Ist die innere Energie U abhängig vom Druck und Volumen?
Hallo,
ich frage hier nicht, ob Volumenarbeit oder chemische Wechselwirkungen die Temperatur des Systems erhöhen könnten (was sie können), sondern ganz einfach: Hat die Systemgröße (Druck und Volumen) eine Auswirkung auf die innere Energie U (in einem geschlossenen System)?
Zuerst dachte ich, dass sich Druckenergie nicht innerhalb des Systems nutzen ließe, aber dann bin ich auf folgende Idee gekommen, die Druckenergie zu nutzen:
Man baut in einem geschlossenen System ein luftdichten Raum, der noch denselben Druck wie im System besitzt. Danach leistet man am System isotherme Volumenarbeit und kann durch Ausgleich der Drücke Energie (bspw. elektrische) innerhalb des Systems teilweise zurückgewinnen.
Demnach müsste die Innere Energie den Druck und das Volumen berücksichtigen, da ja keine Energie mit der Umgebung ausgetauscht wurde und deshalb die innere Energie seit der Volumenarbeit konstant geblieben ist.
Ein anderes Gedankenexperiment führt jedoch zu einem anderen Ergebnis.
Dass die innere Energie nicht den Druck und das Volumen berücksichtigt, kann man folgendermaßen herleiten:
Man hat einen großen Raum voller ideales Gas. Jedes Mal, wenn sich Gasmoleküle von der Systemwand entfernen entsteht eine Einbuchtung an dieser Stelle, ohne dass dem System in irgendeiner Weise Energie zugeführt werden würde. Mit zunehmender Zeit wird die Wahrscheinlichkeit für eine Volumenabnahme immer größer. Man verkleinert das Volumen und erhöht den Druck ohne Energie an das System zu geben. Ab einen gewissen Grad wird es unrealistisch unwahrscheinlich, dass sich das System noch verkleinert. Diese Unstimmigkeit ignoriere ich einfach mal.
Dasselbe kann man mit der Ausdehnung machen. Man hat ein System voller ideales Gas und dehnt die Systemgrenzen aus. Die innere Energie bleibt konstant, zumindest dachte ich das zuerst. Ich bin auf die Idee gekommen, dass sich die potentielle Energie innerhalb des Systems bei Volumenänderung verändern könnte: Wenn man ein Haufen Elektronen hat steigt oder sinkt die potentielle Energie, je nach Verteilung der Elektronen innerhalb des Systems.
Dennoch hat das eine Gedankenexperiment zur Volumenverkleinerung zu einem anderen Ergebnis geführt. Wo liegt mein Denkfehler?
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The first main set of thermodynamics is:
From this context, we see that, for example, an isothermal compaction, in which no change in internal energy occurs, is not possible without heat exchange with the environment. If dU=0 and volume change work is performed, heat must pass over the system limit, because pdV would no longer be equal to zero and thus dU=0.
If I increase the volume of a closed system, then does the internal energy of this system always increase?
Why? If I have a gas in an unbelievably large system and then this has increased even further, this has practically no effect on the particles in the system.
Of course, I am going out of a system with almost vacuum-like pressure, which is completely unrealistic, but thermodynamic laws should always apply, shouldn’t I?
No, in the case of ideal gases, internal energy does not increase. In real gases, however, internal energy is already changing since there are intermolecular interactions. In the case of real gas, volume change work must be performed during expansion.
You’re talking about volume change work all the time, but as I said in the first sentence of my question, I’m not talking about volume change work.
Much more about the question whether the position of the system boundaries has an influence on the internal energy. As an example, I have brought a thought experiment in which the size of the system can be reduced with a certain probability without work against internal pressure. This is of course unrealistic or unlikely, but possible if there are only a few cubic picometers.
In real gas, the system boundaries have an influence because volume change work has to be done to shift the system boundaries. They don’t have that in the ideal gas.
The interior (specific) energy is exclusively a function of temperature. She has nothing to do with pressure and volume.
In order to keep T constant, exactly as much heat must be removed as is supplied to volume change work. Thus, u remains constant. ♪
I can’t. An isothermal compression is not possible without the exchange of heat.
You might add that your first statement only applies to ideal gases with constant particle count.
That we are talking about closed systems is a question and even in real gases, the internal energy is a function of temperature, although not necessarily a linear one.
The phrase “because no energy has been exchanged with the environment” also has an end: “since volume work has remained constant”.
I’m sorry if I don’t make it clear in the question, but I’m talking about the internal energy of a general closed system, not just of gases. Suppose internal energy is generally not dependent on pressure and volume, then it follows that it has nothing to do with potential energy (chemical and electrical energy). Is that correct? Then it would make all sense.
One more question: If I have a closed system in which an electrical device is present that uses electrical charge to generate heat, then this will break the first main set of thermodynamics, as the internal energy is now changing. I mean to have read that one always assumes that a system is in the thermodynamic equilibrium, so that the entropy is maximum. Of course, that would not be the case. Is that correct?
No:
dU = Q + W
When heat is supplied, the inner energy changes and thus the temperature changes. Then a new equilibrium is established with another state.
Heat is the carrier of entropy badly. Once heat is supplied, entropy is also added and the entropy of the system increases. As soon as the new thermodynamic equilibrium has been set, the system has reached the maximum entropy with respect to the existing state variables. Then nothing happens.
In classic thermodynamics, quasi-static state changes are assumed. This means that the system never leaves the proximity of the thermodynamic equilibrium, but that equilibrium formation can follow the process quickly enough so that dynamic processes can be neglected.
By the way, electrical energy is not a potential energy, but a separate energy category.
As soon as dynamic processes are decisive, we leave the classic linear thermodynamics and enter the relatively new area of nonlinear thermodynamics.
“If I had a closed system, in the an electric device is present …”
Here potential electrical energy in thermal energy within of the closed system. According to the Wikipedia page, the concept of internal energy can only be applied if the system is in the thermodynamic equilibrium (= Entropy max). That is why there is no contradiction, did I get that right?
But not only from the temperature, but also from the volume (dU/dV_m) at constant T, a/V_m^2 results in the vdW state equation, since the cohesion forces contribute to the energy.