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Internal energy and enthalpy

  1. Nov 4, 2005 #1
    I figured there must be a relationship between internal energy and enthalpy after doing this physics question sheet which seemed to suggest this because there was a graph (pressure-volume) showing 2 routes to same finial condition....which reminded me of la chatelier's principle (think thats the right guy..where enthalpy change following one route is the same as following another). does anyone know a simple website where a physics a level student can understand how they related, if they are. (plus theres the tempreture factor that seems to be present in both cases)
     
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  3. Nov 4, 2005 #2

    GCT

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    H=U+PV

    dH=dU+VdP+pdV

    dH=q+work+VdP+pdV

    it really depends on the conditions, isothermal, isobaric, isochoric, you'll just need to work with the equations
     
  4. Nov 4, 2005 #3
    but i thought enthalpy was = mass*change in temperature*specific heat capacity*1/moles(scaling factor or something)?
    thats a bit like E = mc(change in)T per mole.
    is that the same energy as Q? Q = delta u + delta W?
     
  5. Nov 4, 2005 #4

    GCT

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    at constant pressure dH=CpdT, which is what you're probably referring to. For your case you can assume that [tex] \Delta H_p =q_p = mc_p \Delta T [/tex] where c is the specific heat capacity (assumed fairly constant at the range the temperature). If it were constant volume then you would have dE=CvdT and [tex] \Delta U= q_v =mc_v \Delta T [/tex]. I'm not quite sure what you're going for here, but if you have a specific question, I'll be glad to help you out.
     
  6. Nov 4, 2005 #5

    Q_Goest

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    Hi Alias, you ask a pretty good question, and unfortunately I'm in a verbose mood! LOL Sorry for the overly explicit description but this is one of those things that has always bothered me and I think I can do a better job of explaining than the typical prof at school. Hell, I have to use it all the time, so here ya go.

    There is a very direct relationship between internal energy and enthalpy, but I don't think most engineers really understand it very well themselves. The text books I've seen don't do a very good job of explaining the concept either, they all seem to resort to equations that make little sense and tend to hide the fundamental idea. It's actually fairly simple. You don't need to make it as difficult as the text books would have you believe.

    I think the best place to start is the simple concept of "conservation of energy". The first law of thermodynamics is all about conservation of energy. Energy can take many forms, and all those forms added together can only switch from one kind of energy to another. They don't simply disappear. Thermal energy can be converted to work, or visa versa. Heat and pressure can also represent energy. For example, we can readily visualize that if a cylinder contains a gas under high pressure, and it has a piston inside that moves as the gas expands, the pressure has a kind of potential energy similar to a spring. Pressure is a kind of 'mechanical energy' which exists in the form of a force that can push through a distance. So thermal energy can be converted to pressure energy and work energy etc…

    Let's look at thermal energy. Take a block of steel for example. Put heat into it and it gets hotter. This is an increase in its thermal energy. Similarly, one could put heat into a gas and that would increase ITS thermal energy. The funny thing about a gas though is that as you increase its thermal energy, it changes shape which is very much unlike a solid. That strange phenomenon creates tremendous confusion unfortunately. I suppose folks had a very hard time figuring out how to separate the concept of thermal energy from potential energy when heating a gas. It's absolutely frustrating. Pressure increases as you heat it at constant volume, or volume increases as pressure stays constant, so there is potential energy as well as thermal energy being stored in that volume. Somehow, you must separate the two concepts of thermal energy and potential energy.

    What we need to do is to separate thermal energy and pressure energy. In short, internal energy is only the thermal energy of matter while enthalpy is the sum of its internal energy and pressure energy. Enthalpy therefore is often writen to describe this as:
    H = U + PdV
    That "pressure energy" is the PdV term.

    The next issue is one of how we calculate the value of enthalpy and pressure energy. If we know the internal energy of some fluid, then all we need is the pressure energy to determine the enthalpy. I assume you're familiar with the ideal gas law. I also assume you realize the ideal gas law becomes much less accurate as pressure increases and/or temperature decreases. The ideal gas law if applicable, or if adjusted to compensate for REAL gas, can be used to determine density of a real gas. Simple enough. . . Assuming real gas density, one can calculate PdV energy very easily. It is simply the pressure times volume per unit mass. Another way of looking at this is it's the pressure divided by density or even the pressure times specific mass. The result is all the same. Add that to internal energy and you have enthalpy. It's that easy.

