# Enthalpy vs helmholtz?

1. Oct 9, 2013

### iScience

Hi, i just wanted to confirm that i had the concepts right.

So, if i was on earth and i created something out of nothing (please just bear with me..), and i supplied all the heat that the created system had, then the total energy i would be supplying would be the enthalpy right?

and if i were to do the exact same thing, create a system out of nothing, except this time i let the environment give the system its heat, then the total energy i would be supplying would just be the work needed to create the system which is the helmholtz energy right?

and gibbs... i've looked and looked online and in my book but i'm still confused conceptually; it just says a process taking place at constant pressure and constant temperature, and then just gives..

ΔG=ΔH-TΔS

what does it mean?.. ie/and what is the gibbs free energy conceptually?

(i already read wikipedia please don't copy and paste)

2. Oct 9, 2013

### UltrafastPED

3. Oct 10, 2013

### Staff: Mentor

Enthalpy is the energy you have to supply to create the system from nothing, considering that you will have to push the atmosphere out to make room for it. So it is the internal energy of the system plus work against the atmosphere at constant P.

The Gibbs free energy is is the energy you have to supply to create the system from nothing, considering that you will have to push the atmosphere out to make room for it and that you can get some extra energy from the environment in the form of heat (i.e., create the system at 0 K and let the enviroment bring it to ambient T). So it is the internal energy of the system plus work against the atmosphere at constant P plus energy supplied by the environment at constant T.

For actual processes, it is of course possible that work will be done on the system and that heat will leave the system to the environment.

As UltrafastPED mentionned, the Gibbs free energy is what is left to do work after expansion/compression work at const. P and heat entering/leaving the system at const. T have been accounted for.

4. Oct 10, 2013

### BruceW

that hyperphysics link is pretty good.
$dG = dU - TdS + pdV$
$dF = dU - TdS$
$dH = dU + pdV$
So I guess the Gibb's free energy is the most general, if you want to know how much energy you need to create the system. But if you assume constant volume or constant entropy, then you can use one of the other free energies (since they will be the same as the Gibb's free energy in that case).

edit: wait, the equations are even better than that:
$G=U-TS+pV$
$F=U-TS$
$H=U+pV$

edit again: There's also the potential that takes into account changes in the number of molecules.

Last edited: Oct 10, 2013
5. Oct 10, 2013

### stevendaryl

Staff Emeritus
Why are there are all these different kinds of "energy"? Here's a rough idea what the point is.

Well, if you have a system of particles, and you let it reach equilibrium, there are several tendencies:
1. It tends to reach a state with lowest energy.
2. It tends to reach a state with highest entropy.
3. It tends to reach a state with largest volume.

However, since energy, entropy and volume are related--if you set any two of them, the third is determined--it's not possible to satisfy all three tendencies at once. So the system will reach some kind of compromise between those three tendencies. Exactly what compromise is found depends on the constraints acting on the system.

Let's look at a couple of examples. Consider the following setup: We have an insulated upright cylinder (no heat goes in or out). We have a moveable insulated plate that can move up and down in the cylinder, dividing it into two chambers, with variable volume. (So basically, we have a piston). We put a certain amount of gas into each chamber. We put pressure on the plate so that it moves slowly in one direction or the other. What is the equilibrium position of the plate?

Well, since no heat can flow in or out of either chamber, we know the entropy can't change (because for reversible processes, change in entropy is given by $\delta S = \delta Q/T$ where $\delta Q$ is heat added and $T$ is temperature). So the composite system (two chambers) will change so as to minimize its total energy. But the total volume can't change--it's constrained to be the volume of the cylinder.

So let $U_1(V_1)$ be the energy of one of the chambers when its volume is $V_1$ and let $U_2(V_2)$ be the energy of the other chamber when its volume is $V_2$.

Total energy : $U_1(V_1) + U_2(V_2) = E$.
Total volume : $V_1 + V_2 = V$

So we want to minimize $E$ subject to the constraint that $V$ is constant.
So to minimize one quantity subject to a constraint, we use the technique of Lagrange multipliers. Have you studied those in calculus? I'm hoping you did. If not, ask about them.

