Enthalpy - published values

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The discussion centers on whether published enthalpy values account for work done against constant pressure, particularly in gas reactions. Enthalpy, defined as H = U + PV, incorporates the work needed to accommodate a system at constant pressure, typically 1 atm. While enthalpy serves as a useful shorthand in thermodynamics, it is primarily a property of the material in thermodynamic equilibrium rather than a specific process. The difference in enthalpy between liquid and gas phases of water is attributed mainly to variations in internal energy, with only a minor contribution from PV work. Careful interpretation of enthalpy values is essential for accurate thermodynamic analysis.
Joseph M. Zias
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Do the published values of Enthalpy include the work done against a constant pressure, e.g., the atmosphere? (I am not a chemist). I am reviewing enthalpy and entropy as part of the statistical mechanics applied to transistors. I assume, from my reading, that the work done would mostly apply to reactions involving gasses.
 
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One finds a useful definition for enthalpy ##H## in the book “Introduction to Thermal Physics” by Daniel V. Schroeder:

Enthalpy

Constant-pressure processes occur quite often, both in the natural world and in the laboratory. Keeping track of the compression-expansion work done during these processes gets to be a pain after a while, but there is a convenient trick that makes it a bit easier. Instead of always talking about the energy content of a system, we can agree to always add in the work needed to make room for it (under a constant pressure, usually 1 atm). This work is ##PV##, the pressure of the environment times the total volume of the system (that is, the total space you would need to clear out to make room for it). Adding ##PV## onto the energy gives a quantity called the enthalpy, denoted ##H##:

$$H\equiv U + PV.$$

This is the total energy you would have to come up with, to create the system out of nothing and put it into the environment (see Figure 1.15). Or, put another way, if you could somehow annihilate the system, the energy you could extract is not just ##U##, but also the work ##(PV)## done by the atmosphere as it collapses to fill the vacuum left behind.
 
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I totally oppose introducing this description in an attempt to assign a physical interpretation to enthalpy. I have seen it accomplish nothing but causing confusion among generation after generation of neophytes to the subject of thermodynamics.

Enthalpy is just a convenient shorthand parameter (for U+PV) that is useful in analyzing many kinds of thermodynamics problems. It is a property of the material being processed (exclusively for thermodynamic equilibrium states), rather than a feature of any particular kind of process (such as an isobaric process, for example). We could solve all the problems we encounter in changes between thermodynamic equilibrium states without ever introducing the enthalpy parameter H=U+PV.
 
I think we can close this thread. I found the enthalpy of H2O l is different that the enthalpy of H2Og so apparently the PV work is included when a gas is involved.
 
Joseph M. Zias said:
I think we can close this thread. I found the enthalpy of H2O l is different that the enthalpy of H2Og so apparently the PV work is included when a gas is involved.
Please provide your reference. One must be very careful when interpreting enthalpy values.
 
Joseph M. Zias said:
I think we can close this thread. I found the enthalpy of H2O l is different that the enthalpy of H2Og so apparently the PV work is included when a gas is involved.
This is not a correct interpretation. The difference between H2O l enthalpy and H20 g enthalpy is primarily the result of the difference in internal energies of the liquid and vapor, with only a veery small contribution to the difference in PV.
 
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