Confused - chemical potential of an ideal classical gas

AI Thread Summary
The discussion revolves around the confusion regarding the chemical potential of an ideal classical gas, particularly its behavior at varying temperatures. It highlights that while adding particles to a system typically increases energy, the chemical potential can become negative at higher temperatures due to the influence of entropy. The conversation emphasizes that the chemical potential is defined by the relationship between energy and the number of particles, considering fixed entropy and volume. Additionally, it notes that the chemical potential is relevant in various contexts, including systems with different chemical species and external potentials. Overall, understanding the interplay between energy, entropy, and the number of particles clarifies the behavior of chemical potential in thermodynamic systems.
Mireno
Messages
3
Reaction score
0
Can anyone help me. I am very confused about the chemical potential.

In the following equation

dU = TdS - pdV + u dN, where u is the chemical potential

it seems to me that if you add particles to a system you are increasing the energy of that system, i.e. the chemical potential is positive.

Why is it then that the chemical potential is negative above a certain temperature (as can be seen by taken the derivative of the canonical partition function with respect to the number of particles (times -1/kT))?

This would seem to suggest that as you add particles to a system, the energy of the system decreases!
 
Physics news on Phys.org
Mireno said:
Can anyone help me. I am very confused about the chemical potential.

In the following equation

dU = TdS - pdV + u dN, where u is the chemical potential

it seems to me that if you add particles to a system you are increasing the energy of that system, i.e. the chemical potential is positive.

Why would you say that ?

Why is it then that the chemical potential is negative above a certain temperature (as can be seen by taken the derivative of the canonical partition function with respect to the number of particles (times -1/kT))?

This would seem to suggest that as you add particles to a system, the energy of the system decreases!

There's nothing wrong with that.

You can put your "zero" for energy wherever you want, what counts are *differences* in energy (at least in the classical setting where we are in this case).

Consider mechanical potential energy, for instance. Suppose that you put the "0" at floor level. That means that below floor level, in the basement, potential energy is negative. If you add more blocks of lead in your basement, the total potential energy decreases. Nothing wrong with it. With chemical potential, it is the same...
 
Mireno said:
In the following equation

dU = TdS - pdV + u dN, where u is the chemical potential

it seems to me that if you add particles to a system you are increasing the energy of that system, i.e. the chemical potential is positive.

Why is it then that the chemical potential is negative above a certain temperature (as can be seen by taken the derivative of the canonical partition function with respect to the number of particles (times -1/kT))?

This would seem to suggest that as you add particles to a system, the energy of the system decreases!

Hi Mireno, welcome to PF. Watch out, you're ignoring the other variables in that equation (most significantly, entropy). And entropy has an increasing influence on system behavior with increasing temperature.

Consider the Gibbs free energy, criterion of spontaneous processes at constant temperature and pressure, defined as G=U-TS+PV. If I add a protruding atom to a formerly flat plane of atoms, for example, the increase in energy could be small compared to the increase in entropy (or rather, TS) because of the many possible equivalent positions of the new atom. The chemical potential \mu=(\partial G/\partial N)_\mathrm{T,P} would therefore be negative for this system. Does this make sense?
 
What if you use dU = TdS - pdV + udN.

Then u = the partial derivative of U w.r.t. N (WITH S FIXED).

Mireno (aka "Morrie")
 
Mireno said:
What if you use dU = TdS - pdV + udN.

Then u = the partial derivative of U w.r.t. N (WITH S FIXED).

Mireno (aka "Morrie")

It's the same chemical potential:

\mu=\left(\frac{\partial U}{\partial N}\right)_\mathrm{S,V}=\left(\frac{\partial G}{\partial N}\right)_\mathrm{T,P}

In this case, to maintain a constant entropy, you're going to need to cool the system to counteract the entropy of adding the particle. Depending on the magnitude of this entropy, the energy removed via cooling may be larger than the energy added by the particle, which will make the chemical potential negative again.
 
Mapes (is that your name or a pseudonym?)

Thanks for your help. By way of background, I am a 48 year old revising my university (college) physics - currently 1/2 way through 3rd year. I kept meticulous notes in my university years. This is one problem that has caused me trouble and that I have not been able to nut out using Reif (my text) or the Internet.

I take it that the chemical potential is also used in circumstances where there are different chemical species or where electrical (potential) or gravitational (potential) effects, etc, need to be taken into account.

Again

Many thanks for your help.
 
Hello everyone, Consider the problem in which a car is told to travel at 30 km/h for L kilometers and then at 60 km/h for another L kilometers. Next, you are asked to determine the average speed. My question is: although we know that the average speed in this case is the harmonic mean of the two speeds, is it also possible to state that the average speed over this 2L-kilometer stretch can be obtained as a weighted average of the two speeds? Best regards, DaTario
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...
Back
Top