Comparing delta(PV) vs P(deltaV)

[ifihadsomebacon]

Going from:

E = q + w

To:

H =

E +

(PV)

I'm confused as to why you add the product of the pressure and volume of the system to the internal energy to get enthalpy. Is it just because "that's what enthalpy is defined as"? I think I understand that when holding pressure constant, the

(PV) becomes P

V and cancels with the w = P

V from

E, giving

H = q. If pressure is not constant, does that mean

H = q + w +

(PV) or

H = q + P

V +

(PV) is true? What does this mean, practically? I just am wondering why are there two "PV" expressions in the first place? What even is

(PV)? It doesn't make sense to me like P

V, what about

(PV) makes it so it can be added on initially?

I know I asked a lot of questions, I'm just trying to make my confusion clear. (Ha)

 

[stevendaryl]

The meaning of $\Delta(PV)$ is the change in the product of $P$ and $V$. For an ideal gas, $PV = NR T$, where $N$ is the number of molecules, and $R$ is a constant, so $\Delta (PV)$ is proportional to $\Delta T$.

The practical reason for the various thermodynamic quantities is because which one is most important depends on what is being held constant. If a system is isolated---no interaction with the environment--then internal energy is constant. If you have two or more subsystems that are in thermal contact (allowed to exchange heat, but nothing else), then the internal energy is a matter of keeping track of the heat flowing in and out of each system.

But now, if your experiment is held in a constant-pressure environment, then there is an additional way that internal energy can change: By the system expanding (doing work on the environment and thus decreasing its internal energy) or contracting (increasing its internal energy). So keeping track of internal energy is more complicated. If we switch to $H$, though, we're back to keeping track of heat in and out of each system.

 

[BvU]

If you consider $U$ as a function of $S$ and $p$, then $$dU = T\;dS - p\;dV\ .$$Changing to another thermodynamic potential is then a Legendre transform (here, p 15 ff). For $H = U+pV$ you naturally get $$dH = dU + d(pV) = T\;dS + V\;dp\ ,$$ thus making $S$ and $p$ the natural variables for $H$.

There's a whole jungle of these transforms and there are tricks to not lose orientation

 

[Chestermiller]

I think you answer "that's what enthalpy is defined as" is the best answer. Unlike internal energy U and entropy S, enthalpy is not a fundamental entity in thermodynamics, but, in many cases it is a convenient function to work with in many problems of practical interest, which you would learn once you begin working many problems. And, in many of these problems, the difference between $P\Delta V$ and $\Delta (PV)$ will be significant. So, just be patient.

 

[ifihadsomebacon]

The meaning of $\Delta(PV)$ is the change in the product of $P$ and $V$. For an ideal gas, $PV = NR T$, where $N$ is the number of molecules, and $R$ is a constant, so $\Delta (PV)$ is proportional to $\Delta T$.

The practical reason for the various thermodynamic quantities is because which one is most important depends on what is being held constant. If a system is isolated---no interaction with the environment--then internal energy is constant. If you have two or more subsystems that are in thermal contact (allowed to exchange heat, but nothing else), then the internal energy is a matter of keeping track of the heat flowing in and out of each system.

But now, if your experiment is held in a constant-pressure environment, then there is an additional way that internal energy can change: By the system expanding (doing work on the environment and thus decreasing its internal energy) or contracting (increasing its internal energy). So keeping track of internal energy is more complicated. If we switch to $H$, though, we're back to keeping track of heat in and out of each system.

So, is enthalpy just a term defined this way in order to simplify the "energy bookkeeping" for constant pressure reactions? I think I saw somewhere that the (+ PV) accounts for the energy required to push the atmosphere out of the way for the volume, which cannot be considered internal energy, but it is useful to consider. Does that sound familiar?

 

[Chestermiller]

So, is enthalpy just a term defined this way in order to simplify the "energy bookkeeping" for constant pressure reactions? I think I saw somewhere that the (+ PV) accounts for the energy required to push the atmosphere out of the way for the volume, which cannot be considered internal energy, but it is useful to consider. Does that sound familiar?

My advice to you is to delay reaching your decision regarding a physical interpretation of the enthalpy function until you have solved some actual problems first? Otherwise, in my judgment, you will come to learn that you've wasted your valuable time.