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Difference between Internal Energy and Enthalpy

  1. Oct 14, 2009 #1
    [tex]\Delta[/tex] U = Q - W = Q - PV (where PV is the work done BY the system).

    [tex]\Delta[/tex] H = [tex]\Delta[/tex] U + PV (where PV is work done BY the system).

    1) Above I defined PV in both equations as work done BY the system.
    I think this must be correct to show that when P is constant we can have [tex]\Delta[/tex] H = Q

    <proof>: H = [tex]\Delta[/tex] U + PV = Q - PV + PV = Q

    Surely both PV in the term above have to be defined equivalently as work done BY the system (in my case, as formulas written above) for them to cancel out?
    Can anyone confirm this?

    2) I really do not understand WHY we need to enthalpy when we already have internal energy? Does the W in internal energy not already take into account any work done due to change in P or V? What is the point of inventing enthalpy when it is just internal energy added by a further PV?

    <example>: Say heat (Q) is added to a machine, and the piston expands (does work).
    So change in internal energy = Q - PV (heat added - work done by system)
    while change in enthalpy = Q + PV (heat added - work done by system + work done by system)

    So whats the point?
     
  2. jcsd
  3. Oct 14, 2009 #2

    Mapes

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    (1) You mean [itex]P\Delta V[/itex], not [itex]PV[/itex]. Yes, it's the same in each equation. But be careful not to mix these up, as you did in the second question also.

    (2) The enthalpy is a useful function for processes that occur at constant pressure, rather than constant volume. The enthalpy is the internal energy plus the energy required to move our atmosphere out of the way to accommodate the system.
     
  4. Oct 14, 2009 #3
    (1) I don't understand what u mean by "not to mix these up"? Why can't I add them and so on when they are both [itex]P\Delta V[/itex]? (as u corrected)

    (2) I still don't understand... isn't "the energy required to move our atmosphere out of the way to accommodate the system", work done by the system? And it is already taken into account in the formula for internal energy?

    PS: Can u give an actual example that demonstrate the difference between internal energy and enthalpy, and likewise clearly shows that the term enthalpy is indeed helpful to the case? Thanks~
     
  5. Oct 14, 2009 #4

    Mapes

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    The difference between a system's energy and enthalpy is [itex]PV[/itex]. The work done by a system at constant pressure is [itex]P\Delta V[/itex]. They are not the same!

    The internal energy function U (or E) doesn't incorporate the energy needed for a system to expand (for example) in a constant-pressure environment. The enthalpy function H does.

    Consider the difference in heat capacity of a gas in constant-pressure vs. constant-volume conditions, or search for "isenthalpic processes."
     
  6. Oct 14, 2009 #5
    First it is important to say that it is incorrect to put these equations in the same context.
    The actual equations are the differential equations
    [tex]\mathrm{d}U=\mathrm{d}Q-p\mathrm{d}V[/tex]
    [tex]\mathrm{d}H=\mathrm{d}Q+V\mathrm{d}p[/tex]
    and also generally valid is
    [tex]H=U+pV[/tex] and therefore [tex]\Delta H=\Delta U+\Delta (pV)[/tex]
    Everything else is a special case derivation.

    If the change in the system is done under constant pressure, then you get [itex]\Delta_p U=Q-p\Delta V[/itex], [itex]\Delta_p H=Q[/itex] and [itex]\Delta_p H=\Delta U+p\Delta V[/itex].

    If the change in the system is done under constant volume, then you get [itex]\Delta_V U=Q[/itex], [itex]\Delta_V H=Q+V\Delta p[/itex] and [itex]\Delta_V H=\Delta U+V\Delta p[/itex].

    It seemsin both cases
    [tex]\Delta U=Q-p\Delta V[/tex]
    [tex]\Delta H=Q+V\Delta p=\Delta U+\Delta(pV)[/tex]
    But in general [itex]\Delta_p U[/itex] and [itex]\Delta_V U[/itex] are different things. Depending on which case you physically have, you can only use one or the other. Actually [itex]\Delta U=Q-\Delta(pV)[/itex] is incorrect and on the other hand mathematically [itex]p\Delta V[/itex] and [itex]V\Delta p[/itex] are different things, so your cancellation doesn't work.

    These last two equations are your working tools now.

    It is very important to not forget all the [itex]\Delta[/itex] otherwise you will make mistakes occationally. But indeed your conclusion is correct here.

