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Difference between enthalpy and internal energy.

  1. Sep 20, 2011 #1

    U is the total kinetic and potential energy of the system.


    Enthalpy is the internal energy of the system, plus how much energy it takes to set up the system.

    I don't understand how the PV term from enthalpy is not included in the internal energy. It seems like it would be a part of the potential energy.

    For example, at constant pressure I heat a gas and it expands. q=H in this case. To calculate U, you have to add the work that it took to change the volume to q. Why? Why is the work used for the w in U=q+w but not in the PV from H=U+PV?

    It looks like w is used in the U calculation because it would be part of the new potential energy of the gas. But that work was done to set up the system, so why isn't it used in calculating enthalpy? Do you see what my confusion is arising from here?
  2. jcsd
  3. Sep 20, 2011 #2
    I'm not sure why you're equating the PV term with the amount of energy required to set up the system. PV doesn't represent any work by itself; the work is path dependent so must be related to a differential: -PdV. Internal energy is a state function: the state is characterized by an internal energy which includes all forms of energy; it doesn't matter how the system got to its current state. The internal energy isn't "missing" an additional energy term.

    No form of energy has an absolute reference point, so mainly we are interested in changes in the thermodynamic potentials. Adding PV to the internal energy is a mathematical technique called Legendre transformation which changes the independent variable from V to P. The differential that you get out seems more intuitively meaningful than the raw expression for H. dU=TdS-PdV, dH=TdS+VdP.

    The primary value of these potentials is that they predict stability of the system, in the conditions where their independent variables are the experimentally controlled variables (eg, set by an external reservoir). You can then notice other physical interpretations from the differentials, like the fact that the enthalpy is the heat transfer at constant pressure.

    I'm not sure how much that answers your question, but it seems like your confusion comes from attaching some intuition to the potentials that they don't necessarily have.
  4. Sep 20, 2011 #3
    That sort of helps but I'm still a little confused. Going back to my example:

    A gas is heated at constant pressure. So we have a constant pressure and a change in volume.

    If I want to calculate the internal energy I have to add the heat+work. So dU=q+w=dq+PdV

    PdV is also contained in the equation for H. dH=dU+PdV

    So to me, because there is only one value for PdV, it looks like when you calculate dH that:


    Which is probably not the right way to think about it. What is the difference between the work in the U equation and the work (because PdV=work) in the enthalpy equation?
  5. Sep 20, 2011 #4

    Andrew Mason

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    You are correct that in a constant pressure process, the change in enthalpy is equal to the heatflow:

    [tex]\Delta H = \Delta U +\Delta (PV) = \Delta U + \int PdV + \int VdP = \Delta Q + \int VdP = \Delta Q + 0[/tex]

    W is used in the calculation of [itex]\Delta U[/itex] from the heat flow because of the first law: dU = dQ - dW = dQ - PdV where W is the work done by the system.

    dH = dU + d(PV) = dU + PdV + VdP. From this you can see that:

    [tex]\Delta Q = \Delta H - \int VdP [/tex]


    [tex]\Delta H = \Delta U + \Delta (PV) = \Delta U + nR\Delta T = \Delta U + n(C_p-C_v)\Delta T[/tex] (assuming constant heat capacities for all T).

    For an ideal gas, therefore, [itex]\Delta H = nC_p\Delta T[/itex]

    From this we can see that the change in enthalpy is equal to the change in internal energy plus the difference between the heatflow in the constant pressure process and the heatflow in the constant volume process for the system in moving between the initial and final temperatures.

  6. Sep 20, 2011 #5

    Andrew Mason

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    That is correct ONLY if dP = 0 (constant pressure). To differentiate, you have to use the product rule: d(PV) = PdV + VdP

    The difference is that to calculate dH from the heatflow dQ you have to add the VdP term to dQ: dH = dQ + VdP. See my post #4.

