Consider a volume separated in two parts by a diathermal wall. The whole volume is isolated from the outside.

In the left volume $V_1=V/2$ I put $n_1$ moles of perfect gas $1$, at pressure $P$ and temperature $T_1$.

In the right volulme $V_2=V/2$, I put $n_2$ moles of perfect gas $2$ at pressure $P$ and temperature $T_2$.

I assume the gas have the same $\gamma$.

Using the first principle, I have :

$$\Delta U_1 + \Delta U_2=n_1 c_v (T_1^f - T_1) + n_2 c_v (T_1^f - T_1) = 0$$

This equals $0$ because the full system is isolated from the outside.

Actually, both system will equilibrate towards the same temperature, and we will find : $T^f=\frac{n_1 T_1 + n_2 T_2}{n_1+n_2}$

My question is : is there a rigorous proof to say they will equilibrate toward the same temperature ?

How can we actually show that those systems won't "equilibrate" at $T_1 \neq T_2$. I know it is obvious physically but how can we show it ?


2 Answers 2


At equilibrium, entropy must be extremal.

So, for a virtual infinitesimal transfer of energy, $dS=0=\frac{\partial {{S}_{1}}}{\partial {{U}_{1}}}d{{U}_{1}}+\frac{\partial {{S}_{2}}}{\partial {{U}_{2}}}d{{U}_{2}}$

As the system is isolated $d{{U}_{2}}=-d{{U}_{1}}$ .

By definition of temperature $\frac{1}{{{T}_{1}}}=\frac{\partial {{S}_{1}}}{\partial {{U}_{1}}}$ and $\frac{1}{{{T}_{2}}}=\frac{\partial {{S}_{2}}}{\partial {{U}_{2}}}$ Finally, $dS=0=\left( \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right)d{{U}_{1}}$ or ${{T}_{1}}={{T}_{2}}$

For more details, you can look at Callen, Thermodynamics....p 43


There is no proof; evolution towards thermodynamic equilibrium is a fundamental assumption in physics, a generalization of experience (experiments nad observation).

Usually, thermodynamic equilibrium requires same temperature everywhere. There are exceptions to this when gravity is involved, but those are usually negligible.

How can we actually show that those systems won't "equilibrate" at $T_1\neq T_2$.

Again from experience: in general, if body A is at equilibrium with body B, then third body T (a thermometer) when brought into contact with both of them, will be at equilibrium with both of them. The state of the body T can be read of as temperature.

Usually different temperatures mean heat will flow from the hotter to the colder region. Such flow means the system is not at equilibrium.

  • 1
    $\begingroup$ Just another question then. How do we know that the state of equal temperature is the thermodynamic equilibrium rigorously ? Does that mean we need to compute all state function and show that they don't evolve anymore when we have equality of temperatures ? $\endgroup$
    – StarBucK
    Mar 3, 2019 at 11:33
  • $\begingroup$ @StarBucK I've changed my answer. $\endgroup$ Mar 3, 2019 at 11:59

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