Let's suppose I put you inside a room filled with a gas. You can measure its pressure directly with a barometer. You can measure its temperature directly with a thermometer. Can you measure its chemical potential directly with a "chemical potentiometer?"

To be clear: You have to measure it directly. You aren't allowed to measure other quantities then use an equation of state or whatever.

  • $\begingroup$ Chemical potential is the degree to which a system's entropy could increase if it were allowed to have more particles. See here for some discussion. This is an interesting post. $\endgroup$
    – DanielSank
    Jan 26, 2016 at 10:47

2 Answers 2


No, this is not possible at least in the way you are implying.

Chemical potential as temperature are abstractions that apply to large systems as a whole, because they depend on equilibrium conditions, or in other words, they rely on fluctuations being much smaller than average values.

You don't measure temperature directly either, once your thermometer is in equilibrium with its surroundings you measure the change in volume (or other directly measurable property) which you can relate to temperature because you have a theory establishing such inter-dependences.

So if you are understanding direct temperature measurement in such way, then yes, you can use a thermometer in such a way, and scale it in terms of average particle energy, making assumptions on the surroundings the same way you do for thermometers.

How concretely can you do this?

First how does the typical thermometer work? The standard Mercury thermometer would contains isolated liquid mercury, which is very incompressible but has a good thermal expansion coefficient. Thus it responds visibly and linearly to temperature changes by expanding/contracting.

Since is in a closed environment it can only exchange heat with the exterior, and the number of particles will be constant. Therefore the changes in $T$ due to heat exchange with the exterior

In these conditions the Helmholtz Free Energy is the thermodynamic potential, or the state function for the liquid here. And in any change of temperature or volume, the systems's Free Energy remains 0, since if it were different from zero, it means it can potentially exert some work and expand/contract further until this value was zero. So this explains already shows how $\Delta T\propto \Delta V$, since $$dF = 0 = -SdT - PdV$$ and we know pressure in the liquid stays constant (it does not expand nor contracts given the freedom to do so) so entropy has to be constant.

The relevant property here is $\Delta T\propto \Delta V$, which allows to relate volume changes to temperature changes, and hence measure the equilibrium conditions established between the thermometer and the measured environment.

For the Chem. Potentiometer, using this same reasoning and writing the full expression of $dF$ for a system according to Gibbs: $$dF = -SdT -PdV +\mu dN$$ we can use this expression to calibrate our instrument. Since in the is the same thermometer, $dN$ will be zero, but knowing the values of $T$ and $V$ we can know the state of the system by evaluating $F(T,V,N)$, and the derivative of this function w.r.t. $N$ will give the chemical potential values $$\mu = \frac{ \partial F}{\partial N}$$

  • 3
    $\begingroup$ Ok so how would you do that for chemical potential? $\endgroup$
    – DanielSank
    Jan 26, 2016 at 17:11
  • 1
    $\begingroup$ What @DanielSank said. Ok, so I can build a mercury thermometer because I know how to relate changes in volume with temperature. How would I apply the same concept to build a "chemical potentiometer?" $\endgroup$
    – Osport
    Jan 29, 2016 at 9:19
  • $\begingroup$ @rmhleo First you say the chemical potentiometer is the same physcal instrument as the thermometer, but then you say we let the number of particles vary. This seems inconsistent. $\endgroup$
    – DanielSank
    Jan 29, 2016 at 17:56
  • $\begingroup$ @DanielSank the instrument would be the same, with a fixed $N$. To draw a scale on it, we need to know its $F$ values, by having studied a separate equal system, in which can also vary $N$ and have measurements of $\mu$. But once calibrated, and drawn the scale, we don't have to vary $N$, we just put it in contact with the system to measure and let it equilibrate, and read the scale. $\endgroup$
    – rmhleo
    Jan 29, 2016 at 21:21
  • $\begingroup$ If the instrument is the same, then you're saying we just use a thermometer. Can I read chemical potential from my thermometer? I still don't get it. $\endgroup$
    – DanielSank
    Jan 30, 2016 at 0:57

This is a very interesting question, I'll try.

First let's compare to pressure measurement. Let's forget our microscopical knowledge that pressure is also force over area. Let's focus just on what thermodynamics tells us.

That P is the thing that is equalized between two systems separated by a movable barrier. And volume is what will change to accommodate that.

So we take a reference system (for example a box of air at the pressure of the atmosphere), and put an unknown system next to it through a movable (adiabatic and impermeable) barrier.

If the volume of the system increases, it has less P than 1 atm. If it decreases, more. If it stays the same, it was the same.

We can repeat this between our system and another one, and thus create an ordered scale of the P of every system we want.

Now we repeat that reasoning with chemical potential and it's conjugate, N.

We take a reference system of known N which will be our chemical potential reference standard. We connect an unknown system through a membrane permeable only to our particle in question.

In N of our system increases, it has less chemical potential than the reference, if it decreases, more, and if constant, the same.

And like before, we can compare against other systems to create an ordered scale of the chemical potential of any system.

That's it!

I'm not sure, however, how could we show that $P_1$ is "twice as much" as $P_2$. Same with chemical potential. I think it involves creating an arbitrary degree scale that relates changes in V with changes in P. Maybe someone else can clear that part.


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