# What exactly is enthalpy?

What does the term enthalpy really mean and what is its physical interpretation? What does the wikipedia definition of enthalpy mean, which says that $pV$ work in the enthalpy definition is the energy required to place the system in its surroundings? And why is the heat of a reaction measured in enthalpies and not internal energy?

• Welcome on Physics SE :) I am happy that you had a look at wikipedia before asking here - however, please keep your questions limited to one issue and not a whole number of them and detail what exactly you did not understand - this will help people writing answer that actually adress your question. – Sanya Dec 5 '16 at 12:26
• This is a duplicate question, but of a closed question. In my opinion, this one should be left open and the other left closed because this version is a much better worded question. – David Hammen Dec 5 '16 at 13:03
• Thank you for your answers . Now I can intuitively understand enthalpy. – user138269 Dec 5 '16 at 13:04
• No offence and you might know this already, but when you have received an answer that you are happy with, whenever that is, you can upvote and then accept it. As a new user you may not know this, thanks – user108787 Dec 5 '16 at 13:53
• Possible duplicate of What is enthalpy? – user36790 Dec 5 '16 at 17:03

Enthalpy is all energy associated with something at its current position; it is internal energy and the work (found as $pV$) required to place it where ever it is.

$$H=U+pV$$

To create a rabbit out of nothing and place it on the table, the magician must summon up not only the energy $U$ of the rabbit, but also some additional energy, equal to $pV$, to push the atmosphere out of the way to make room.

(Thanks, @user115350, for link to this illustration.)

It is a useful parameter when you for example compare different parts in a pipe system. A system with flow through pipes is e.g. What we have in the back of our fridge, or in a power plant etc. Enthalpy takes into account not only the energy stored in each part/point of the fluid in this moment, but also the energy required to "place" it there.

If there are no pumps or turbines etc. along the way, the enthalpy is therefore constant within a pipe even if the pipe narrows in and pressure and volume change.

However, in many textbooks I've read the reason for defining this parameter is out of convenience, since internal energy plus $pV$ is a very commonly used expression in energy systems. But I personally like to think of it in the above way for a intuitive picture.

• But still I am not getting why is energy required to place a system in its surroundings . Also why is this state function useful for heat changes in a chemical reaction . – user138269 Dec 5 '16 at 12:43
• @user138269 For heat changes in reactions, remember that one thing is the change of temperature, but another thing is e.g. expansion. Expansion does work on the surroundings. Both of these are taken into account in enthalpy, while only temperature and other internal features are included in internal energy. – Steeven Dec 5 '16 at 13:00
• @user138269 Energy is always required to place something somewhere where it rather wouldn't be. It's like placing a balloon under water; it requires some work done by you, otherwise it would scoop up and out of the water again. Also, while this balloon is under water, the water tries to compress it. Increases pressure and tries to reduce volume. The balloon holds back, and this is what the $pV$ term is about. – Steeven Dec 5 '16 at 13:01
• +1 i think your answer adds to the pppqqq one, in physical intution terms – user108787 Dec 5 '16 at 13:49
• you can check out this quora.com/… – user115350 Dec 5 '16 at 19:42

Enthalpy is a useful quantity when the pressure is constant. Indeed, for a $PVT$ system, taking the differential of $H=U+pV$, and using the first and second law of thermodynamics: $$T\text d S=\text d U +p\text d V$$ we obtain: $$\text d H = T\text d S +V\text dp.$$This shows that the natural choice is to consider $H=H(p,S)$. This also shows that, if the pressure $p$ is constant: $$\text d H =\text d Q _p$$is the exchanged heat at constant pressure, which is why $H$ is often used in the characterization of chemical reactions, when they are studied at, say, a constant $p=1\,\text {atm}$.

However, the enthalpy is also useful in other context. One which surprised me when I learned about it, is fluid-dynamics of ideal, isentropic flows. This exploits the fact that the entropy is constant and is therefore somewhat parallel to the precedent case.

In fluid-dynamics, it is common to consider thermodynamic quantities per unit mass. It is also assumed that each unit mass of the fluid is a copy of the same thermodynamical system, which can be considered in thermodynamical equilibrium at every instant of time. If $h,s,p$ are the local (specific) enthalpy, entropy and pressure, then the above relation for $\text d H$ becomes: $$\text d h = T\text d s + \frac {1}{\rho} \text d p$$($\rho$ is the density). Here, now, it is the entropy which is constant over all the fluid. This allows us to reformulate Euler's equation: $$\frac{\text D \mathbf v}{\text D t}=-\frac{\nabla p}{\rho}$$ in the useful form: $$\frac{\text D \mathbf v}{\text D t}=-\nabla h.$$ If, for example, the flow is stationary, the LHS reads:$$\frac{\text D \mathbf v}{\text D t}=\nabla{\frac{v^2}{2}}-\mathbf v \times (\mathbf \nabla\times \mathbf v),$$ so that the quantity: $$h+\frac{v^2}{2}=u+\frac{p}{\rho}+\frac{v^2}{2}$$ is constant along the trajectories of fluid particles. This is Bernoulli's theorem generalized for a compressible perfect fluid. If the density is constant, then $u$ is also constant in an isentropic flow, so that Bernoulli's integral becomes: $$\frac{p}{\rho}+\frac{v^2}{2},$$ but in the general case we have to include the whole enthalpy in the integral.