    It is incorrect to say it like this. Work in general is
    [tex]W=-\int p\mathrm{d}V[/tex]
    In the case of constant pressure this reduces to
    [tex]W_p=-p\Delta V=-\Delta(pV)[/tex]
    or for constant volume in fact
    [tex]W_V=0\neq -\Delta(pV)[/tex]

    The important point here is, that we are concerned about the free energy, i.e. which energy we will extract by the process. Hmm, have to think myself what that means :) Anyway

    One can prove that to find the final state of the system one needs to minimize a certain thermodynamical energy. The final state is what is reached if the system is left on its own (with its surrounding environment).

    For a process at constant volume one has to minimize the internal energy given all other contraints.
    [tex]V=\text{const}\qquad\to\qquad U\to\text{min}[/tex]

    However, for a process at constant pressure one has to minimize the enthalpy
    [tex]p=\text{const}\qquad\to\qquad H\to\text{min}[/tex]
    So it's not always the internal energy that reaches a minimum!

    The point is that to keep your contraints (constant pressure) the system will move around energy between itself and the environment. This energy will be inaccessible for you for "extraction".

    It surely does, but note all the details mentioned above.

    To consider the free extractable energy the change in one or the other quantity is important. Depending on the case.

    In fact all the cases (constant volume or pressure) you have considered so far are processes that are fast enough (adiabatic) so that entropy is constant! So it is more exact to write

    For slow processes however the temperature would be constant. In that case you would need even new potentials
    [tex]p=\text{const}, T=\text{const}\qquad\to\qquad G\to\text{min}[/tex] where Gibbs energy [itex]G=U-TS+pV[/itex]
    [tex]V=\text{const}, T=\text{const}\qquad\to\qquad F\to\text{min}[/tex] where Helmholtz energy [itex]F=U-TS[/itex]

    With the correct notation you can also deduce
    [tex]Q_{Tp}=\Delta H[/tex], [tex]W_{Tp}=\Delta F[/tex]
    [tex]Q_{TV}=\Delta U[/tex], [tex]W_{TV}=0[/tex]
    [tex]Q_{Sp}=0[/tex], [tex]W_{Sp}=\Delta U[/tex]
    [tex]Q_{SV}=0[/tex], [tex]W_{SV}=0[/tex]
    where the subscript denotes what is held constant respectively.
    That is another reason to introduce H and F. When you have processes at constant pressure and temperature, then the heat exchange is given by the change in enthalpy.

    [tex]\Delta U=Q-p\Delta V=Q+W[/tex]
    [tex]\Delta H=Q+V\Delta p=Q+W+\Delta(pV)[/tex]
    Note the difference.
     
    Last edited: Oct 15, 2009
  7. Oct 15, 2009 #6

    Mapes

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    Very nice, Gerenuk!
     
  8. Apr 7, 2010 #7
    Hi, could you pls explain whats difference b/w PdV and VdP? Are both work? if so how are they different.

    Also

    mc(p)T = mc(v)T+PV

    This is the equation my lecturer told for enthalpy(mc(p)T).

    So how is c(p) and c(v) related?

    i just dnt get the equation
     
  9. Apr 7, 2010 #8

    Mapes

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    [itex]P\,dV[/itex] is work (specifically, the differential form, an infinitesimal amount of work). [itex]V\,dP[/itex] is not work; as far as I know, it doesn't have a common name.

    [itex]C_P[/itex] is the amount of energy needed to increase the system's temperature one degree at constant pressure; [itex]C_V[/itex] is the amount needed at constant volume. [itex]C_P>C_V[/itex] because the material can change size at constant pressure (it can't at constant temperature), but the values are similar for condensed matter, which doesn't expand or contract much with temperature compared to how much gases expand.

    Does this help answer your question?
     
  10. Apr 7, 2010 #9
    hi Mapes, I guess i understood the specific heat thing. So specifiv heat is inversely proportional to density, so with constant pressure when volume increases, density decreases, so specific heat increase. ie Cp increase with increase in volume- is that the concept?
     
  11. Apr 7, 2010 #10

    Mapes

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    No, this is not right. Specific heat is not necessarily proportional or inversely proportional to volume. It is the amount of energy needed the heat the system by one degree at some constraint (e.g., constant volume or constant pressure).
     
  12. Apr 7, 2010 #11
    I do understand the definition of specific heat, but I still dont understand y Cp is greater than Cv.
     
