Thanks Jos. Valuable video. But I think Shannon's entropy is defined as: S = - sum p(j) log2 p(j), which a measure of information (gain), while what you wrote: S = - kB sum p(j) ln p(j) is the Gibbs entropy, the definition for the thermodynamic entropy for a system at equilibrium. The difference is the Boltmann constant and the logarithmic base. The first needed to define temperature from statistics, the natural logarithm because we are dealing with physical properties, not information properties. They have similarities, but I agree with A. Ben-Naim, Shannon misused the term entropy; information is not a state of equilibrium.
On the chemical potential, how do we get the value of it? Is it empirically determined? And what would the implication be if the chemical potential is positive or negative?
Dear Geoffry, thanks for your question. The concept of a chemical potential is not as clear-cut as many other thermodynamic parameters. However it is measurable: whenever you use a volt meter, you measure the chemical potential difference of electrons! For neutral particles, the measurement principle would be equivalent to that of a volt meter: the difference in chemical potential between two reservoirs is determined as the work needed to transfer a particle from one reservoir to the other. In practice that is not always easy to measure. A typical example of exchanging neutral particles is when you are sitting in an office and the temperature is the same as the outside temperature, but the relative humidity is different. That relative humidity is one-to-one related to the chemical potential of the water molecules, which itself is difficult to measure. Note that I only mention 'difference': as with all energies, the difference in chemical potential is what drives diffusion, there is a freedom in setting their absolute scale (i.e. where the zero chemical potential is).
@@josthijssen6782 Oh i see. The concept is kind of mysterious to me because I can't really tell how the chemical potential is determined for a given substance. Lets say there are two rooms (with the same temperature) separated by a partition. Room A is just filled with hydrogen gas. Room B is filled with the same quantity of hydrogen gas and n moles of (H2O vapor + CH4 gas + Xe gas). If the partition is removed, how would those extra gases diffuse related to their chemical potential? Which one has the higher/lower chemical potential? or are they all the same? can the value be zero? On the case of the solid, you were able to model the quantum mechanical potential well (like in the video). I wonder if we can derive the chemical potential of a substance from its microscopic property (intermolecular forces?) without having to directly measure it.... allowing us to make predictions
@@GeoffryGifari Dear Geoffry, The chemical potentials in your example all depend on the partial densities. For ideal gas mixtures, the chemical potential of species i can be calculated as mu_i = k T ln (P_i lambda^3/k T), where lambda is the thermal wavelength and P_i is the partial pressure: P_i = N_i k T/N, with N is the number of all particles.
@@josthijssen6782 I see. I pick those examples because water is polar, methane nonpolar, and xenon is the closest to ideal gas to see whether or not intermolecular forces have an effect. I was thinking in cases where both solid and vapor phases are involved (like in the video) that might cause a difference. It makes sense that for ideal gas it depends on partial pressure. Thank you for your insight.
@@GeoffryGifari Yes, I made it easy considering ideal gases. If you take interactions into account then expansions, mean field theory, fluid density theory, computer simulations.... will be necessary.
You left an error there, lucky it didn't cause any problems (because the derivative of 1 is 0). When using Lagrange multiplies each one has to be (lambda)(x-y), where x-y = 0. One of those is correct k*b(sum of energies) is indeed equal to (mew*k*b*num of particles). Now it's clear that mew = chemical potential. But the other one clearly isn't. -lambda(sum p) = -lambda and not 0! it should've had a 1 in there. So lambda(1-(sum of p)). It helped me, thanks, but when physics professors not only rush through math but also leave mistakes it really confuses me.
Dear Misteretc, I do not see your point, I'm afraid. Is your point that when we have a constraint sum_r,N p_N,r = 1, we should include a term - lambda (sum_r,N p_N,r -1) instead of just -lambda sum_r,N p_N,r? In that case, I should also have included a -kB beta and similar for . But because all these terms do not depend on the variables p_r,N I leave them out.
Sorry, yes that was my point, I haven't seen them left out before (I haven't been doing them for a long time, but I have gotten marks removed when I left them out in a function itself rather than the derivative). For the other case of and case, I thought you left them out for a different reason that is define mew as mew * N = -E, then it's clear that the 2 constraints are the same apart from 1 is multiplied by a constant. P.S Didn't expect a reply, thanks I appreciate it.
