# 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?

• en.m.wikipedia.org/wiki/Thermodynamic_square I had not seen this until just recently, but I think it can be useful. – Tyberius Apr 19 '17 at 1:31
• Did you say that you teach thermodynamics to students? – Chet Miller Apr 19 '17 at 11:42
• Hi @ChesterMiller, yes, introductory, I do it from the kinetic molecular theory, the aim is to have that mentioned list to all the students with internet access. – Another.Chemist Apr 19 '17 at 12:05
• It seems to me that some of the equations are incorrect. – Chet Miller Apr 19 '17 at 12:46
• @ChesterMiller please, feel free to edit the post – Another.Chemist Apr 19 '17 at 12:53

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) $$\Delta H=\delta Q$$

• Thank you so much Chester, now I see clearer how to introduce the enthalpy concept :) – Another.Chemist Apr 20 '17 at 3:41

I believe that motivating the definition of enthalpy via the Legendre transform is indeed useful, and would like to attempt to do that here.

We start with the differential form of the first law, $\mathrm{d}U = T\,\mathrm{d}S - p\,\mathrm{d}V$. We see that $S$ and $V$ are the natural variables for $U$, in that the differential $\mathrm{d}U$ takes a simple "natural" form. In addition, if we're considering constant-volume processes, then $\mathrm{d}V = 0$, and $\mathrm{d}U = T\,\mathrm{d}S = \delta q_\text{rev}$, which is even simpler to work with.

Many processes, however, occur at constant pressure rather than at constant volume, and we'd like to work with a thermodynamic variable that undergoes the same simplification as does $U$ at constant $V$. This can be done by the Legendre transform*: $$\mathrm{d}H \equiv \mathrm{d}(U+PV) = \mathrm{d}U + \mathrm{d}(PV) = T\,\mathrm{d}S - p\,\mathrm{d}V + \mathrm{d}(PV) = T\,\mathrm{d}S + V\,\mathrm{d}P.$$ Adding the differential $\mathrm{d}(PV)$ on either side of the first law leads to a new thermodynamic variable $H$ that is a natural function of $S$ and $P$, instead of $S$ and $V$. We call $H$ the enthalpy. It easily follows that, for a constant-pressure process, $\mathrm{d}H = T\,\mathrm{d}S = \delta q_\text{rev}$.

*One must be careful about adding arbitrary differential elements and calling it a Legendre transform. To be precise, the reason that adding the differential $\mathrm{d}(PV)$ works is because $P(V)$ is monotonic; equivalently, the inverse function $V(P)$ exists; equivalently, a value of $P$ defines a single value of $V$ and vice versa. The Legendre transform works only between variables satisfying such a criteria, for otherwise information is lost during the transformation.

Incidentally, that $P(V)$ is monotonic comes from a variational treatment of the second law. The definition of enthalpy thus relies on the second law, even though it is commonly introduced after solely the first law.