Enthalpy change with standard conditions

  1. Why is it that change in enthalpy (h=e + pv) depends upon pressure as they say, we should measure at standard pressure, while pv=nrt so no matter da presure nrt is the same the product p by change in volume is constant so accordin 2 me we should specify only da temperature change for an isothermal process since only change in temperature gives the workdone! Help me out people
  2. jcsd
  3. DrDu

    DrDu 4,643
    Science Advisor

    From the rules of the forum:
    "In the interest of conveying ideas as clearly as possible, posts are required to show reasonable attention to written English communication standards. This includes the use of proper grammatical structure, punctuation, capitalization, and spelling. SMS messaging shorthand, such as using "u" for "you", is not acceptable."
  4. Why is it that change in enthalpy (h=e + pv) depends upon pressure as they say, we should measure at standard pressure, while pv=nrt so no matter the surrounding presure nrt is the same the product of external pressure by change in volume is constant as long as change in temperature is the same! So according to me i think we should specify only da temperature change for an isothermal process since only change in temperature gives the workdone! Help me out people! The book standardises measurements should be done at 25C and 1atm
  5. DrDu

    DrDu 4,643
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    I still don't understand completely your question, but the relation pV=nRT only holds for ideal gasses and not e.g. for liquids. Also, the internal energy alone depends on p and T for systems other than ideal gasses. And pV is not the work done, not even for an ideal gas.
  6. The question and the problem arises because (in physical chemistry) no clear distinction is made between the internal pressure and the external pressure. Even if they happen to be equal it us usefull to maintain the distinction for conceptional reasons. p in (pV=nRT) is an internal pressure, but the p in (H=U+pV) is the external pressure. The latter is obvious for solids, and because the definition of enthalpy should be the same for all substances and phase, this must also be true for liquids and gases. If p(in) = p(out) it is allowed to substitute one for the other, but it must be remembered the p(out) is an independent parameter, and p(in) is a dependent variable of state.
  7. DrDu

    DrDu 4,643
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    "p in (pV=nRT) is an internal pressure, but the p in (H=U+pV) is the external pressure." That is not true. The variables in the definition of the thermodynamical potentials are always the internal variables of the system. I also don't see why your statement should be obvious for solids.
  8. 1. take a system of 1 mol (55.8g) of iron (Fe). [any other solid will do]
    What would be the "internal pressure" of this system ? How do you measure it ? Is it tabulated somewhere ?. Does it change with T ?
    2. The standard enthalpy of formation of this system = 0 kJ (298K, 1 atm).
    Why is the condition 1 atm chosen if H is independent of the external pressure, ie depends only on internal variables ? (see also original question of Godwin Kessy above).
    3. (dH/dT)p = Cp, where the index p refers to the constant external pressure, since we do not know how to measure or control the inside pressure of iron.
    4. As stated: most textbooks use the symbol p for both internal and external pressure: they define p = pressure. Only in the text is specified what is meant specifically.
    5. Look at ENTHALPY in wikipedia:
    what is written under 'History' is complete fantasy, but the paragraph 'Difference between H and U: the additional term pV' is almost correct and p is identified there as atmospheric pressure = external pressure.
  9. DrDu

    DrDu 4,643
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    To 1: The pressure in a solid is the isotropic part of the strain tensor, which is extensively tabulated ( ask an mechanical engineer of your choice). It changes with T and density. Anyhow, in a solid body the expression pV in the definition of enthalpy should be replaced by stress*strain. The reference of choice of the physicist is Landau Lifshetz, elasticity theory.
    To 2: Because 1 atm refers to internal pressure.
    To 3: See 1
    To 4: Not everything in Wikipedia is correct.

