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 '16 at 10:47

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 $$\mu = \frac{ \partial F}{\partial N}$$

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    $\begingroup$ Ok so how would you do that for chemical potential? $\endgroup$ – DanielSank Jan 26 '16 at 17:11
  • $\begingroup$ 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 '16 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 '16 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 '16 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 '16 at 0:57

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