Let $\lambda$ be the multiplier for the volume constraint. We minimize

$U_1 + U_2 - \lambda(V_1 + V_2)$

with respect to $V_1$ and $V_2$ by taking derivatives and setting them to zero:

$\frac{\partial U_1}{\partial V_1} - \lambda = 0$
$\frac{\partial U_2}{\partial V_2} - \lambda = 0$

So using thermodynamics, $\frac{\partial U_1}{\partial V_1} = -P_1$, the negative of the pressure of the first chamber, and similarly, $\frac{\partial U_2}{\partial V_2} = -P_2$, the negative of the pressure of the second chamber. So these two equations tell us that the pressures are equal at equilibrium, and the multiplier $\lambda$ turns out to be $-P$, the negative of this common pressure.

So minimizing the total energy of the two chambers turns out to be equivalent to minimizing the quantity
$U - \lambda V = U + PV$ for each chamber separately, with $P$ held constant.

So in such a circumstance where the pressure is shared between the two subsystems, but heat is not, the thing that is minimized is enthalpy = $H = U+PV$ for each subsystem separately. Equilibrium between the two chambers is achieved when $H$ is minimum for each system, and $P$ is the same for the two systems.

Now, suppose we change the problem so that the total energy is held constant (no energy comes into or out of the two-chamber system), but the plate is now no longer an insulator--heat can flow between the two chambers. In this case, you can't minimize total energy, since that's constant. But you can maximize the entropy. So we want to maximize the total entropy $S$ subject to the constraints that the total energy and total volume are held constant.
Again using Lagrange multipliers, we want to maximize:

$S_1 + S_2 -\lambda_1 (U_1 + U_2) -\lambda_2 (V_1 + V_2)$

where $\lambda_1$ is the multiplier for the total energy constraint, and $\lambda_2$ is the multiplier for the total volume constraint. So maximizing gives us these equations:

$\frac{\partial S_1}{\partial U_1} - \lambda_1 = 0$
$\frac{\partial S_2}{\partial U_2} - \lambda_1 = 0$
$\frac{\partial S_1}{\partial V_1} - \lambda_2 = 0$
$\frac{\partial S_2}{\partial V_2} - \lambda_2 = 0$

Again, using thermodynamics, $\frac{\partial S_1}{\partial U_1} = 1/T_1[itex] where [itex]T_1$ is the temperature in the first chamber. $\frac{\partial S_2}{\partial U_2} = 1/T_2[itex] where [itex]T_2$ is the temperature in the second chamber. $\frac{\partial S_1}{\partial V_1} = P_1/T_1[itex] where [itex]P_1$ is the pressure in the first chamber. $\frac{\partial S_2}{\partial V_2} = P_2/T_2[itex] where [itex]P_2$ is the pressure in the second chamber. So our four equations tell us that at equilibrium:

$T_1 = T_2 = \lambda_1$
$P_1/T_1 = P_2/T_2 = \lambda_2$

So at equilibrium, the pressures and temperatures of the two chambers must be equal. So maximizing the total energy of the two chambers subject to the volume and total energy constraint is equivalent to maximizing the quantity:

$S - \lambda_1 U - \lambda_2 V = S - U/T - PV/T$

for each chamber separately, with $P$ and $T$ held constant. So let's define a new quantity, $G$, the Gibbs Free Energy, which is defined by:

$G = U+PV-TS$

Then maximizing the entropy subject to the volume and total energy constraints is equivalent to maximizing $S - U/T - PV/T = -G/T$ for each chamber separately, with $T$ and $P$ held constant. Maximizing $-G/T$ with $T$ held constant is the same thing as minimizing $G$.

So in such a situation, the two systems are in equilibrium if $G$ is minimum for each system separately and the temperature and pressures are equal.

6. Oct 14, 2013

### iScience

i'm really sorry guys, especially to stevendaryl, but i didn't understand the lagrange multiplier arguement, i don't learn that until a few weeks later though in my math class i'm currently in.