  7. Sep 20, 2011 #6
    Thanks I finally see my mistake. Pointing out the product rule REALLY helped me. For some reason I was thinking that d(PV) was the same as work from w= PdV.


    dH = dU + d(PV) = q - PdV + PdV + VdP = q (at constant pressure)

    dH= q - PdV + PdV + VdP= q + VdP (at constant volume)

    At constant pressure, the terms -PdV and PdV cancel out. So if I have this right, PdV represents an amount of work. But since the internal system (I don't know what to actually call it, but you get the point) is doing work on the surroundings, by the third law the surroundings will "do work" at an equal but negative value. And because the surroundings are included in the enthalpy term (in the sense that they are a part of the energy it takes to set up the system) it is an internal change that results in zero net work.

    At constant volume, no work is done so q=U. There is an energy difference that arises from setting up a system with a different pressure though, so you have to account for that in dH.

    The above is my mathematical and qualitative understanding of the subject. I'm not sure if my explanations make sense but I think it makes sense to me now.
  8. Sep 20, 2011 #7

    Andrew Mason

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    Not quite. They always cancel out (ie. not just in an isobaric process). The following is always true (not just at constant volume):

    dH = dU + PdV + VdP = (dQ - PdV) + PdV + VdP = dQ + VdP

    If the gas does positive work on the surroundings. you could say that the surroundings do the same magnitude of negative work on the gas so long as you understand that negative work is done by reducing the internal energy of the gas.

    I am having difficulty following you here.

    The system is a quantity of gas confined to a particular volume. Its internal energy is a combination of the total kinetic energy and the total potential energy of the molecules of the gas. In that sense, the internal energy represents the energy required to take the molecules from zero degrees Kelvin and 0 volume to the present temperature and volume.

    When looking at the internal energy, you can also look at the product of P and V. This product, which has dimensions of energy, is included in the internal energy term. For example, for an ideal gas enthalpy would be:

    [tex]H = U + PV = nC_vT + nRT[/tex]

    For an ideal monatomic gas, Cv = 3/2R so:

    [tex]U = PV + nRT/2[/tex] so

    [tex]H = (PV + nRT/2) + PV = 2PV + nRT/2 = 5PV/2[/tex]

    [Or, you could say that in bringing the ideal gas from 0 Kelvin and 0 volume to temperature T and volume V, the change in enthalpy would be:

    [tex]\Delta H = \Delta U + \Delta (PV) = \Delta U + nR\Delta T = nC_v\Delta T + nR\Delta T = n(C_v+R)\Delta T = nC_p\Delta T[/tex]]

    Enthalpy is just a mathematical quantity that happens to be useful. It does not really represent anything tangible or physical in itself. It happens to represent the heat flow in a constant pressure process. It can also be thought of in the terms set out in my previous post #4. But apart from that don't try to imagine what it is in terms of something physical. Look at it as a way of thermodynamic bookkeeping where, instead of keeping track of internal energy (which includes the PV energy) you want to keep track of U AND the PV energy (effectively counting the PV energy twice).

  9. Sep 20, 2011 #8
    So I can't assign actual physical significance to the cancellation like I was trying to do? (more explanation of this below where I link the wikipedia article if you don't want to respond to this twice)


    What I am trying to say is related to what the wikipedia article on this says:

    Chemists routinely use H as the energy of the system, but the pV term is not stored in the system, but rather in the surroundings, such as the atmosphere. When a system, for example, n moles of a gas of volume V at pressure P and temperature T, is created or brought to its present state from absolute zero, energy must be supplied equal to its internal energy U plus pV, where pV is the work done in pushing against the ambient (atmospheric) pressure. This additional energy is therefore stored in the surroundings and can be recovered when the system collapses back to its initial state.

    My thinking is that U= just the gas but H= the gas + the energy stored in the surroundings to get the gas in the state it is in.

    This thinking leads me to my point that with internal energy, I need to count expansion work. But because enthalpy includes the energy stored by the surroundings due to the system, doing expansion work against the surroundings will result in 0 change. The only type of work that you count in enthalpy is therefore the energy change that arises from changing the pressure at constant volume.

