Is enthalpy just the sum of internal energy and work against external pressure?

[Aurelius120]

Homework Statement

Gas expands from (1L, 10atm) to (4L, 5atm) against constant external pressure of 1 atm. The initial temperature is 300K and heat capacity is 50J/°C. Find change in enthalpy

Relevant Equations

$$1L.atm=100J\ (approx.)$$
$$\Delta U=Q+W$$
$$\Delta H=\Delta U+\Delta(PV)$$

This was the question

This is my solution

The problem arose after reading this post on PhysicsSE and this answer given

But first he must push away all the air that is in the way. This requires some work, $W=pV$
. In total, the energy he must spend is U+pV
. Let's call that enthalpy $H$
:$$H=U+pV.$$

So If I remember correct work done is $-P_{ext}\Delta V$
I don't understand why $$\Delta H=\Delta U+(5×4-1×10)L.atm$$
If that answer (the answer on the PSE post) is correct, the extra work to "push all that air that is in the way" should be $$P_{ext}(\Delta V)=1×(3)L.atm$$
The change in internal energy accounts for extra energy to change the volume, pressure and temperature of the gas so only extra work to push the air out should be as I calculated above(their sum being change in enthalpy) and the solution would be wrong?

 

[Chestermiller]

Your solution is correct, and, the work calculated in PhysicsSE is incorrect.

 

[Aurelius120]

Your solution is correct, and, the work calculated in PhysicsSE is incorrect.

But this answer on physics SE was the best definition/analogy for enthalpy. How do I understand enthalpy now? (logically/intuitively)

 

[Chestermiller]

It is not necessary to assign a physical interpretation to enthalpy. This description in SE is just a feeble attempt to do so (in my judgment). Again, in my judgment, enthalpy is just a convenient shorthand combination U and PV which occurs frequently in many analyses of thermodynamics. Unlike U and S, H is not a fundamental property of materials at thermodynamic equilibrium.

Of course, if you want to think of enthalpy the way they have interpreted it, there is nothing wrong with that.

 

[Aurelius120]

Again, in my judgment, enthalpy is just a convenient shorthand combination U and PV which occurs frequently in many analyses of thermodynamics. Unlike U and S, H is not a fundamental property of materials at thermodynamic equilibrium.

But $$\Delta G=\Delta H-T\Delta S$$
$\Delta G$ and $T\Delta S$ are well defined
$\Delta G$ is the amount of useful work done or the energy released/gained after both enthalpy and entropy changes are accounted for.
$T\Delta S$ is the amount of work/energy provided by the environment

From the equation we may conclude that $\Delta H$ is the net energy change/energy required for the process to happen.

Then my question becomes:
Why is the net energy change in a reaction/process equal to $\Delta U+\Delta (PV)$ ?
It is only logical that the energy change in a process is $\Delta U=Q+W$. It is not understood (by me) where the extra $\Delta(PV)$ change in energy comes from or goes to

 

[Chestermiller]

But $$\Delta G=\Delta H-T\Delta S$$
$\Delta G$ and $T\Delta S$ are well defined
$\Delta G$ is the amount of useful work done or the energy released/gained after both enthalpy and entropy changes are accounted for.
$T\Delta S$ is the amount of work/energy provided by the environment

From the equation we may conclude that $\Delta H$ is the net energy change/energy required for the process to happen.

Then my question becomes:
Why is the net energy change in a reaction/process equal to $\Delta U+\Delta (PV)$ ?
It is only logical that the energy change in a process is $\Delta U=Q+W$. It is not understood (by me) where the extra $\Delta(PV)$ change in energy comes from or goes to

You can interpret these things in any way that works for you. But the only way to really test your understanding is to solve some problems. I hope you are solving actual problems, rather than speculating forever about what these functions mean.