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Internal energy and enthalpy. I am finding it hard to distinguish between the two. Confused. Can someone explain me the two terms and difference between them? I tried learning from wikipedia but it said for both the above terms that it is the total energy contained by a thermodynamic system, so Iam confused?

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    $\begingroup$ Enthalpy is internal energy + energy to create the system (fighting the 'atmospheric' pressure) $\endgroup$ – Danu Jan 7 '14 at 13:27
  • $\begingroup$ Enthalpy is internal energy + $PV$. The term $PV$ does not require any deep interpretation; it is added to energy for convenience, in descriptions of isobaric processes and processes that begin and end at the same pressure (while in between, the pressure may vary). @Danu: When we consider how thermodynamic systems are prepared, often the atmospheric pressure does not need to be fighted (any isochoric process). $\endgroup$ – Ján Lalinský Jan 7 '14 at 18:06
  • $\begingroup$ The wizard analogy is great for understanding enthalpy intuitively. physics.stackexchange.com/questions/356412/… $\endgroup$ – Steeven Nov 23 '20 at 9:28
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In pure thermodynamics, the internal energy is defined operatively: given a state of reference of our system, the internal energy of another state it's defined as the work needed to bring it to the new state: $$U=W,$$ provided the system is thermally isolated. The empirical observation behind this definition is that “this definition makes sense”, i.e. there exist substances such that, if the system is entirely contained in some recipient made of these, then the amount of work $W$ is the same for all transformations that bring the system in the new state, which is the precise meaning of the phrase “being thermally insulated”.

Enthalpy is $$H=U+PV.$$

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    $\begingroup$ You definition of internal energy is almost right, but you confused the sign in front of $W$; energy in state A is the work that was supplied to get to the state A. Also it should be said that the system has to be isolated, otherwise the work $W$ would depend on how the state A was approached (with heating, without...). $\endgroup$ – Ján Lalinský Jan 7 '14 at 17:57
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When you create something 'out of nothing' you need the energy $U$ to make it, but you also have to add additional energy to create room for it i.e the volume displacement. Or if you annihilate a system, this additional energy is the energy you would get when the atmosphere collapses. Enthalpy $H$ is then this total energy required for this process.

$H=U+PV$

Notice, that this is at constant pressure.

As you see, internal energy $U$ is energy without this displacement. It's the energy cointained in the system without the volume displacement.

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  • $\begingroup$ A statement says'the energy stored within the substance or the system that is available for conversion into heat is called the enthalpy' confused o_O $\endgroup$ – Uzair Jan 7 '14 at 14:59
  • $\begingroup$ Okay. For example a chemical reaction $H_2+1/2 O_2->H_2O$. Now the enthalpy change for this is according to my reference table $\Delta H=286kJ$. This is the amount of heat that you would get if you burned a one mole of hydrogen. $\endgroup$ – Mathma Jan 7 '14 at 15:16
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The internal energy of a system is simply the sum of its potential and kinetic energies i.e.

E(int) = U + K

The enthalpy is then the total internal energy of the system as well as how much pressure the system exerts on the volume it inhabits i.e.

E(enthalpy) = E(int) + PV

I hope this helps!

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  • $\begingroup$ It should be said that $K$ is the kinetic energy of the constituents of the system, not of the system's center of mass. Similarly, $U$ is total potential energy of the constituents of the system, not potential energy of the system in some external field. $\endgroup$ – Ján Lalinský Jan 7 '14 at 17:50
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Well now if you don't mind abusing mathematical notation and formulas, perhaps a more hand-wavy answer can help illustrate the difference more intuitively.

Let's say we have some ideal gas in some sealed bag at absolute zero -- i.e, $T = 0$. Then, according to the ideal gas law, $PV = nRT = 0$, so we must have either $P = 0$ and/or $V = 0$. However, it actually follows that $V = 0$, since $V \ne 0 \Rightarrow P = nRT/V = 0$ and so the bag will be crushed into oblivion ($V = 0$) by the external atmospheric pressure $P_{atm} > 0$ anyway!

