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Offline orthoformate

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Free Energy
« on: August 13, 2015, 08:22:07 PM »
For the question:

One mole of an ideal gas expands isothermally until it's volume is doubled. what is the change in Gibbs energy for the process?

My first thought was:

ΔG=ΔH-TΔS and ΔH= ΔU-PV so ΔH=-PV

ΔH=-(nRT/V)dV-(nrT/P)dP

integrating

ΔH=-nRTln(V)-nRTln(P)

ΔH=-nRT(2)-nRT(1/2)

ΔS=dQrev/T and q=w so:

nRTln(2)/T= -nRln(2)

so ΔG= -nRT(2)-nRT(1/2)+nRln(2)

I'm getting lost! ??? can anyone point me in the right direction?

Thanks!

Offline Corribus

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Re: Free Energy
« Reply #1 on: August 13, 2015, 09:11:12 PM »
I hate problems like this because it's never clearly defined what you're supposed to use as a starting point.

But, if we start with ΔG = ΔH - TΔS. The enthalpy change for expansion of an ideal gas is 0 because the internal energy change is zero and the Δ(pV) is also zero. This is a good thing to remember: enthalpy change for isothermal process is equal to zero. This just leaves the entropy term. ΔS = nR ln (Vf/Vi). So ΔG = - nRT ln (Vf/Vi). n = 1 and the volume fraction is 2, so the answer is - RT ln (2).

Are you supposed to derive the ΔS term or is it ok just to remember it? *shrug*

(Another useful thing to remember, as shown here, is that the expansion of an ideal gas is completely entropic. This is not the case, by the way, for a real gas, where intermolecular interactions make the enthalpic change nonzero as the gas particles are compressed or decompressed.)
« Last Edit: August 13, 2015, 09:22:51 PM by Corribus »
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Offline orthoformate

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Re: Free Energy
« Reply #2 on: August 13, 2015, 10:59:59 PM »
Hey Corribus,

Thanks for the response  :)

Would using the equation ΔG=RTlnK also be acceptable? In which case:

ΔG=-RTln(Q2/Q1)

Q= mol/L

mol/L=P/RT

Assuming no Δn in the reaction and since PV=constant

ΔG=-RTln(1/2)

Does this make sense?

I also wanted to ask you about Enthalpy at constant temperature

ΔH=U+PV we know in this case that U=0 so;

ΔH=PV and PV=nRT so ΔH=nRΔT which =0?

My instinct is telling me:

if U=0 then ΔH=PV; meaning the enthalpy is equal to the pressure volume work done on the system. Why is this not the case?

Thanks for your help. I am studying for the Chemistry GRE, and I am having some difficulty to say the least!


Offline Corribus

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Re: Free Energy
« Reply #3 on: August 14, 2015, 01:17:52 AM »
Quote
Would using the equation ΔG=RTlnK also be acceptable?
Of course, all these equations are ultimately related to each other and derived from each other in some way. That said, this would seem to be an unconventional way to go about solving this problem, since this equation is most often employed in the process of managing a chemical equilibrium - that is, a chemical reaction or process. What exactly is the equilibrium here, and what exactly is the reaction quotient in the context of an isothermal expansion? I'm not really seeing it.

Also, bear in mind that it is not DG that equals RT ln K, but DG0, the standard Gibbs energy. It is important to remember this distinction.

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if U=0 then ΔH=PV; meaning the enthalpy is equal to the pressure volume work done on the system. Why is this not the case?

Work is applying a force over a certain distance. An example of pressure-volume work is expanding a volume of a certain quantity of a gas against an external pressure - such a process requires energy input into the system, often in the form of heat, because you are working against the external pressure. In an expansion of a real gas, there's an extra element of "work" being done against the intermolecular forces of attraction, say, that are trying to hold gas particles together. But for an ideal gas, there are no such interactions, and the internal energy only depends on the temperature - i.e., kinetic energy.

With that in mind, there are two basic ways an ideal gas can expand.

1. Imagine a case where the system is perfectly insulated from the surroundings. In this case, there can be no transfer of energy between particles in the system and the outside environment. When the volume expands against the external pressure, it takes work to do this, and that energy has to come from somewhere - so it comes from the particles in the system. Internal energy is lost from the particles because kinetic energy (the only form of energy available in an ideal gas) is lost. And therefore the temperature goes down. In an adiabatic expansion, the temperature decreases - here the gas (the system) does work against the surroundings. 

The only exception to this is if the expansion occurs into vacuum. Because the system is not working against a pressure here, no energy is lost, and therefore the temperature does not change.

Note that in an adiabatic expansion, there is no heat exchanged or lost from the system to or from the environment, and therefore the entropy doesn't change. An adiabatic expansion is therefore a fully reversible process.

(Of course, there's no such thing as an adiabatic expansion, because no container is perfectly insulated. Heat is always exchanged. Entropy always increases. So true reversibility is a pipe dream.)