    But the most important point: The easiest way to measure the internal pressure is in mechanical equilibrium situations, where internal and external pressure have to coincide. This is done e.g. in diamond anvil cells up to very high pressures. So it is easy to measure the pressure of a solid body e.g. as a function of density and temperature. This equation of state between pressure, density and temperature you may use to calculate internal pressure in more general situations, just like you use pV=nRT to calculate the internal pressure of an ideal gas e.g. during the explosion in a motor.
  10. 1a None of the textbooks on thermodynamics or physical chemistry I have ever seen, has the audacity to either define, explain or refer to the quantity p in (U+pV) as "strain", or as "strain tensor" let alone as "the isotropic part of the strain tensor", and there is little or no support for this position in the litterature. Personally I would not look for a definition of enthalpy in a book on elasticity theory, nor would I ask a mechanical engineer about it.

    1b. If pV is replaced by (stress*strain) we arrive at a completely different quantity, which may be unrelated to enthalpy. We surely will not get out of the woods this way.

    2a In the calorimetric experiments used to measure enthalpy, (1 atm) refers to the external pressure as an experimental condition.

    2b Apparently the "internal pressure" (the isotropic part of the strain tensor) of iron = 1 atm. Now we have to assume that the same holds for all other solids, because the condition holds for all solids in their standard state, hence p(in) = p(out) = 1 atm. However this condition is independent of temperature, where the "the isotropic part of the strain tensor" apparently is not, so there is no common ground for discussion left.
  11. DrDu

    DrDu 4,643
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    I meant pressure is the isotropic part of the stress tensor, not strain tensor (I translated it wrong).
    Most text on physical chemistry don't spend much time on enthalpy of liquids and solids, as in usual reactions at ambient pressure the difference between H and U is minute. This is not so in geophysics where pressure may drive the transformation e.g. graphite into diamond.

    Enthalpy is a thermodynamic potential, like internal energy. As such it is defined primarily for substances in thermodynamic equilibrium and depends only on the intrinsic variables of the system, like its volume, temperature, entropy or pressure.
    That usually the external pressure is specified is due to the fact that you have to assure that the internal pressure is in equilibrium with the external pressure. If you don't make sure that this is the case then you will measure something different.
    For example, if you use a bomb type calorimeter, internal pressure may be anything else than the external pressure, but then you will measure Delta U and not Delta H.
  12. (McGraw-Hill Dictionary of Engineering)
    enthalpy: The sum of the internal energy of a system plus the product of the system's volume multiplied by the pressure exerted on the system by its surroundings.
  13. DrDu

    DrDu 4,643
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    "Personally I would not look for a definition of enthalpy in a book on elasticity theory, nor would I ask a mechanical engineer about it." Aha, but now you cite the "McGraw-Hill Dictionary of Engineering".

    I wonder how you would define the enthalpy of the gas inside a steel cylinder. This is obviously a system in mechanical and thermal equilibrium.
    Is the external pressure in your definition of the enthalpy of the gas the ambient pressure outside the bottle?
    Which pressure do you use to define the enthalpy of the steel cylinder? The "external pressure" on the inner side of the cylinder or the ambient pressure on the outer side of the cylinder?
  14. The one discipline that uses enthalpy on a daily basis is called chemical engineering: so there is no 'aha' here.
    With respect to the gas-filled cylinder: there is nothing obvious about it, since you have said nothing about a movable piston. I assume this was implied, but the obvious equilibrium condition is not necessary. (with one exception, see below).
    The dictionary is clear: for the system as a whole: p in (U+pV) is the pressure exerted on the system by its surroundings, ie the atmospheric pressure. This holds for the subsystems (cylinder, piston, gas) as well, independent of any mechanical or thermal equilibrium. (Both cylinder, piston and gas have enthalpy as a variable of state). In equilibrium thermodynamics where p(in)=p(out), it is allowed to substitute one for the other. (this is the exception)

    Indeed, if you want total confusion you should call the 'internal pressure' by the name 'external pressure'. Confucius (no pun intended) wrote 2500 years ago: "If words don't agree with reality, then language becomes meaningless and common action impossible". (leaving alone discussion).
    This being said it must be admitted that the difference can be subtle. If you take as a system one liter of water somewhere in a lake or ocean, both the internal and external pressure increase with depth and so does enthalpy. If you replace 1 liter water by a cylinder of 1 liter, the difference between the inside and outside pressure will re-emerge.
  15. DrDu

