Weird and Deep question about enthalpy...

[Skirmish]

Supposedly the enthalpy would only depend on the temperature and pressure "h (T, P)" or at least this is how commonly enthalpy is expressed in partial derivatives...

So why in an isothermal and isobaric transformation is it different from 0?
*We are taliking of real subtances here, nor ideal gases.
*We know that is non zero!, beacuse we see it all the time on state changes and some chemical reactions.

That equation must be incomplete... or just misleading.
Maybe i'm failing to grasp a mathematical concept that I do not remember.

Please help or guidance.

 

[zdas04]

This is a joke, right? If both temperature and pressure are constant, then enthalpy also has to be constant. Neither a wierd, nor a deep question. Just a naive one.

[bold]David Simpson, PE[/bold]
MuleShoe Engineering

In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist

 

[Skirmish]

Yes, to this point I could classify it as a dumb question of mine. But i'm stuck..

There is the catch...

In many phase changes, and in many Chemical Reactions we consider P and T constant. But energy change is obviously non-zero and you really have some enthalpy difference. Work is done by the volume change of the substance.

So that means the h(P;T) derivative notation does not hold all the information? Because dH=0 if I use that partial derivative equation... but there is work done by the change of volume, that do not appears on that equation.

So... V is a totally independent variable of T and P. For a real substance.

 

[IRstuff]

How do you get volume change without pressure change or temperature change? Isn't V ∝ T/P ?

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[Compositepro]

The equations you posted assume no chemical or nuclear reactions are involved.

 

[Skirmish]

I dunno...

Assume a constant P calorimeter where liquid water is at 100 ºC and is brought to water vapor also to 100ºc with the mixture of a hot fluid. Calculate Q needed. (enthalpy)

*T and P are constant on this process.
*We know also that Hv-Hl=L where "L" is the Latent heat, that indeed is different from 0.
BUT
The problem remains on:

It certainly does not work here, but ... why???. If there is only a cuastiestatic work (Volume expansion) that equation should work.

 

[moltenmetal]

I'm not sure I understand your question.

Is your question what is the mechanistic underpinning of a change in enthalpy due to a change of phase? That's really easy to understand actually, if you understand the forces which act between molecules.

The way you've stated your last question puzzles me. If you want to take saturated liquid water to saturated water vapour at constant temperature and pressure, the thing you're adding is energy (i.e. heat), not "mixture with a hot fluid". The difference in enthalpy between the liquid and the vapour state in this case is precisely the heat energy you need to add to make this transition, which we call the enthalpy of evaporation. Where does the heat go? It goes into overcoming the attractions between the molecules in the liquid- in the case of water it's both van der Waals attraction and hydrogen bonding.

 

[Latexman]

Skirmish,

Are you talking an open or closed system? It's kind of important.

You will have an easier time thinking about this with a more fundamental equation with no assumptions (open or closed system) attached to it, such as:

dH = TdS + VdP

Good luck,
Latexman

To a ChE, the glass is always full - 1/2 air and 1/2 water.

 

[Skirmish]

The specific issue is with that equation, is a closed system.

The thing comes to dependence between P and T. When there is a phase change or a chemical reaction they become dependent and they do not follow the State Postulate that allows one to write h(T;P).
If I define h(T;v) then I can use that equation on phase changes and chemical reactions without drama.the

 

[georgeverghese]

If you look further back in the basic premises for this State Postulate in your thermo text, it would state that this equation is for single phase systems with no chemical reactions ?

 

[davefitz]

if it is a single phase gas with no chemical changes then roughly PV=ZRT, if P and T are constant , then T is fixed as well, and H is constant . In 2-phase ( water/steam) situation T and P are constant but H varies as x ( steam by weight ) changes. Multiphase or multicomponent ( as with ammonia -water or binary hydrocarbon) cases will have H vary as phases or chemical componengts change. So it looks as if you are readsing a thermo book and may have forgotten the preface stating the elementary discussion assumes single phase or without chemical reactions.

"...when logic, and proportion, have fallen, sloppy dead..." Grace Slick

 

[Skirmish]

The state postulate says:

“The state of a simple compressible system is completely specified by two independent, intensive properties
”
A system is considered to be a simple compressible one in the absence of certain effects which are uncommon in many engineering applications. These are electromagnetic and gravitational fields, surface tension, and motion. For such a system, only two independent intensive variables are sufficient to derive all the others by use of an equation of state. In the case of a more complex system, additional variables must be measured in order to solve for the complete state. For example, if gravitation is significant then an elevation may be required.

Two properties are considered independent if one can be varied while the other is held constant. For example, temperature and specific volume are always independent. However, temperature and pressure are independent only for a single-phase system; for a multiphase system (such as a mixture of gas and liquid) this is not the case. (e.g., boiling point (temperature) depends on elevation (ambient pressure))

So that seems to say to me that even if I have a mulphase/multicomponent system enthalpy h of the whole system can always be calculated has h(T,v)? Is that OK?

*I know that h "natural variables" are h(s,P) but it seems that h(T,v) should also work.
**There is no reference to chemical nor phase changes on the book I'm using.

 

[moltenmetal]

"So that seems to say to me that even if I have a mulphase/multicomponent system enthalpy h of the whole system can always be calculated has h(T,v)? Is that OK?"

No. In a multi-phase system you have another degree of freedom (x, i.e. mol fraction liquid or vapour) and hence you need another state variable. In fact, you will have many more such fractions in multicomponent systems, and each such fraction is another degree of freedom.

 

[IRstuff]

"only two independent intensive variables are sufficient to derive all the others by use of an equation of state. In the case of a more complex system, additional variables must be measured in order to solve for the complete state."

Isn't that precisely what your citation says?

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[urgross]

I think you need to define what you mean by "closed system". Does this closed system include volume? Is the ideal gas law not to be applied? Perhaps you should think how a steam boiler operates, and why the steam is transported at relatively high pressures as to its end usage after going through a pressure reducing valve, operating in a constant volume system to end use (steel boiler, steel pipe, relatively negligible change in volume owing to thermal expansion as compared to energy of condensate).

If you do not consider adding energy to a closed volume of water to create a phase change in water, you appear to be proposing an energy free boiler. Changing volume against a real opposing force requires work, sort of how your car engine needs gasoline, air and a spark before the piston moves.

To the mechanical engineer, it doesn't matter if the glass is half full or half empty, it only matters if the bottom half is mine.

 

[chicopee]

Before you get too embroiled into your OP, please refer to the attachment so as to dispel any misconception.