  13. Apr 7, 2010 #12

    Mapes

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    It's easiest to explain for the case where a material expands when heated (which is the case for most materials). At constant pressure, the material is free to expand with increasing temperature, which means it does work against the pressure of the atmosphere. This energy is lost, so we need to input more energy to get the material to a certain temperature.

    In contrast, a material heated at constant volume is not free to expand, so it does no work and less energy is needed to reach a certain temperature. Thus, [itex]C_V<C_P[/itex]. Does this make sense?
     
  14. Apr 7, 2010 #13
    Oh yes, thats a really nice explanation, thank u mapes :-)
     
  15. Apr 8, 2010 #14

    Andrew Mason

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    Change in enthalpy and heat flow are similar. Both represent heat flow from or to the system. Enthalpy is a state function (ie it is path independent), whereas [itex]\Delta Q[/itex] is the heat (energy) that actually flows into or out of the system in a thermodynamic process. The change in enthalpy between two thermodynamic states represents the heat that flows out of or into the system in moving between those two states on a particular path (quasi-static at constant pressure).

    In a thermodynamic system, energy exists in the form of heat flow (into or out of the system), the internal energy of the system, and the work done by (or on) the system. The change in internal energy + the work done by the system has to equal the heat flow into/out of the system (first law).

    The change in enthalpy is the heat that flows when the transition between the two states is by way of a particular path (constant pressure). The change in internal energy + work done by the system is equal to the actual heat flow in moving from one state to the other on the actual path taken. The difference in enthalpy between two states tells you how much heat content can be released in moving from one state to the other.

    A change in internal energy does not require heat flow (eg. adiabatic expansion or compression). A change in enthalpy signifies a flow of heat.

    AM
     
  16. Nov 28, 2010 #15
    the attachment of this thread could be usefull:
    https://www.physicsforums.com/showthr...errerid=219693 [Broken].
    The conclusions of 'enthalpy changes at decompression' are:
    a. It would be useful for sciences that employ energy balances, to recognize 'pressure energy' (Epv=penV) as a separate form of energy, where pen is the external pressure of the system with volume V.
    b. It would be advantageous to define enthalpy as the sum of internal energy and pressure energy, and to list enthalpy as a form of energy. This allows directly for the formulation of the First Law as ΔH=Q-Ws i.e. as an energy balance equation.
    c. If there is a need to define a quantity (U+pinV), we propose to call this quantity 'inthalpy'. For the energy-balance of the throttle process this quantity is not needed.
    d. If the Joule-Thompson experiment is carried out with an ideal gas and the gravitational energy of a pistons providing the shaft-work needed, then the gravitational energy is completely converted into pressure-energy, while the internal energy is constant. Hence the passage across the porous plug is not isenthalpic as is commonly assumed, when the work is supplied externally. If this energy is supplied by a large source of pressure, the process is isenthalpic, even if the gas is not an ideal gas.
    e. Since heat Q is not a variable of state, hence not a form of energy, the general mantra that 'heat is a form of energy' should be replaced by 'enthalpy is a form of energy' in line with the replacement of the concept 'total heat' by 'enthalpy' in 1922. Romer pronounced: "Heat is not a noun", but it can be expressed more convincingly: "Heat is not a noun, enthalpy is".
     
    Last edited by a moderator: May 5, 2017
  17. Feb 26, 2013 #16
    I read all the things you expain. However, shortly to say? How many points are different between internal energy and enthalpy? What are they? Please, help me. Those thing make me confuse
     
  18. Feb 26, 2013 #17
    The difference between U and H is pV. This pV term stands for the energy of displacement, equal to the work needed to displace the atmosphere from the space occupied by the system: p is thus the atmospheric pressure. For all systems where internal and atmospheric pressure are equal, one can substitute one for the other if you want.
     
  19. Feb 26, 2013 #18
    Another powerful application of enthaply is in steady state flow systems. If you separate out the work needed to push the material into and out of the system, you can express the total work per unit mass flow as:

    [tex]W=W_s + \Delta (PV)[/tex]

    where W_s is the so called Shaft Work.

    If you substitute this into the first law, you get:

    [tex]\Delta U = Q - W_s-\Delta (PV)[/tex]
    or
    [tex]\Delta H = Q - W_s[/tex]

    This relationship is very useful in practical applications.
     
  20. Feb 26, 2013 #19
    I agree with Chestermiller, except that it would be better to adhere to the same sign convention for all work (and heat) terms: work in >0, work out <0.
    Then the relationship becomes DH=Q+Ws.
     
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