2
$\begingroup$

I've been trying to understand the conceptual idea and purpose of enthalpy for the past two days.

From what I've learned, enthalpy is defined as:

$$E = U + PV$$

This leads me to my first question. Was this formula created or was it derived?

For example, the formula for kinetic energy can be derived from the equation for work (F*dx). So for enthalpy, is this formula derived from something else, or is it just a brand new idea that was simply created by someone as U + PV, because it's easier to solve this than find the internal energy for some processes (like in an open lab, where PV work is hard to measure)?

Next, the change in enthalpy is expanded as:

$$ΔE = ΔU + Δ(PV)$$ $$ΔE = Q + W + Δ(PV)$$ $$ΔE = Q - PΔV + Δ(PV)$$

From what I read, unlike internal energy, enthalpy becomes very useful when the pressure is constant. The explanations all say that this is because when pressure is constant, the -PΔV and Δ(PV) terms become equal to each other and cancel out, leaving only Q (heat).

This makes me ask: why does pressure need to be constant? If pressure is non-constant, couldn't they still cancel out? Just as a random example, if P = 2x, then wouldn't it be

$$ ΔE = Q -2xΔV + 2xΔV$$ thus cancelling out?


Secondly, why do we even need to measure enthalpy? If enthalpy is simply the heat transferred (Q) when pressure is constant, we're obviously simply ignoring the PV work done. So why introduce a new idea at all, instead of just continuing to use internal energy and heat (Q)? Many websites say things along the lines of:

"Sometimes it's hard to measure changes in internal energy because it's hard to measure the PV work done in a lab setting, since beakers and test tubes are open. So, we introduce enthalpy, which accounts for the work, and is simply heat transfer (when pressure is constant)"

Isn't the above statement just the same as saying:

"It's hard to measure work, so let's ignore work and measure heat instead"?

If so, then what benefit does measuring enthalpy give us, when we could just always measure heat transfer regardless of whether pressure is constant or not?


Please let me know if I should elaborate on my question if something is unclear. At this point, I'm so confused that I'm not even sure how to word my questions well. I spent the past 2 days trying to figure it out, but ended up getting even more confused and frustrated!

$\endgroup$
4
$\begingroup$

From what I've learned, enthalpy is defined as:

$$H = U + PV$$

This leads me to my first question. Was this formula created or was it derived?

I have taken the liberty of changing $E$ to $H$ as $H$ is the most common letter used for the enthalpy.

Enthalpy is a mathematically derived property. Other examples of mathematically derived properties are Gibbs Free Energy and Helmholtz Free Energy. Internal Energy $U$ is a fundamental thermodynamic property that comes to us from the first law of thermodynamics. Pressure $P$ and volume $V$ are physical thermodynamic properties. Since enthalpy is derived from other properties, it is also a thermodynamic property.

Enthalpy is one of four so called thermodynamic properties and is considered to be the energy needed to create a system (U) plus the work needed to make room for it (PV). It is a convenient property for, among other things, evaluating chemical reactions and for doing open systems analysis (e.g., evaluating control volumes in a Rankine steam cycle.)

From what I read, unlike internal energy, enthalpy becomes very useful when the pressure is constant. The explanations all say that this is because when pressure is constant, the -PΔV and Δ(PV) terms become equal to each other and cancel out, leaving only Q (heat). This makes me ask: why does pressure need to be constant?

Pressure does not have to be constant. Enthalpy is not only used for constant pressure processes, though one of its common applications is calculating heat transfer for phase changes where pressures and temperatures are constant. In such applications enthalpy is referred to as "latent heat".

Secondly, why do we even need to measure enthalpy? If enthalpy is simply the heat transferred (Q) when pressure is constant, we're obviously simply ignoring the PV work done. So why introduce a new idea at all, instead of just continuing to use internal energy and heat (Q)?

Because enthalpy is not just for measuring heat transfer. Examples are calculating the work done by adiabatic turbines, pumps, and compressors in steam cycle applications. In these applications no heat transfer occurs and there are PV and internal energy changes occurring. One example is calculating the output power of an adiabatic steam turbine in a steady flow open system

$$W_{T}= \dot m (h_{i}-h_{e})$$

Where $\dot m$ is the steam mass flow rate in kg/s, $h_i$ is the enthalpy of the steam entering the turbine in kJ/kg, and $h_{e}$ is the enthalpy of the steam exiting the turbine in kJ/kg. $W_{T}$ is then the power output in kilowatts. For convenience, values of enthalpy can be found in steam tables avoiding the need for calculations, except for liquid-vapor mixtures where calculations using steam quality are required.

If so, then what benefit does measuring enthalpy give us, when we could just always measure heat transfer regardless of whether pressure is constant or not?

The example I gave above for the steam turbine demonstrates that enthalpy does not "always measure heat transfer, regardless of whether the pressure is constant or not". In the steam turbine application the energy extracted from the steam to run the turbine comes from a combination of the internal energy of the steam and the PV changes that occur when the steam expands and pressure drops as the steam drives the turbine blades.

Hope this helps.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Yes. The most important thing to remember is that enthalpy (per unit mass) is a physical property of the material being processed, and is not directly related to any specific process that the material is experiencing. This is what we mean by a function of state. It really is just a very convenient property to work with in many types of problems. For that reason, it is often tabulated for many substances such as water/steam as a function of T, P, and phase. $\endgroup$ – Chet Miller Oct 1 '19 at 12:26
5
$\begingroup$

Pressure does not need to be constant, but unless you are doing an experiment in a close/pressurized (or depressurized) environment, the pressure at which a reaction occurs will be the ambient atmospheric pressure (which is effectively a constant). Moreover, if you are doing a reaction in an open vessel, the energy that you can usefully extract typically comes in the form of the heating of the material, which is what makes the enthalpy $H$ important.

As to why we define the enthalpy $H$, rather than just talking about heat transfer $Q$—the reason is that $H$ is a state function, while $Q$ is not. You can talk about the total enthalpy $U+PV$ of a system, just like you can talk about the total energy. You cannot talk about the "heat" of a system, or the "work" of the system. This makes enthalpy a more powerful tool.

(Your equation for what happens at a variable pressure is not really accurate, in part because it does not adequately distinguish which quantities are state variables—which are determined by the state of the system—and which are the ones, like heat and work, that can only be defined as transfers into or out of the system.)

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Upvoted for the very well explained first 2 paragraphs! However, I'm a little bit confused about the third paragraph. Could you explain further (maybe show some basic math) about what happens with variable pressure? Also, why do the terms even cancel out? I read that the PV term (not from W) is the energy required to push air out of the way for the system to exist. So, isn't this something completely different from the work done by/on the system after it exists, therefore they can't be equal? $\endgroup$ – F16Falcon Oct 1 '19 at 0:22
  • $\begingroup$ Enthalpy is also a very important function in flow (open) systems. When the OP learns about the open system (control volume) form of the first law of thermodynamics, he will learn that, for steady adiabatic flow through a porous plug or valve, the enthalpy per unit mass of the fluid does not change across the device.. And for steady adiabatic operation of compressors and turbines, the change in enthalpy per unit mass is equal to the shaft work. For all these different types of flow devices, the inlet pressure is not equal to the outlet pressure, so the pressure is not constant. $\endgroup$ – Chet Miller Oct 1 '19 at 1:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.