What is pv in enthalpy's equation? if we transfer energy to a system in the form of work does it reflect in its internal energy fully and if it does why do we need a term called enthalpy which seperates pv term in its equation?if pv is not included in internal energy then why on doing work internal energy of the system changes?
[Physics] the difference between enthalpy and internal energy
energythermodynamics
Related Solutions
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.
$$H=U+pV$$
Daniel V. Schroeder, "Thermal physics", Addison-Wesley, 2000 (page 34)
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.
It's not entirely wrong to think of enthalpy as the heat content of a system1, but due to the definition of heat as a certain type of energy transferred between systems, there are subtle differences.
Enthalpy is a state function, a priori only well-defined after equilibrium has been established, whereas heat is a characteristic of the specific process $p$ that takes you from initial state $i$ to final state $f$.
Let $$ \Delta H_{i\to f} = H(f) - H(i) $$
In general, we have $$ \Delta H_{i\to f} = Q(p) + \Delta H^\text{uncomp}(p) $$ where we can have an additional change in enthalpy due to entropy production within the system (eg when friction is involved) that is not compensated by entropy loss of the environment due to heat transfer.
For quasistatic processes where state variables are well-defined at all times, we also have $$ \Delta H_{i\to f} = \int_p T \mathrm dS $$
In that case, we may go to an infinitesial description $$ \mathrm dH = T\mathrm dS = \delta Q + T \mathrm \delta S^\text{uncomp} $$ where the $\delta$ forms need only be defined along the specific curve through phase space.
In case of reversible processes, we additionally have $$ \delta S^\text{uncomp} = 0 $$ and thus $$ \mathrm dH = \delta Q \qquad\qquad \Delta H_{i\to f} = Q(p) $$ which was your starting point.
1 or rather, as Wikipedia puts it, "the capacity to do non-mechanical work plus the capacity to release heat"
Best Answer
Enthalpy is a state property and is defined as
$$H=U+pV$$
Where $H$ is the enthalpy, $U$ internal energy, and $pV$ is the product of pressure and volume. So enthalpy is a state property derived from internal energy and other properties, $p$ and $V$.
Enthalpy is simply a useful derived property for analyzing certain type of thermodynamic problems. It is not a fundamental property such as internal energy, entropy, pressure, volume and temperature.
Hope this helps.