is proportional to the natural log of the mole fraction, but the zero number of moles trumps the natural log term in the limit.

    Chet
     
  20. Jan 8, 2014 #19
    Now I understand a bit more. I can see why abs(ΔG)=wE,rev=wE,max where w is the non-expansion work (that will be done by the system as ΔG comes to 0, as G falls to minimum) and abs(ΔG) is the magnitude of ΔG. I can't see a situation which would require work to be done on the system, except to do work to bring abs(ΔG) to a larger value than it is (whereas if it is falling in value as the system goes to equilibrium, with either forward or reverse direction of the reaction, you then get work out of the system).

    A couple of linked questions? 1: I saw ΔG defined as the derivative of G with respect to extent or progress of reaction. I'm guessing ΔH and ΔS also have the same meaning (derivative of H and S respectively with respect to extent). Do ΔH and ΔS reach 0, like ΔG, at equilibrium, or do they reach some other arbitrary value that cancels out so that ΔG=ΔH-TΔS=0 at equilibrium? Given an isothermal system, it's impossible for them to be constant... but then what does the first law of thermodynamics mean, as ΔU (a derivative that changes value as reaction occurs) is related to q and w (two values that represent changes over the whole reaction) - is it that, like wE with ΔG, the values q and w refer to heat gain or work done on the system between this stage in extent (this value of ΔU=∂U/∂ε) and when equilibrium is reached?

    2: What is ΔG°? I've got the idea that it is some "limiting form" of ΔG, which applies to equilibrium rather than at any point - possibly ΔG° is ΔG when the extent of the reaction is 0. Does ΔG° have any relationship to maximum non-expansion work - I saw them equated? Similarly I see ΔH°=q at constant pressure (or ΔU°=q at constant volume) all the time; I never thought about it before but is this the limiting case of completion, where ΔH=q represents heat change as you get nearer to equilibrium, whereas ΔH°=q represents heat change for one mole of reaction going to completion?

    Then how is the ΔG (not just G) for the whole system defined with respect to the different reactions occurring, i.e. in terms of ΔG as well-defined for each reaction in the mixture, if in a mixture? Or is ΔG only defined for a reaction (as derivative wrt extent of that reaction) whereas G is defined with respect to the whole mixture - so then, would we sum over all ΔG from each reaction, to find maximum non-expansion work that can be obtained from the mixture under its current composition?
     
    Last edited: Jan 8, 2014
  21. Jan 8, 2014 #20
    Unfortunately, I don't understand what you are saying here, so i can't help with the answer to this question, but I can answer your other two questions.
    ΔG° is the change in free energy at a fixed reference temperature and pressure (usually 1 atm) in going from stoichiometric amounts of separate (pure) reactants to corresponding stoichiometric amounts of separate (pure) products.

    ΔH° is the amount of heat that has to be added to at a fixed reference temperature and pressure (usually 1 atm) in going from stoichiometric amounts of separate (pure) reactants to corresponding stoichiometric amounts of separate (pure) products.

    The way you can solve a problem involving multiple reactions, multiple reactants, and multiple products for the final equilibrium state at a fixed temperature and pressure is to find the set of species concentrations that minimizes the free energy of the whole final mixture [itex]G = \Sigma μ_in_i[/itex]. G being minimized is always the criterion for equilibrium of a reaction mixture at fixed temperature and pressure.
     
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