    Now one more point that might help is to explain the term I'm sure you've heard called "state". The state of a fluid is simply the condition it is in. If you have the pressure and temperature of a gas or liquid, and as long as it's not saturated (a mixture of both), the state of that fluid can be completely defined by its pressure and temperature. Its pressure and temperature tell you exactly what the density is. Similarly, those two characteristics tell you exactly what the internal energy, enthalpy and entropy is. In fact, you could reverse it and say if you knew what the density and pressure was, you could find the fluid's temperature, internal energy, enthalpy and entropy. If you knew the enthalpy and pressure, you'd also know the temperature, density, internal energy and entropy. Get it? The state of a fluid is simply a definition of what condition it is in and what properties it has. So the internal energy and enthalpy are both simply states of a fluid.

    Finally, it is worth mentioning that even if a fluid is saturated and is a mixture of both a liquid and a gas, it still has a state. We can define that state by defining the pressure and vapor fraction (also called quality) of the fluid. If we know the fluid is saturated and if we know its pressure, we know its temperature, but it can have any vapor fraction, so we still need to define that. Once defined though we similarly know the internal energy, enthalpy, entropy, and many other properties of the fluid.

    So the state of a fluid defines the various properties and one of those properties is its internal energy and one of those properties is its enthalpy. Internal energy and enthalpy are simply "state properties" or "state variables" of a fluid.
     
  7. Nov 5, 2005 #6
    thanks!.......I never thought of pressure that way before being someform of potential energy.
     
  8. Nov 5, 2005 #7
    Q_Goest: I don't see how a system can possess "pressure energy". My interpretation of enthalpy is that the PV term simply accounts for work done in expansion, and this energy is itself negated from the internal energy of the system. Does anyone else agree/disagree with this?
     
  9. Nov 5, 2005 #8

    Q_Goest

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    I'll admit I'm taking some liberties in calling the PV term "pressure energy" or "potential energy". Those are not terms widely used, but I can't think of any term that is used to define PdV. It is generally refered to as just a "term" and P and dV is pointed out. It rarely is given any real meaning. I feel however, the terms I've provided are very descriptive and aid in understanding that the PV term is energy that comes about from the fluid's pressure. The tricky part is that this term is not the energy needed to compress it, nor is it the energy obtained on expansion. Those are processes, and those processes result in considerable conversion of thermal to mechanical energy, so the internal energy also changes as those processes occur.

    I don't think "negated" is the right term. Try converted. Certainly an expanding gas converts thermal energy to mechanical energy. In the case of an expanding piston/cylinder arrangement, there is considerable conversion of thermal AND pressure energy to work energy. I think that's what makes it so difficult to visualize and understand - the conversion of thermal to mechanical energy is not simple and straight forward, there is considerable exchange between thermal energy and mechanical energy in just about any process that involves a gas expanding, and even liquids do this also but to a much smaller degree.

    I think what helps to understand the PdV term is how it is calculated.
    PdV = P V/mass = P / density

    Take a gas at pressure P and temperature T. Calculate the density. The PdV equals the pressure divided by the density. Add that to the internal energy, and you get enthalpy. So the PdV term is a calculation of energy that comes from the pressure and density of the fluid. Remember that these are simply properties of the state of the fluid and not representative of what will happen under any given process. To determine what happens in a given thermodynamic process, we need to apply the laws of thermodynamics - most commonly, the first law.
     
  10. Nov 6, 2005 #9
    sorry made a wrong deduction

    ill say it anyway...
    i thought maybe internal energy per mole = enthalpy?

    because from AS chemistry i know that enthalpy = mc(change in)T /moles
    and on another physics question sheet it asked for internal energy of a block and gave me values of specific heat capacity and mass and i already obtained change in temperature...so that suggests internal energy = mc(change in) T
    since enthalpy = mc(change in)T /moles then would it be right to say internal energy per mole = enthalpy?
    and i know internal energy is the sum of KEs so..sum of KEs per mole = enthalpy?

    but i think there may be a slight difference in calculating internal energy for a solid and gas ..so i think im wrong can anyone confirm?
     