But, let’s say we want our bag to actually have some $V' > 0$. Then, by the ideal gas law, our new temperature $T'$ would have to be $$T' = \frac{P_{atm}V'}{nR} > 0$$

But how in the world do we accomplish this? Clearly, we must supply some amount of energy to the system to raise its temperature, and then even some more on top of that to make it expand against the atmosphere in order to occupy the newly desired $V’ > 0$.

Without changing the system’s volume (i.e., have any mechanical work done on or by the system), we can simply supply some energy just to heat up the system to our desired $T’ > 0$. How much do we need? Well, simply just $E_1 = nC_vT'$, where $C_v$ is the specific heat of the gas at constant volume.

So now we’re at the target temperature $T'$, but how do we expand from $V = 0$ up to our desired new volume $V' > 0$? We don’t want to use the system’s own internal energy for this (since that will in turn lower the temperature!), so, once again, the energy required must come from its surroundings! Since the atmospheric pressure $P_{atm} > 0$ is relatively constant, we’ll just need to also add $E_2 = P_{atm}V'$ in the form of heat to the system.

And so, we have now arrived at the concept of enthalpy! That is, $H = E_1 + E_2 = U + PV$ — the amount of energy needed to give a system some internal energy $U > 0$ (all the way from absolute zero), and also to expand it against its surroundings to occupy some desired positive volume as well.

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The internal energy and the enthalpy are thermodynamic potentials. I will explain them as in the book A Modern Course in Statistical Mechanics by L. Reichl, which, I think, is a very intuitive and simple way to do it. "In conservative mechanical systems, such as a spring or a mass raised in a gravitational field, work can be stored in the form of potential energy and subsequently retrieved. Under certain circumstances the same is true for thermodynamic systems. We can store energy in a thermodynamic system by doing work on it through a reversible process, and we can eventually retrieve that energy in the form of work. The energy which is stored and retrievable in the form of work is called the free energy (sic)."

There are as many different free energies as there are combinations of constraints. For example, for a reversible process carrried out at fixed entropy ($S$), volume ($V$), and number of moles ($N$), the change in internal energy (U) [which simply is the energy stored in a thermodynamic system] is equal to the total amount of work that can be done on or by the system. In other words, for a reversible process carried out at fixed $S$, $V$ and $N$ work can be stored in the form of internal energy and can be recovered completely. Therefore, the natural variables of the internal energy are $S$, $V$ and $N$, i.e., $U=U(S,V,N)$.

In systems which are thermally isolated and closed but mechanically coupled to the outside world, we can only stored work in the form of enthalpy ($H$). In other words, for a reversible process at constant $S$, $P$, and $N$, work can be stored as Enthalpy and can be recovered completely.The natural variables of the enthalpy are $S$, $P$ (the pressure of the system), and $N$, therefore $H=H(S,P,N)$.

Answering your question, enthalpy and internal energy are both thermodynamic potentials, which means that both of them can be used to store useful work. They only differ in their natural variables, $U=U(S,V,N)$ and $H=H(S,P,N)$. We want to express the thermodynamic potentials in many different variables because some of them are easier to measure than the others. For example, we do not measure entropy directly, we measure temperature. Sometimes it is easier to measure pressure instead of volume, so the enthalpy is a good option in theses cases. To change from one variable to another, we use the Legendre transformations, that's why the enthalpy is defined as $H=U-PV$, in this way we change the variable $V$ in $U$ to $P$ in $H$.

Note that thermodynamic potentials have different names depending on their natural variables. The internal energy is the thermodynamic potential which has $S$, $V$, and $N$ as its natural variables. A thermodynamic potential, which natural variables are $S$, $P$, and $N$ is called the enthalpy. The Gibbs free energy ($G$) has $T$, $P$, and $N$ as its natural variables, and so on. For a more detailed discussion about this, I strongly encourage you to read the second edition of A Modern Course in Statistical Mechanics by L. Reichl. Hope this help you to figure out what is the difference between enthalpy and internal energy.

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