U=(Q-∫PdV)

H=(Q-∫PdV)+∫PdV

the '+∫PdV' term at the end refers to how much pressure work the system has to do against the environment and the '-∫PdV' term refers to the amount of work being done on the system. i guess the thing that's bugging me is, why do the two pressures have to be held constant?
can someone dive a little deeper into this? this is confusing me because if i have to keep both P's at constant pressure, how would the system be able to expand? one pressure has to be greater than the other and so if the pressures are both equal, there should be no net pressure making the work zero. but then if it's zero how is the system being expanded?...

i know i'm looking at this the wrong way. i just don't know what way to look at it from i guess? please help me i'm pretty lost

thanks

7. Oct 15, 2013

### Staff: Mentor

Note that you are simply removing the expansion/compression work term from U. The fact that it is work being done on or by the system is taken care of by the sign of ΔV.

P doesn't have to be constant. It is inside the integral, so it can vary. For instance, when you have an isothermal process in an ideal gas, you can substitute $P = n R T / V$, take $nRT$ out of the integral (since it is constant) and integrate what is left, namely $1/V$.

8. Oct 15, 2013

### iScience

but why do people say that enthalpy only applies for constant pressure?

9. Oct 15, 2013

### Staff: Mentor

It doesn't. The definition of enthalpy is simply $H \equiv U + PV$. What you can do when P is constant is to write
$$\Delta H = \Delta U + P \Delta V$$
That equation is only valid at constant pressure. When you look up tables of standard enthalpies for different processes, it is that change $\Delta H$ at constant pressure that is reported.

10. Oct 15, 2013

### stevendaryl

Staff Emeritus
It's not that enthalpy only applies for constant pressure--you can compute U+PV for any system at equilibrium. It's that when a system is held at constant pressure, the system is in equilibrium when the enthalpy is at its minimum.

11. Oct 15, 2013

### stevendaryl

Staff Emeritus
It's a little incorrect to write U as U = Q - ∫PdV. That's appropriate if you suddenly dump heat Q into the system, and then slowly change volume without adding or removing any heat. That's not the way things happen. The usual way that things are written is:

$\delta U = \delta Q - P \delta V$

which is only valid for tiny changes to the system's thermodynamic variables. If you want to write it as an integral, the correct way to write it is:

$U = \int (dQ - P dV)$ or even better $U = \int (T dS - P dV)$.

The equation for $H$ has a different problem: $H = U + PV$ so in terms of differentials,

$\delta H = \delta Q - V \delta P$

so in terms of integrals,

$\delta H = \int (dQ - V dP)$

This means that if the pressure is held constant (so $dP = 0$), all the heat put into the system goes towards increasing the enthalpy.

12. Oct 15, 2013

### stevendaryl

Staff Emeritus
Okay, well let's go back to the problem of two systems, $A$ and $B$, that are confined to share the same total volume $V$. The total internal energy for both systems is $U = U_A + U_B$. In delta form,

$\delta U = \delta U_A + \delta U_B = \delta Q_A - P_A \delta V_A + \delta Q_B - P_B \delta V_B$. At equilibrium, the total internal energy is at a minimum. But that doesn't mean that $U_A$ and $U_B$ are at a minimum. Because of the minus sign on $-P_A \delta V_A$, system $A$ can minimize its energy by expanding (increasing $V_A$). But if $V_A$ increases, then $V_B$ must decrease, since the two volumes add up to $V$. So the two systems can't both minimize their internal energy at the same time. So what to do? How do you minimize the total energy $U$?

Well, let's do a little trick of adding and subtracting $P V = P (V_A + V_B)$:

$U = U_A + U_B = U_A + P V_A + U_B + P V_B - (P V)$

$\delta U = \delta(U_A + P V_A) + \delta(U_B + P V_B) - \delta(P V)$

The latter quantity, $\delta(P V)$ is ZERO, if the both the pressure and total volume are held constant. So in the case of constant pressure and constant total volume, we have:

$\delta U = \delta H_A + \delta H_B$

where $H_A = U_A + P V_A$ and $H_B = U_B + P V_B$

So minimizing the total internal energy $U$ is the same, in this case, as minimizing $H_A$ and $H_B$, separately. So we conclude that in a case of constant pressure, the total internal energy is minimum when each subsystem's enthalpy is minimum (for that pressure).

So in the case of constant pressure and total volume, the total internal energy $\delta U$ is the sum of the

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