    Last edited: Nov 6, 2005
  11. Nov 6, 2005 #10
    The word "negated" would really only apply to a system undergoing expansion. "Converted" i agree is more correct because during compression you are indeed doing work ON the system.. (rather than the system doing work during expansion)

    This "thermal AND pressure energy" thing is totally wrong IMO. I'm not sure how you propose this pressure energy is actually locked up in the system? To me it is clear that thermal energy is locked up in one or more degrees of freedom (depending upon the substance). Try looking at an ideal gas. It has pressure and it has internal thermal energy but it in no way has "pressure ENERGY". The only energy in this system is due to the kinetic energy of each molecule. When the gas expands and does work on a piston, this energy is coming from the internal energy only.

    Enthalpy is not some magical mysterious property imo, its just useful when looking at flow works and expansions of gases. The PdV term really doesn't have anything to do with current energy possessed by the system in the way in which you propose.

    Dan
     
  12. Nov 6, 2005 #11

    Q_Goest

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    Hi alias. You're right in that enthalpy (and internal energy) can be expressed in terms of energy per mole. Converting to energy per unit mass is simply dividing the energy per mole by mass per mole. You can use either term, it just depends on what's more useful for a given calculation. They are equivalent ways of looking at it.

    Suni, I have no doubt you understand the first law and how to use it to calculate the thermodynamic relationships of a process. I'd like to think that after doing it for almost 20 years, I know how to do it too. As you know, the first law is generally written with dU on one side and all other terms such as Q, W and H on the other. We could change that to Q, W, U and PdV. The term PdV has units of energy, or if divided by mass or moles, then specific energy meaning energy per unit mass (or mole as alias points out), would you agree? I've found doing analysis using specific energy is generally more usefull, though not always. So if the U term is the thermal energy term, how would you describe the PdV term?

    I'd agree with this statement. The question then is, when a gas expands (ex: through a valve or out of a pressurized container) do we have sufficient information about the process if we consider only the kinetic energy and other forms of thermal energy to determine what happens to that gas during expansion? Obviously, specific internal energy is not constant inside a tank that is venting, nor is it constant as it expands across a valve. We need to add in the PdV term which also represents energy per unit mass or mole.

    For example, a gas expanding across a valve generally results in a decrease in the thermal energy U, and temperature drops. But this isn't always true, sometimes internal energy increases and sometimes it doesn't change at all. It all depends on how any given molecule or atom rearranges itself as the space between the molecules increases. If we neglected the PdV term, the internal energy of the gas inside a venting tank and the internal energy of a gas expanding across a valve wouldn't change. We can see this by applying the first law and simply neglecting the PdV term. So how do we define this PdV term? When a gas undergoes an expansion without doing work, such as the expansion of a gas across a valve, internal energy can, and usually does change.

    I think it comes down to semantics, but it's obviously an energy term we can't argue that. And obviously it is a function of pressure and volume, we can't argue that. So I'd suggest that it helps to think of it as a pressure energy term. I suppose to a purist, that may rub them the wrong way.
     
  13. Nov 6, 2005 #12

    Q_Goest

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    Suni, how about this - the PdV term is an energy term which quantifies the energy added to or liberated from the thermal or internal energy of a real gas undergoing free expansion from one pressure to a lower one. Perhaps that would better describe what is going on and how it is arrived at than simply "pressure energy".

    (Edit: No, that doesn't work because that could also be viewed as a completely contained expansion as opposed to the free expansion across a valve…)
     
    Last edited: Nov 6, 2005
  14. Nov 6, 2005 #13

    GCT

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    one way to ascertain internal energy

    [tex]dU= ( \frac{dU}{dT} )_{v} dT + ( \frac{dU}{dV} )_{T} dV [/tex]
    the [tex] ( \frac{dU}{dV} )_{T} dV [/tex] corresonds to [tex]pdV[/tex], [tex]p=( \frac{dU}{dV} )_{T} [/tex]
    applies for the most part to non-ideal gases and other states for first order transitions, pressure term describes the trend in internal energy with respect to volume
     
  15. Nov 7, 2005 #14

    Q_Goest

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    Thanks GCT, that actually got me thinking. I'd be interested in your description of the pV term.

