Temperature scale from equation of state 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
$$\frac{p_2}{p_1} = \frac{T_2}{T_1}$$
 A: 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.
A: 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.
A: 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 $$c_{T}=nRT=Nk_{B}T$$
