thermodynamics - Equation of enthalpy

This question arises because: by giving classes in thermodynamics, I have observed that students are often confused between the different definitions (or applications) of the enthalpy concept.

The enthalpy expression is obtained as follows:

From the first law of thermodynamics: \begin{align*} U=Q+W \end{align*} Recalling the definition of work: \begin{align*} W=-PV \end{align*} Variations added: \begin{align*} \delta U=Q - \delta (PV) \end{align*} Its obtained that: \begin{align*} \delta U&=Q-(V\delta P+P\delta V) \end{align*} Constant volumen: \begin{align*} \delta U=Q-V\delta P \end{align*} Constant pressure: \begin{align*} \delta U=Q-P\delta V \end{align*} Variations enlarged: \begin{align*} (U_f-U_i)&=Q-P(V_f-V_i)\\ (U_f-U_i)&=Q-(PV_f-PV_i)\\ Q&=(U_f-U_i)+(PV_f-PV_i)\\ Q&=(U_f+PV_f)-(U_i+PV_i)\\ Q&=H_f-H_i \end{align*} Therefore, the definition of enthalpy is: $$ H = U - W $$

Is this properly proposed?

Answer

Definition of internal energy: U is a function of state, representing the total kinetic and potential energy of the molecules. U = U(T,V)

First Law of Thermodynamics:$$\Delta U=\delta Q+\delta W$$where the symbol $\Delta$ is used to represent the change in a (path-independent) function of state (like U) between an initial and final thermodynamic equilibrium state of a closed system and the symbol $\delta$ is used to represent the change in a parameter that depends on the process path between and initial and final thermodynamic equilibrium state of a closed system. $\delta Q$ is the heat added to the system over the path, and $\delta W$ is the work done on the system over the path.

Relationship for the Work:$$\delta W=-\int{P_{ext}dV}$$where, for both for reversible and irreversible process paths, $P_{ext}$ is the force per unit area exerted by the gas on the piston face, and, by Newton's 3rd law, the force exerted by the piston face on the gas. For an irreversible process path, the pressure typically varies with spatial location within the cylinder, so that the average gas pressure does not match $P_{ext}$ at the piston face. For a reversible process, the gas pressure is uniform within the cylinder, so $P_{ext}=P$ where P is the gas pressure calculated from the equation of state of the gas (such as the ideal gas law), based on the number of moles in the cylinder, the gas pressure in the cylinder, and the gas temperature in the cylinder.

Combining the First Law with the Relationship for Work:$$\Delta U=\delta Q-\int{P_{ext}dV}\tag{rev and irrev processes}$$ $$\Delta U=\delta Q-\int{PdV}\tag{rev processes}$$

Constant Volume ($V_i=V_f)$:$$\delta W=0$$ $$\Delta U=\delta Q$$

Constant Pressure ($P_{ext}=P_i=P_f=P$):$$\delta W=-P_{ext}\Delta V=-P\Delta V$$ $$\Delta U=Q-P\Delta V$$ So,$$\Delta U+P\Delta V=\delta Q$$But, since P is constant, $$\Delta U+\Delta (PV)=\delta Q$$ The definition of enthalpy is $$H\equiv U+PV$$ Therefore, for a constant pressure process (a process in which $P_{ext}$ over the entire process path is equal to the equilibrium pressures in both the initial and final equilibrium states of the system) $$\Delta H=\delta Q$$