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### Topic: Isothermal enthalpy  (Read 3423 times)

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#### orthoformate

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##### Isothermal enthalpy
« on: August 15, 2015, 02:30:56 PM »
Hey everyone,

considering the change in enthalpy of an isothermal process brought up a question:

ΔU=0

q=w

work in an isothermal process is nRTln(Vf/Vi)

but when considering enthalpy

ΔH=ΔU+Λ(PV)

ΔH=pδV+vδP

pδV+vδP≠ nRTln(Vf/Vi)

ls this because, at constant temperature, enthalpy refers to the work done against a changing pressure plus VδP?

what is VδP in the physical sense? change in energy per L of expansion?

#### Corribus

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##### Re: Isothermal enthalpy
« Reply #1 on: August 17, 2015, 11:08:53 AM »
At first, I wouldn't get too caught up in what each term means, since to some degree it's artificial - resulting just from the derivative chain rule. The way I would think of it is this:

Enthalpy of a system is essentially a measure of how much energy it takes to create the system from nothing. The reason that H = U + PV is that if you want to create some system of particles, you have to supply (A) enough energy to create all the particles in the system at the specified conditions (for an ideal gas, particles only have kinetic energy, so this is the only energy you have to supply) plus their interactions with each other (for an ideal gas, there are no interactions) and (B) enough energy to create the space for the system of particles to occupy. (A) is U and (B) is PV. It goes without saying that the more space a system is going to occupy, and the higher pressure the surrounding environment has, the more energy it's going to take to create the system.

(Note that, of course, this isn't a full picture of what it takes to create a system, because entropy must invariably be considered.)

Calculating the enthalpy change for a process, is in some ways the same thing, conceptually, that we do when calculating heats of formation: if a system is changing from state A to state B, then the amount of energy it takes about to process this change is the same as the difference in energy that it takes to create the system in state A and in state B, each from nothing.

If a process of transitioning a system composed of an ideal gas between two states is done isothermally, the amount of straight-up energy it costs is equal to zero. Consider the two terms (A) U and (B) PV separately.

(A) In an ideal gas, the only energy involved in creating the system of particles itself goes toward supplying their kinetic energy. The kinetic energy is only a funciton of temperature. Therefore if two systems of identical particles have the same temperature, they have the same internal energy. ΔU = 0.

(B) If the process is an expansion, then the system after expansion takes up more space than the system before expansion. It takes more energy to create a larger system. Therefore all things being equal, we might be inclined to think we have to use some of the system's energy to increase its volume. Yes, this would be the case with some types of expansions. In the event the system is insulated from its surroundings (adiabatic process), the energy would come from the moving particles - and therefore their velocity, temperature and internal energy would all drop. However, if the system is NOT insulated, then the environment and the system can achieve thermal equilibrium. And what happens is that while it does take energy to effect the expansion, this energy can be supplied by the environment rather than system. If you want to think of it as a sequential process, you could imagine that the system expands, which causes the particles in the system to slow (and lose energy/temperature), but then heat from the environment will spontaneously flow back into the system to compensate. In effect, therefore, it costs no energy to bring about the expansion: whatever is lost to create the volume needed to bring about the expansion, is paid back by the environment. Insofar as enthalpy is a type of thermodynamic potential, the only thing that determines whether an isothermal process will happen spontaneously is the entropy, because the enthalpy change is a wash.

(As far as the equation goes, I think it's important to realize that when the system expands, the environment changes as well. If the system is insulated, then while the temperature of the system decreases (volume expanding), the temperature of the environment must be increasing (volume shrinking). If the system is NOT insulated, then the temperature of the system and surroundings are in equilibrium and unchanging - at least on long timescales. The volume of the system increases (and its pressure decreases - because temperature is constant) but the volume of surroundings decreases (and its pressure increases - again because the temperature is constant). Since both systems are ideal, then the volume / pressure changes of the system match exactly the volume / pressure changes of the surroundings. Therefore while it might be tempting to view this all as unfair from the environment's point of view - it has to supply energy to help the system expand - you might also realize that the environment is losing volume, which is itself a sort of energy (enthalpy) gain, since it took energy to occupy that extra volume. So another way to view it might be that the energy it takes for the system to gain volume is paid for the by the energy savings the environment makes by losing volume. It should be easy then to see, at least in a conceptual way, how PdV and VdP exactly cancel out in an isothermal expansion - that is, a situation that allows free energy flow between the system and the surroundings. PdV and VdP cancel out, so that the enthalpy change for an isothermal process is zero. Of course, this all presumes that both the system and the environment behave ideally. Do note that this is sort of the conceptual understanding I've formulated for myself over the years. There may be better ways to visualize these things, so alternative opinions/explanations are welcome.)
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

#### orthoformate

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##### Re: Isothermal enthalpy
« Reply #2 on: August 17, 2015, 07:49:29 PM »
Thank you, that was helpful. The math makes perfect sense now
As for the physical understanding, I think the picture will get clearer and clearer with time...I hope