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How do I measure chemical potential of a gas?

  1. Dec 12, 2011 #1

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    How can I measure the chemical potential of a gas (not necessarily ideal), using calorimetry alone? I mean, without knowing any equations of state, being able to measure pressure, temperature, volume, and number of particles. Also, I can add a measured amount of heat energy to the gas using a paddlewheel or an electrical resistor.

    I can't think of a device like a thermometer or pressure gauge that would measure chemical potential, so I suppose it has to be measured by calculations involving the second law:

    [tex]dU=T\,dS-P\,dV+\mu\,dN[/tex]
     
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  3. Dec 14, 2011 #2
  4. Dec 15, 2011 #3

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    This link tells me how to calculate the chemical potential using a linear approximation, assuming certain derivatives of the chemical potential are constant. This is not what I asked for. I can measure temperature without calculating linear approximations assuming certain derivatives of temperature are constant. I measure it with a thermometer. I measure pressure with a pressure gauge. I measure volume with a ruler. I measure particle number with a scale (assuming I know the mass per particle).

    I can accept that only differences in chemical potential can be measured, so its ok to specify the chemical potential of some standard state. I cannot accept that one of the "constant" coefficients [itex]\beta=(\partial V/\partial n)_{T,p}[/itex] is constant. It is generally not.

    So again, how do I measure the chemical potential, without assumptions? I have no problem making a series of measurements and then doing some calculations with those measurements to find the chemical potential (or chemical potential difference). But no approximations, no tables of previously measured values for the gas, etc.
     
  5. Dec 15, 2011 #4

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    You probably know how to calculate Gibbs free energy G=U-TS+pV. U and S can be calculated by integrating over heat capacity C_V (/T) from T=0 to T. Then use that Gibbs energy is a homogeneous function of particle number to get [itex]G=\mu N [/itex].
     
  6. Dec 15, 2011 #5

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    I assume you mean T=0 C, or T=some standard temperature, not 0 Kelvin. I was hoping for a method that did not require a whole set of measurements integrating from a standard state to the state in question.

    I've been thinking about this and this is about the best I have so far:

    If I have a container of gas with a particular pressure, temperature, volume, particle number, I can measure its temperature with an ideal gas thermometer. Thermally connect a small volume of ideal gas to the test gas, keep the pressure and particle number constant, and the volume of the gas will be proportional to the temperature by PV=NkT. To measure the pressure, mechanically connect a small volume containing an ideal gas to the test gas with a piston that disconnects the two gases and moves until the pressure is the same on either side. Set the temperature and particle number of the small volume, and the volume will be inversely proportional to pressure. Its a pressure gauge.

    I guess to make a device to measure chemical potential, you have to have a small volume connected by a membrane that is permeable to the gas particles, but does not move nor conduct heat. The problem is, the gas in the measuring device cannot be selected, it has to be the same as the gas being tested. But if you set the temperature very high and pressure very low, it will behave as an ideal gas. So you fix the temperature and pressure in the measuring volume, and the volume will change to be proportional to the number of particles. The chemical potential of the measuring gas will be equal to the chemical potential of the test gas, and it can be calculated from T, V, P using the equation for chemical potential of an ideal gas.
     
  7. Dec 16, 2011 #6

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    No, I really meant T=0K.
    Obviously it is easier to define the chemical potential relative to a standard state which corresponds to an infinitely dilute gas, extrapolated to some standard pressure. In fact that is how fugacity is defined. From [itex]d\mu=vdp-sdT [/itex] where v and s are the partial molar volume and entropy, respectively, you get [itex] \mu(T,p)=\mu_0(T,p_0^*)+\int_{p_0^*}^p vdp[/itex]. You can chose [itex]p^*_0[/itex] to be so small that it falls in the region where the gas behaves like an ideal gas.
    In much the same way you can get the temperature dependence, although it is easier to formulate it in terms of h than of s.
    The harder part is to get the constant [itex]\mu_0[/itex], that's what my last post was about, but it has been tabulated for all elements, so you don't have to bother to much about it.

    Btw. it is not much easier to determine the thermodynamical temperature once you don't assume the ideal gas law a priori.
     
    Last edited: Dec 16, 2011
  8. Dec 16, 2011 #7

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    You wrote "but it has been tabulated for all elements, so you don't have to bother too much about it." This is not my question. If I was asked to describe a technique for calculating the logarithm of a number, the correct answer would not be "punch the number into a calculator and hit the log button". I'm not looking for a specific answer, I am looking for a technique which does not rely on tables of previous measurements. e.g., how were those tabulated values measured?

    [itex]\mu(T,p)=\mu_0(T,p_0^*)+\int_{p_0^*}^p v\,dp[/itex], that's more like what I am looking for. I am not too worried about [itex]\mu_0(T,p_0^*)[/itex], I think its like entropy, you can't really measure it in general without worrying about the situation near T=0 K. Only differences can be measured otherwise. Even the equation for entropy of an ideal gas screws up near T=0.

    I am wondering if there is a conceptual device like a thermometer or a pressure gauge that will measure chemical potential. The idea I had above of a small measuring system with a barrier permeable to the gas particles, but not to pressure and temperature is not quite right, I now realize. The chemical potential of an ideal gas in terms of pressure and temperature is [itex]\mu=-kT\,\ln(T^{5/2}/p)+\varphi[/itex] where [itex]\varphi[/itex] is some constant. (Not using molar quantities). No volume is involved. So I guess you just fix the volume and set the temperature of the measuring system very high and measure pressure, or set the pressure very low and measure temperature and calculate the chemical potential of the measuring system, which will be equal to the chemical potential of the measured system.
     
  9. Dec 16, 2011 #8

    DrDu

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    I understood that, but as I showed you you can calculate it from measured C_V values via S and U.
    There is no problem to determine S as C_V/T goes to 0 in the limit T to 0 and S(T=0)=3 by the third law so that integration is not a problem.

    Quite a direkt way to measure chemical potential is via an electrolytic cell like e.g. the lambda sensors to measure the chemical potential of O2 in car exhausts.
     
  10. Dec 16, 2011 #9

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    Sorry if I sound aggravated, but I have been discussing this with people and I have difficulty getting the point across that I am not trying to do thermodynamics, I'm trying to understand it.

    I wonder if you could give a quick explanation of why S(0)=3 by the third law, I don't understand that. I think I understand the integration process you have mentioned, thanks for that insight.

    I will also look into lambda sensors, thanks for that lead.
     
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