Isotherms don't intersect? Is it true that the isotherms of a system will never intersect? In other words, in the equation of state
$$f(X_1,X_2,\cdots,T)=0$$
where $X_i$ are some macroscopic parameters that can be defined outside statistical mechanics,
is the temperature uniquely determined by $X_i$ so that one can always write it as
$$T=g(X_1,X_2,\cdots)?$$
So let's take a fixed amount of gas (not necessarily an ideal gas) with EOS $f(p,V,T)=0$ as an example, and put my question simple. Is it possible for two gases with the same $(p,V)$ to have different temperatures? If no, then $T$ is determined uniquely by $(p,V)$ and one can write the EOS as some $T=g(p,V)$. So the textbooks' keeping the general form of EOS as $f(p,V,T)=0$ instead of $T=g(p,V)$ seems to hint that it is possible for two gases with the same $(p,V)$ to have different temperatures. Is that right?
EDIT: As an example of an EOS in which the same $(p,V)$ does not determine a $T$ uniquely, consider
$$f(p,V,T)=T^2-pV=0$$
 A: Crossing isotherms are rare but appear in some real fluids in regions where
the thermal expansivity changes its sign. For example, due to the density anomaly of water, liquid water has at atmospheric pressure a negative thermal expansivity for T<3.983 degree C, while it is positive at higher temperature. Thus it exhibits crossing isotherms. See
A. Neumaier and U.K. Deiters,
The characteristic curves of water,
Int. J. Thermophysics, published online July 23, 2016.
DOI: 10.1007/s10765-016-2098-1
http://arnold-neumaier.at/ms/amagat.pdf
To see this with a simple argument: At atmospheric pressure, the density of water has a maximum at around 4 degree Celsius. The existence of this maximum is called the density anomaly of water. Looking at the temperature-density curve, water at 2 degree has lower density than the maximum, hence must have the same density as water some corresponding temperature at (not much) higher than 4 degree. Thus there are two different temperatures corresponding to the same pressure and density, hence to the same pressure and volume. Thus the isotherms at these temperatures intersect.
A: Your question seems to be: given $f(p,V,T)=0$ for some real system is it true that this always corresponds to a unique function $T=g(p,V)$? If you think as a mathematician, then it is not true in general. You can cook up a function, say $pV=C(T-T_0)^2$, so that isotherms $T_0+\Delta T$ and $T_0-\Delta T$ intersect (actually worse, they are completely coincident). However it is a $postulate$ of thermodynamics that a state is completely determined by specification of any two thermodynamic variables. So if the system has particular $p$ and $V$, then the postulate asserts that it has a unique value of $T$ (and any other thermodynamic variable). It is only a postulate, but it has not been proven wrong so far. In fact if you see that iostherms intersect in any experiment, it is an indication that some other thermodynamic variable besides $p,V$ has come into play.
A: Well for almost every given $f(p,V,T)$, one can use implicit function theorem to write $T=g(p,V)$ for some $g(p,V)$ locally at almost everywhere. But mathematically there is nothing to prevent $T$ not able to be written as $g(V,T)$. In fact, some empirical equations of states such as Redlich–Kwong equation of state, where 
$$P=\frac{nRT}{V-bn}-\frac{an}{\sqrt{T}V(V+bn)}$$
might not be able to be written as T=g(P,V) globally. To see this, consider given $P$, $V$, one has a cubic polynomial of $T$, whose discriminant is 
$$\frac{a^2 n^5 R^3 \left(-27 a^2 n^3 R-4 p^3 V^2 (b n-V) (b n+V)^2\right)}{\left(V^3-b^2 n^2 V\right)^4}$$
which could be positive when $V$ is large. 
So you might ask what the intersection really means. Even though the two isothermal lines intersect, one can't go from one isothermal line to the other, otherwise the second law of thermodynamics could easily be violated. The intersection is merely an artifact- in the $p,V,T$ space, the two isothermal lines do not intersect, and represent different physical processes, but they seem to intersect after projecting to $P,V$ plane. 