    DrDu 4,643
    Science Advisor

    How do you describe then the Joule-Thompson experiment? I.e. the expansion of a gas from (internal) pressure p1 to pressure p2 trough a throttle?
    According to e.g. the book of Max Planck "Thermodynamik" (Walter de Gruyter, Berlin, 1930) (the argument can be found in numerous more modern treatments on thermodynamics, however, Planck certainly understood what he was writing about as he did his thesis on thermodynamics and got a formulation of the second law named after him.) one choses the enthalpies before and after passing through the throttle as H1=U1+p1V1 and H2=U2+p2v2. The process is called "isenthalpic" as H1=H2.

    In my opinion, at best one of the two pressures can coincide with the external pressure, which may however even be distinct of the two pressures in the experiment.

    PS: In my previous post I didn't assume any piston but only that of gas enclosed under pressure in a steel cylinder as is used for storage of gas.
  16. in my opinion the proper handling of the joule-kelvin experiment depends on the proper definition of the system, ie system-environment interface. It is not a system in equilibrium however and it brings us far from the problem that started this discussion. To me it seems rather important to the whole of science what the physical meaning is of p in (H=U+pV): I wonder if there are only 2 or 3 people in this forum that are interested in the subject ??
  17. DrDu

    DrDu 4,643
    Science Advisor

    I was just trying to find a situation where the difference between internal and external pressure might matter. While the expansion itself is certainly not an equilibrium state (like a chemical reaction taking place in a calorimeter), the initial and final state are equilibrium states which suffices to define the enthalpy difference.
    I would be glad when you could propose another example, togehter with your assignment of internal and external pressure, where this difference might matter. Then we can analyze it together.

    PS: I found the more general definition of enthalpy in terms of the stress and strain tensor also in Buchdahl, Concepts of classical thermodynamics, Cambridge UP, 1966,which seems to have a high reputation.
  18. I note that Godwin talks of change in enthalpy, but quotes the absolute form, not the differential form.

    The straightforward answer to his question is easily provided by differential form.

    As regards Dr Du's question, the relationship between the stress state and enthalpy is 'not necessarily'.

    Any system has internal energy. When this energy is changed the above equation is merely an accounting for what happens to it.

    Some of the energy released or absorbed appears as heat, some as mechanical work due to volume change, some in other forms eg strain energy.
    It has to be observed that if bond energies are involved they dwarf stress energies, which involve mere bond distortions.

  19. DrDu

    DrDu 4,643
    Science Advisor

    Dear studiot,
    it is nice to have someone else in this discussion. You are certainly right that, once one has several mechanical degrees of freedom, there may be several ways to define enthalpy (or enthalpies). I was bringing up the stress and strain mainly to show that also in solid bodies, internal pressure is a well defined quantity, while Zeppos10 denied that internal pressure can be defined for a solid body and one has to use external pressure.

    I do not quite agree with your last statement about bond energies vs. stress energies.
    Obviously the latter are much smaller, but in chemical reactions bonds are not only broken but also formed, and the difference of bond energies compares usually well with the mechanical energy, be it pressure or stress.
  20. "I have checked many textbooks and websites and it seems that the constant pressure associated with the expression, delta H=q, is the external pressure in which we carried out our experiments and not the pressure of the system". (Chemical Forum,Eutectic6002, 31/8 2006)
    This implies that most thermodynamic data available are compatible with H=U+p(env)V.
    The throttle-example may not be as bad as it first appeared to me: the problem there is that it assumes the existance of two large 'pressure-bath'. If you assume that the high-pressure side has to be maintained by the operator who runs the throttle, then the proces is not isenthalpic, but H increases, in line with the point of view on enthalpy defended by me.
    It is best to compare the throttle with an (irreversible) isothermal expansion. If you use H=U+p(in)V, H=constant, but if you use dH=Q it increases. It seems that not both position can be true.
  21. Any reputable table of data states the conditions, which as you point out is often STP.
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