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How does one define a temperature scale given an equation of state $f(p,T,V)$? From what I have found, one must decide on another property which can be measured more directly than temperature, such as volume or pressure, and define the temperature change in terms of the measured quantity.

As an example: in OpenStax University Physics Volume 2, they give the following problem (problem 19 from chapter 3).

A gas follows $pV=bp+c_T$ on an isothermal curve, where p is the pressure, V is the volume, b is a constant, and c is a function of temperature. Show that a temperature scale under an isochoric process can be established with this gas and is identical to that of an ideal gas.

As far as I can tell, if we want to define a temperature scale with respect to some isochoric process, we can take the ratio of the system at two different pressures, which yields $$ \frac{p_2}{p_1} = \frac{c(T_2)}{c(T_1)}$$

This doesn't seem equivalent to what one finds for the ideal gas $$\frac{p_2}{p_1} = \frac{T_2}{T_1}$$

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  • $\begingroup$ I realize that I mention a specific problem, but my actual question at the bottom is more general. I don't believe this should have been closed. $\endgroup$
    – scmartin
    Commented Nov 7, 2022 at 19:02

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It is very correct of you that the scale is not unique. If you define it this way, it relies on a certain material. To avoid this we can consider ideal gas or the Efficiency of the heat engine.

And the scale varies if you use a different material.

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    $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented Nov 8, 2022 at 7:57
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The thermodynamic definition of temperature is not based on the equation of state, $p = p(N,V,T)$ but on the fundamental equation $S=S(E,V,N)$. According to the entropy equation temperature is $T$ is $$ \frac{1}{T} = \left(\frac{\partial S}{\partial E}\right)_{V,N} \tag{1}$$ and pressure is $$ \frac{p}{T} = \left(\frac{\partial S}{\partial V}\right)_{E,N} \tag{2} $$ If we eliminate $S$ between (1) and (2) we obtain the equation of state.

The main point is that the fundamental equation is the relationship between $S$, $E$, $V$ and $N$, which generates both the temperature and the equation of state.

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  • $\begingroup$ I know how to define the thermodynamic temperature in terms of the fundamental equation. I believe the question from the University Physics book that prompted my question is asking for an operational way to define a temperature scale. $\endgroup$
    – scmartin
    Commented Nov 8, 2022 at 11:57
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    $\begingroup$ You can always define an operational temperature scale, but whether such scale makes thermodynamic sense, you need to check against the fundamental equation. If the operational temperature is not linked to entropy, it will not work. It works for the ideal gas because the ideal-gas equation is consistent with the fundamental equation, i.e., it can be derived from it assuming that molecules are point masses that do not interact. If you can obtain the entropy from your equation of state, you can recover temperature. $\endgroup$
    – Themis
    Commented Nov 8, 2022 at 12:44
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    $\begingroup$ Aren't the more standard definitions of $T$ and $p$ given by $(\partial S / \partial E)_{V,N} = 1/T $ and $(\partial S / \partial V)_{E,N} = p/T $? $\endgroup$
    – scmartin
    Commented Nov 8, 2022 at 16:57
  • $\begingroup$ @scmartin oops.... of course you're right. I corrected the typos. $\endgroup$
    – Themis
    Commented Nov 8, 2022 at 19:02
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    $\begingroup$ I think @JohnDoty 's point is that the question I'm asking is clearly not trying to define temperature from the fundamental equation, but to come up with a scale for temperature based on an equation of state (some experimentally observable relationship between system properties). $\endgroup$
    – scmartin
    Commented Nov 9, 2022 at 13:05
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Thermometric scale:

  • Consider a system characterized by two independent variables, $x$ and $y$ for example. When the values ​​of $x$ and $y$ remain constant, as long as the external environment is not modified, the system is said to be in thermal equilibrium. The temperature $\theta$ is then assigned to the system, which obviously depends on the pairs $x$ , $y$ .

  • A thermometer is a system where one of the variables is kept constant (y for example), the thermometer is brought into thermal equilibrium with the system whose temperature $\theta$ is sought. The equilibrium temperature $\theta$ only depends on the variable $x$ (which can be the volume of a fluid, the pressure, the resistance of a wire,...). The relation $\theta (x)$ defines the temperature scale.

-centisimal scale (with two fixed points):

The linear centisimal scale is defined by the thermometric function: $\theta(x)=ax+b $ $;\;\;\;(a,b)$ constants.

From the equation with $v=\mathrm{const.}$ we have the thermometric scale :$$c_{T}=(v-b)p=\alpha p$$

If $p\rightarrow 0$, the equation $pv=bp+c_{T}$ becomes:$$pv=c_{T}$$

this is the ideal gas equation, so $$c_{T}=nRT=Nk_{B}T$$

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  • $\begingroup$ There are several parts of this that I don't follow. From the top of your answer, do you mean that if we hold one of the two independent variables (e.g. $y$) fixed, then we can then come up with any arbitrary function $\theta(x)$ and call that the temperature of the system at the state with the fixed value of $y$ and whatever value of $x$ we are interested in? How do we then relate such an arbitrary assignment to the thermodynamic temperature? What is $v=cst$? $\endgroup$
    – scmartin
    Commented Nov 9, 2022 at 14:58
  • $\begingroup$ If we invert the equation $\theta(x)=ax+b$, we have an equation of the form $x=x_{0}(1+k\theta )$, at constant pressure=cst, $v=v_{0}(1+\alpha\theta )$, and at $v=cst$, $\;\; p=p_{0}(1+\beta\theta )$, if p-->0, the experiment gives $\alpha=\beta=1/273,15$, so $lim(x/x_{0})=1+\frac{\theta}{273,15}$, let's put $T=\theta+2,72,15$ ==>$ T=273,15 lim (x/x_{0})$, if we designate $x_{0}=x_{t}$ the value of the thermodynamic variable at the triplet point of water, in the case of a thermometer with $v=cst$, $T=273,15\; lim (p/p_{t})$. $\endgroup$
    – The Tiler
    Commented Nov 9, 2022 at 18:21

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