    Alias, Suni I couldn't find much on the typical web sites about enthalpy and first law but I ended up finding a pretty good description on an MIT web site that I'll quote:

    So here, they point out that there's energy involved with ". . . push(ing) the surroundings out of the way to make room for itself." and that energy is the pV term we've been discussing.

    There is an example just a bit farther down that helps clarify that the pV term is the work the gas is doing in expanding across a valve or restriction in a pipe. In this case, the pressure is dropping and from the perspective of the gas, the gas is doing work against the air it is expanding to make room for itself. That work is exchanged with the internal energy, but that work is not taken out of the gas. That energy remains in the gas and can be calculated from the pV energy term. Note also that pV and internal energy can increase, decrease, or stay the same upon expanding in a pipe, but the sum of them both stays constant. Clear as mud?
     
  16. Nov 7, 2005 #15
    If we're looking at a gas flow across a valve or something then yeah the energy does remain in the gas.. just not as internal energy anymore (it'd be converted to KE or PE or something - all assuming no heat loss). Agreed? I think that gets back to the rate form of the 1st law of thermo u were talking about.
    But if we had something like a solid block just expanding against the atmosphere then some internal energy would be totally lost in the work done on atmos.

    "Thus the total energy of a body is its internal energy plus the extra energy it is credited with by having a volume V at pressure p. We call this total energy the enthalpy, H." (quote from site)

    I actually dont see how this can be so. To me the total energy of a system when using 1st law rate form is U + PE + KE. How is this "total energy" H? I really do not get why they put PV into the total energy term.
     
  17. Nov 7, 2005 #16
    Im gunna be a physicist!! sent off my application.
     
  18. Nov 7, 2005 #17

    GCT

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    if you're interested in the actual conversion of the pV term to energy you just need to multiply the pV term (within the enthalpy term, for instance [tex]p \Delta V [/tex] by (8.314J/molK)/(.08206latm/molK).

    The state of an ideal gas is at times described by its PV state, PV=nRT. A change in P or V implies a change in energy, if you're taking physical chemistry, you should be very familiar with the PV diagram and the derivation of the work of expansion. You should also note that the integral under the PV function is deemed to be the work associated with a system, at times the PV function can be complex depending on how the process was carried out. The term enthalpy is frequently employed to take into account the PV dynamics as well as the changes in internal energy, which are not quite independent from each other.

    for instance, the process may involve changes in pressure as well as volume, which would involve the pdV and VdP terms in enthalpy. Assuming inadequate information, and ideal gas, you can equate [tex] \Delta pV [/tex] (at times it reduces to this) with [tex] \Delta nRT [/tex] which is [tex] nR \Delta T [/tex] for constant composition.

    Q_Goest, I couldn't possibly argue with the MIT definition
     
  19. Nov 7, 2005 #18
    ohhh OK the quote is implying the addition of its ORIGINAL internal energy.. my mistake. makes total sense then.
     
  20. Nov 7, 2005 #19

    GCT

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    no, to sum it up there are various ways to represent changes in the energy of the system in relation to the surroundings, no simple concept will apply for all situations. You'll just need to work and make sense of the equations and the derivations. When and if you take physical chemistry, you'll understand the complexity of the matter.

    First try researching the various forms of internal energy, the PV term of the enthalpy, for the most part, describes the interaction of the system with its surroundings.
     
  21. Nov 7, 2005 #20
    Then can you explain why this total energy function does not include [tex]PV[/tex]? Seems almost a contradiction on their own site. Again i would only find enthalpy useful in its changes i.e. [tex]h1-h2[/tex]. (you were commonly making use of this when using [tex]PdV[/tex].. if we're just looking at the original enthalpy definition it is obviously only [tex]PV[/tex]). Which total energy are we talking about exactly? :bugeye:

    Not sure why you mentioned the conversion factor for energy as [tex]PV[/tex] clearly is already given as Joules when using SI. [tex]PV = Pa*m^3 = N/m^2*m^3 = N*m = J[/tex]
     
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