    What do you mean by counting the PV energy twice?
    Last edited: Sep 20, 2011
  10. Sep 21, 2011 #9

    Andrew Mason

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    Wikipedia is often unreliable on difficult and nuanced scientific matters. It is simply wrong here.

    PV is not the work done on the surroundings by n moles of an ideal gas in expanding quasi-statically from 0 volume, 0 Pressure, and temperature absolute 0 K to volume V, pressure P, and temperature T = PV/nR. The work done by a monatomic ideal gas in such an expansion is determined as follows:

    [tex]PV = \int d(PV) = \int PdV + \int VdP = W + \int VdP = nRT[/tex] so

    [tex]W = PV - \int VdP[/tex]

    That Wikipedia paragraph suggests that the work done by the gas is being stored in the surroundings and available to the system to return it to its original state. That is generally not the case. The work done by the expanding gas is generally lost as low grade heat - such as in a car engine pushing a piston to turn wheels whose energy is lost to road friction. That energy is somewhere in the universe, of course, but it is not being "stored" locally or somewhere in the atmosphere as potential useful work that is available to the system.

    Be careful about using Wikipedia as a source of reliable scientific information. (You can often determine the reliability of the Wikipedia article by the quality of the discussion on the article's discussion page). My suggestion: find good text books and study them.

  11. Sep 21, 2011 #10
    Here's how I suggest you look at this:

    First, you have implicitly chosen a particular sign convention for heat and work by writing the first law as dU = dQ + dW (I am rationalizing your notation somewhat to correspond to standard practice.) (Note the "d" in dQ and dW is usually written with a horizontal slash to indicate that dQ and dW are not exact differentials -- but ASCII doesn't have a slash-d so just keep that in mind.) (Note dU is an exact differential... but I digress.) Anyway, if dQ PLUS dW is the change in the internal energy, this means you have chosen a sign convention such that POSITIVE heat flow is "INTO" the system and POSITIVE work flow is also "INTO" the system. In engineering it is commonplace to flip the sign of the work term (writing dU = dQ - dW) because heat engines accept heat input and produce work output.

    But if you write dU = dQ + dW, then you have to be consistent and write dW = - P dV for pressure work. The minus sign is important. But as long as you use the right sign convention for everything, it will all work out.

    Second, heat and work are TRANSFERS. You only need to bookkeep them once for a given process because (almost always) all the terms in your analysis will refer to the system and not the surroundings. So don't try to cancel the work coming out of the system with the work going into the surroundings. It's the very same work.

    So, in the constant pressure process, in words we would say "heat is added to the gas, increasing its internal energy, but it is allowed to expand at constant pressure, and thereby a work transfer to the surroundings also takes place". In symbols we write

    dQ = dU - dW = dU + P dV

    Note how the minus sign works out. Now, since

    H = U + PV

    we have

    dH = dU + P dV + V dP

    For a constant-pressure process, dP = 0 so, merging with the result for dQ, we get

    dH = dU + P dV = dQ for an isobaric process.

    One last thing: you're obviously just starting out, but it is important to note that these equations apply only if the process is reversible. If the process generates entropy, then some of these equations become inequalities.
  12. Sep 27, 2011 #11
    LogicX: it can be observed that in the discussion sofar, no proper distinction is made between the internal pressure and the external pressure. The p in the def of enthalpy (U+pV) is the external pressure, and the lack of distinction causes your confusion. Elswhere in this forum the same issue came up:
    the attachment included in one of my comments there tries to get this straight: it follows that the first law can also be formulated as dH = q +w" where w" stands for the shaft-work = useful-work. This makes dH=q when w"=0, ie when the only work done is work against the external pressure.
  13. Sep 27, 2011 #12
    in previous message, the last 'external pressure' must be understood as ' atmospheric pressure' .
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