Thanks Jos. Valuable video. But I think Shannon's entropy is defined as: S = - sum p(j) log2 p(j), which a measure of information (gain), while what you wrote: S = - kB sum p(j) ln p(j) is the Gibbs entropy, the definition for the thermodynamic entropy for a system at equilibrium. The difference is the Boltmann constant and the logarithmic base. The first needed to define temperature from statistics, the natural logarithm because we are dealing with physical properties, not information properties. They have similarities, but I agree with A. Ben-Naim, Shannon misused the term entropy; information is not a state of equilibrium.
Great explanation!
On the chemical potential, how do we get the value of it? Is it empirically determined?
And what would the implication be if the chemical potential is positive or negative?
Dear Geoffry, thanks for your question. The concept of a chemical potential is not as clear-cut as many other thermodynamic parameters. However it is measurable: whenever you use a volt meter, you measure the chemical potential difference of electrons! For neutral particles, the measurement principle would be equivalent to that of a volt meter: the difference in chemical potential between two reservoirs is determined as the work needed to transfer a particle from one reservoir to the other. In practice that is not always easy to measure. A typical example of exchanging neutral particles is when you are sitting in an office and the temperature is the same as the outside temperature, but the relative humidity is different. That relative humidity is one-to-one related to the chemical potential of the water molecules, which itself is difficult to measure.
Note that I only mention 'difference': as with all energies, the difference in chemical potential is what drives diffusion, there is a freedom in setting their absolute scale (i.e. where the zero chemical potential is).
@@josthijssen6782 Oh i see. The concept is kind of mysterious to me because I can't really tell how the chemical potential is determined for a given substance. Lets say there are two rooms (with the same temperature) separated by a partition. Room A is just filled with hydrogen gas. Room B is filled with the same quantity of hydrogen gas and n moles of (H2O vapor + CH4 gas + Xe gas). If the partition is removed, how would those extra gases diffuse related to their chemical potential? Which one has the higher/lower chemical potential? or are they all the same? can the value be zero?
On the case of the solid, you were able to model the quantum mechanical potential well (like in the video). I wonder if we can derive the chemical potential of a substance from its microscopic property (intermolecular forces?) without having to directly measure it.... allowing us to make predictions
@@GeoffryGifari Dear Geoffry, The chemical potentials in your example all depend on the partial densities. For ideal gas mixtures, the chemical potential of species i can be calculated as mu_i = k T ln (P_i lambda^3/k T), where lambda is the thermal wavelength and P_i is the partial pressure: P_i = N_i k T/N, with N is the number of all particles.
@@josthijssen6782 I see. I pick those examples because water is polar, methane nonpolar, and xenon is the closest to ideal gas to see whether or not intermolecular forces have an effect. I was thinking in cases where both solid and vapor phases are involved (like in the video) that might cause a difference.
It makes sense that for ideal gas it depends on partial pressure.
Thank you for your insight.
@@GeoffryGifari Yes, I made it easy considering ideal gases. If you take interactions into account then expansions, mean field theory, fluid density theory, computer simulations.... will be necessary.
You left an error there, lucky it didn't cause any problems (because the derivative of 1 is 0). When using Lagrange multiplies each one has to be (lambda)(x-y), where x-y = 0.
One of those is correct k*b(sum of energies) is indeed equal to (mew*k*b*num of particles). Now it's clear that mew = chemical potential.
But the other one clearly isn't. -lambda(sum p) = -lambda and not 0! it should've had a 1 in there. So lambda(1-(sum of p)).
It helped me, thanks, but when physics professors not only rush through math but also leave mistakes it really confuses me.
Dear Misteretc, I do not see your point, I'm afraid. Is your point that when we have a constraint sum_r,N p_N,r = 1, we should include a term - lambda (sum_r,N p_N,r -1) instead of just -lambda sum_r,N p_N,r? In that case, I should also have included a -kB beta and similar for . But because all these terms do not depend on the variables p_r,N I leave them out.
Sorry, yes that was my point, I haven't seen them left out before (I haven't been doing them for a long time, but I have gotten marks removed when I left them out in a function itself rather than the derivative).
For the other case of and case, I thought you left them out for a different reason that is define mew as mew * N = -E, then it's clear that the 2 constraints are the same apart from 1 is multiplied by a constant.
P.S Didn't expect a reply, thanks I appreciate it.
Thanks for the derivation and the example. Could you also please provide reference for the example.
Thank you.
second! also great!
or should i say: Grand!