Why does the enthelpy equation include work (PV term) twice?

[JeweliaHeart]

Hello. I am a thermodynamics novice trying to gain a better understanding of state functions, particularly enthalpy.

I understand that enthalpy is defined as

"A measure of the total energy of a thermodynamic system, including internal energy, which is the energy required to create a system, and the amount of energy required to make room for it by displacing its environment and establishing its volume and pressure."

The equation:

ΔH=ΔU(internal energy) + ΔPV

confuses me b/c


ΔU= q(heat added) - w(work done by system on environment)

so

ΔH really means:

ΔH=q - w + ΔPV


There are two terms of work (w and ΔPV) and b/c of the opposite sign, they cancel out, leaving only q. This means ΔH= q which is at odds with the accepted definition of enthalpy. Where did I mess up?

 

[DrDu]

That not at odds with the usual definition. Mostly enthalpy is determined by measuring heat.
However $\Delta(PV)=P\Delta V=w$ only if p is constant and volume work is the only kind of work the system is performing.

 

[JeweliaHeart]

That not at odds with the usual definition. Mostly enthalpy is determined by measuring heat.
However $\Delta(PV)=P\Delta V=w$ only if p is constant and volume work is the only kind of work the system is performing.

Okay,
so you are saying that enthalpy is only equivalent to heat if pressure is held constant?

Meaning,

ΔH=q - PΔV + PΔV= q (only when pressure is constant and only PV work is being exerted)

Otherwise, when pressure is not constant the equation should like this, perhaps?:

ΔH=q - w + PΔV

And the work defined by the 'w' above includes all forms of work, whether PV or mechanical, etc?

If so, that makes a little more sense. It's just that all the example problems I've encountered with ΔU only use PV work and no other form.

 

[DrDu]

ed)

Otherwise, when pressure is not constant the equation should like this, perhaps?:

ΔH=q - w + PΔV

In general, $d(PV)=PdV+VdP$. The second term will not vanish when P is not constant while the first term gives the volume work done in an infinitesimal step.
Hence $\Delta H=q-w+\int PdV +\int V dP$ in general.
If there is no work done other than volume work, this reduces to
$\Delta H=q+\int V dP$.
An example of non-volume work is e.g. the work done when stirring a viscous fluid.

 

[LudusRex]

Hello! First of all, great job on trying to understand enthalpy and the enthalpy equation. It can be a bit confusing at first, but with some clarification, I'm sure you'll have a better understanding in no time.

To answer your question, the reason why the enthalpy equation includes work (PV term) twice is because enthalpy is a state function, meaning it is dependent only on the current state of the system and not on the path taken to reach that state. This means that the enthalpy of a system is the same regardless of whether the work is done by the system or on the system.

Let's break down the enthalpy equation to better understand why this is the case. The first term, ΔU, represents the change in internal energy of the system. This includes the energy required to create the system and the energy required to make room for it by displacing its environment. The second term, ΔPV, represents the work done by the system on its surroundings or vice versa. This term is included twice because it takes into account both the work done on the system and the work done by the system.

Now, you may be wondering why the two work terms have opposite signs in the equation. This is because the work done on the system (ΔPV) is considered negative, while the work done by the system (w) is considered positive. This is due to the convention of sign conventions in thermodynamics, where work done on the system is considered negative and work done by the system is considered positive.

So, in summary, the reason why the enthalpy equation includes work (PV term) twice is because enthalpy is a state function and takes into account both the work done on the system and the work done by the system. I hope this explanation has helped clarify things for you. Keep up the good work in learning about thermodynamics!