2. Now instead of insulating the container, let's let heat flow freely. In this case, the expansion occurs as before, but because heat can flow, the energy to do the work no longer comes from the expanding gas, but rather it comes from the external environment. In this case, the internal energy doesn't change (the energy to do the expansion does not come from the expanding gas particles - or, the heat that is lost from the expanding particles is immediately compensated by transfer of energy from the environment to the particles). As such, the temperature also doesn't drop, because temperature is just a measure of the average kinetic energy of the gas particles, which isn't changing. As long as the heat transfer rate is fast, and the environment has a large reservoir of heat, the temperature in the system will remain constant. Because the energy to expand doesn't come from the system, the system does not do work against the surroundings. There is no pressure-volume work done. Δ(pV) is equal to zero, because the temperature isn't changing - if ΔU = 0, then ΔH = Δ(PV) as you said, but Δ(PV) = nRΔT, and ΔT isn't changing because we allow the heat to flow freely, and therefore Δ(PV) = 0 as well.

That's my take on it, anyway.

(The interesting thing is that it should be apparent that for an isothermal process, the force responsible for expansion is purely entropy. There are no physical forces between gas particles. The only reason a gas expands is basically because of a statistical drive for particles to fill the largest volume possible. But still, unless the expansion is into a vacuum, there will be a pressure holding those expanding gas particles back. So, you have gas particles that want to expand, but there are other particles pushing them back. There is of course an equilibrium point where the drive to push outward is balanced by the pressure pushing inward, at which point there is no more spontaneous expansion. This makes sense - if gas in one container is expanding outward, gas in the surrounding container is being compressed. In the absence of any enthalpic forces of consideration, the point of balance is where the entropy of the metasystem - or universe, if you want to think that big - is maximized. As you may imagine, this point of "entropic equilibrium" is linked closely to the concept of "thermal equilibrium" - because at the same point that neither "container" of gas is expanding or contracting, no thermal energy is any longer being exchanged.)

Quote
I am studying for the Chemistry GRE, and I am having some difficulty to say the least!
If it makes you feel better, I hold the rather unconventional opinion that thermodynamics is 100x harder to understand than quantum mechanics. I feel thermodynamics is just one long circular chain of mutually interdependent equations, such that I still can't figure out what the starting point is supposed to be. My usual approach is just to pick one at random and hope I can get to an equally random ending point somehow. :)
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

Offline Corribus

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Re: Free Energy
« Reply #4 on: August 14, 2015, 10:26:44 AM »
*Update*

Amusingly enough today's comic at SMBC was pertinent to the discussion:

http://www.smbc-comics.com/index.php?id=3828

While there is no such thing as a perfect insulator, we can get pretty close to an adiabatic condition if we can allow a gas to expand fast enough that the heat transfer from the environment to the expanding gas is relatively slow. Of course, temperature will eventually equilibrate, but in the short term, the expansion can be practically done without heat flow into the system, in which case the adiabatic condition prevails momentarily. One way that this can be done is to force a (real) gas through a small, insulated opening very rapidly... such as, say, blowing air out of a small opening in your lips. Try it - if you just breathe on your hand, it feels hot. But force the air out through a small opening, it feels cold. This is called the Joule-Thomson Effect. Note that this only occurs for a real gas, and the temperature loss is due to the energy required to overcome intermolecular attractive forces between gas molecules. It is distinguished from the (reversible) isentropic adiabatic expansion described above, where the temperature change of the gas derives purely from the work the gas particles do against the external pressure.

Anyway, it obviously takes energy to force the gas through a small opening, and the cooled gas also eventually draws heat away from the surrounding environment, because once it is throttled, it no longer is insulated. Hence superman's cooling breath.
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

Offline orthoformate

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Re: Free Energy
« Reply #5 on: August 14, 2015, 09:34:30 PM »
That is very interesting, and the cartoon is very silly haha.

I have been thinking for a really long time today about enthalpy at constant temperature, and here is what I have:

in the case of constant Temperature w=q since ΔU=0

assuming that the external pressure=0 (perfect vacuum) then w=0

in this case then entropy is the only driving force: ΔS=dqrev/T

As you said before: ΔS=nRln(2) so ΔG=-nRTln(2)

For some reason the answer to this question (question 62: http://hky.njnu.edu.cn/jpkc/wzattach/103741_858626.pdf) is ΔG= RTln(1/2)

This makes me think that enthalpy must be making some contribution to ΔG, which implies that the box is not expanding into perfect vacuum.

so H=pdV

H=(nRT/V)V

H=nRTln(2)

this doesn't match the answer either...do you think they made a mistake?
« Last Edit: August 14, 2015, 09:45:29 PM by orthoformate »

Offline orthoformate

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Re: Free Energy
« Reply #6 on: August 14, 2015, 09:43:26 PM »
A thought occurs:

Since H=U+PV then δH=δU+PδV+VδP

which at constant temperature..

ΔH= nRTln(2)+nRTln(1/2)

and we know ΔS=nRln(2)

so ΔG=nRTln(2)+nRTln(1/2)-nRTln(2)

ΔG=nRTln(1/2)

The only reason I have a problem with this is that I was always taught to calculate the work (and in this case q) at constant temperature using this method:

w=pdV

w=(nRTV/V)

integrating

w=nRTln(2)

which is correct?



Offline Corribus

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Re: Free Energy
« Reply #7 on: August 17, 2015, 09:56:46 AM »
As you said before: ΔS=nRln(2) so ΔG=-nRTln(2)

For some reason the answer to this question (question 62: http://hky.njnu.edu.cn/jpkc/wzattach/103741_858626.pdf) is ΔG= RTln(1/2)

The two answers are equivalent. -ln(2) and ln(1/2) are the same thing. You can ignore the n in my answer, since it is equal to 1.
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

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