Chemical potential in thermodynamics


It's intuitive if you are careful in remembering what is held constant.

$$\mathrm{d}U = T\mathrm{d}S - p\mathrm{d}V + \mu \mathrm{d}N$$


$$\mu = \left(\frac{\partial U}{\partial N}\right)_{S,V}$$

means that the entropy and the volume are kept constant. Remember that entropy is a measure of how many microstates you have available to realize the macroscopic state. Usually if you add more particles the entropy increases. You can interchange the (n+1)th particle to with one of the other $n$ particles and have the same macrostate, therefore entropy must increase. So, just adding particles would naively increase the entropy. To keep it constant we need to change something else. Volume is kept constant, too. So the only option is Energy.

Energy has a positive derivative with respect to the entropy (i.e. $T$) which is intuitive because the number of microstates on increases with the energy (think how the circumference of the ellipse-shaped phasespace curve of a simple harmonic oscillator expands if you go to higher energies). So in order to keep the entropy constant, even though we added an additional particle (giving us more combinatorics), we need to decrease the energy.

Another way to see this (and explain why it's called chamical potential) is considering a reaction of particles of type 1 into particles of type 2. The total entropy, which is the sum of the entropies of the particle 1 and 2 systems, must increase in a spontaneous process of the total (isolated system).

$$\mathrm{d}S = \mathrm{d}S_1 + \mathrm{d}S_2 \geq 0$$

If energy and volume are kept constant, this expression becomes using $dN_2 = -dN_1$

$$-\mu_1\mathrm{d}N_1 -\mu_2\mathrm{d}N_2 = (\mu_2 - \mu_1)\mathrm{d}N_1 \geq 0 $$

i.e. the number of particles of type 1 only increases (which means the number of particles of type 2 decreases, as total particle number is conserved) if the "chemical potential" of particle 2 is larger. That means the chemical potential measures how much a system wants a certain particle type to react (and thus disappear, because it reacted into a different particle which in turn wants to react less)


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Updated on June 23, 2022


  • user45959
    user45959 less than a minute

    In many scenarios, on computing the partial derivative of the internal energy ($U$) with respect to mole number ($N$) is negative. This implies that adding more moles of the substance decreases the $U$ of the system.

    I find this strongly counter-intuitive. Shouldn't the derivative always be positive?