Why is entropy an extensive property?

Question

Question. Why is entropy an extensive property? How can we prove that for the general case? Is that why $S(k N)=kS(N)$? I am interested in answer based on classical thermodynamics.

Actuality. Extensionality of entropy is used to prove that $U$ is homogeneous function of $S, V, N$ (like here Why internal energy $U(S, V, N)$ is a homogeneous function of $S$, $V$, $N$?) and that is used to prove Why does $U = T S - P V + \sum_i \mu_i N_i$?.

My attempt to solve this question.

I saw a similar question Why is entropy an extensive quantity?, but is about statistical thermodynamics. I want an answer based on classical thermodynamics.

I can answer on a specific case of my question. Extensiveness of entropy can be shown in the case of constant pressure or volume.

1. From third law of thermodynamics $S(T=0)=0$

2. $dS=\frac{dq_{rev}}{T} $ is the definition of entropy.

3. $S=\int_0^T dS $ from 1, 2 steps.

4. $S_p=\int_0^{T_1}\frac{dq_rev(0->1)}{T}+\int_{T_1}^{T_2}\frac{dq_{melt} (1->2)}{T}+\int_{T_2}^{T_3}\frac{dq_{rev}(2->3)}{T}+... $ from 3 using algebra. Here $T_1=T_2$

5. $dq_{rev}(0->1)=m C_p dT $ this way we measure heat, there is no phase transform, pressure is constant.

   $dq_{rev}(1->2)=m \Delta H_{melt} $ this way we measure heat in isothermic process, pressure is constant.

   $dq_{rev}(2->3)=m C_p(2->3) dT $ this way we measure heat, there is no phase transform, pressure is constant.

...

6. $S_p=\int_0^{T_1}\frac{m C_p(0->1)dT}{T}+\int_{T_1}^{T_2}\frac{m \Delta H_{melt} (1->2)}{T}+\int_{T_2}^{T_3}\frac{m C_p(2->3)dT}{T}+...\ $ from 4, 5 using simple algebra. Here $T_1=T_2$

7. $S_p=m \left( \int_0^{T_1}\frac{ C_p(0->1)}{T}+\int_{T_1}^{T_2}\frac{ \Delta H_{melt} (1->2)}{T}+\int_{T_2}^{T_3}\frac{ C_p(2->3)}{T}+{}... \right) \ $ from step 6 using algebra. Here $T_1=T_2$

8. $S_p(T;k m)=kS_p(T;m) \ $ from 7 using algebra. So entropy is extensive at constant pressure.

9. $S_V(T;k m)=kS_V(T;m) \ $ similarly we can prove this for constant volume case.

Summary. So extensiveness of entropy at constant pressure or volume comes from intensiveness of specific heat capacities and specific phase transform heats. Is there a way to prove that theoretically? We can consider nanoparticle specific heat capacities or specific phase transform heats. Are they intensive too and why?

P.S.: I am chemist, so things that are obvious to physicists might not be obvious to me. So I prefer proofs. Proof is sequence of formulas where each of them is an axiom or hypothesis, or derived from previous steps by inference rules. I prefer Fitch notation. It is very good if the proof comes from a book or publication.

Answer 1

There is some ambiguity in how entropy is defined in thermodynamics/stat. physics, as, e.g., discussed in this answer. To take the two most common definitions:

• In thermodynamics entropy is defined phenomenologically as an extensive quantity that increases with time - so it is extensive by definition
• In statistical physics entropy is defined as a logarithm of the number of microstates. Thus, if we have two systems with numbers of microstates $\Omega_1$ and $\Omega_2$, the total number of mcirostates is $\Omega_1\Omega_2$, whereas its logarithm is additive: $$ S=k_B\log(\Omega_1\Omega_2) = k_B\log(\Omega_1) + k_B\log(\Omega_2) = S_1 + S_2 $$ You may notice that here we again neglected the interaction between the two systems, as we discussed in relation to your other question.

Answer 2

Let's say one particle can be in one of $\Omega_1$ states. Then two particles can be in $\Omega_2 = \Omega_1^2$ states (because particle 1 can be in one of $\Omega_1$ states, and particle 2 can be in one of $\Omega_1$ states). Carrying on this logic, $N$ particles can be in \begin{equation} \Omega_N = \Omega_1^N \end{equation} states. Since the entropy of the $N$ particles is $k$ times the log of the number of microstates, we have \begin{equation} S = k \log \Omega_N = N k \log \Omega_1 \end{equation} which scales like $N$.

From a classical thermodynamics point of view, starting from the first law, \begin{equation} dU = T dS + p d V \end{equation} since $dU$ and $dV$ are extensive, and $T$ is intensive, then $dS$ is extensive.

Answer 3

If I understand your question correctly, you are asking:

1. You define entropy as $S=\int\frac{\delta Q}{T}$.
2. Clearly, $T$ is an intensive quantity, as is $\frac{1}{T}$.
3. If $\delta Q$ is extensive, then so is $\frac{\delta Q}{T}$, since a product of an intensive and an extensive quantity is extensive.
4. So, if $\delta Q$ is extensive, then $\int\frac{\delta Q}{T}$, being a sum of extensive quantities, is extensive.
5. But you don't see why $\delta Q$ should be extensive. After all, when we put a hot block of metal into a cold place, the heat is emitted from the surface; there is at least a speed-of-light delay until the interior "knows" that the exterior is being chilled.

I think this is somewhat definitional. If you have a slab of metal, one side of which is cold and the other is hot, then either:

• Your system is not in (internal) thermodynamic equilibrium, so that entropy is not defined, or
• You really mean you have two adjacent slabs of metal, one cold and one hot (but otherwise indistinguishable, so they we mistook them for a single slab).

But then we expect two slabs at different temperatures to have different thermodynamic states. So an extensive quantity will differ between the two of them.

More generally,

1. Suppose a system $S$ is composed of identical components. (These components can have distinguishable subcomponents; thus we might describe a sulfuric acid solution as a collection of drops of fluid (the components), each droplet containing some water and some oleum (the subcomponents).) These are not statistical-mechanical microstates; they must be large enough for the thermodynamic limit to apply.
2. $S$ will undergo some thermodynamic process.
3. By conservation of energy, $$\delta Q_S=\sum_{s\in S}{\delta Q_s}\tag{1}$$ (This is also somewhat definitional: suppose component $s$ gives off some heat, but then that heat got turned into work "as it travels" to the outside, in component $t$. Then we'd call it "$t$ doing work" instead.)
4. $S$ is in thermodynamic equilibrium iff each component of $S$ is indistinguishable. (This is the definition of thermodynamic equilibrium for composite systems.)
5. In particular, all $s$ start out and end indistinguishable, since $S$ begins and ends in equilibrium. (This is the definition of a thermodynamic process.)
6. Thus the internal energy at the start and at the end are both independent of $s$.
7. Subtracting, $dU_s$ is independent of $s$.
8. Likewise, if components performed different amounts $\delta W_s$ of work on the exterior, then we could "soup up" our process so that it separates them. But that clearly won't leave the separated components indistinguishable!
9. So $dU_s$ and $\delta W_s$ are independent of $s$.
10. Subtracting, $\delta Q_s$ is independent of $s$ (by the First Law).
11. Substituting into (1) and picking any fixed $s_0$ for the common value, $$\delta Q_S=(\delta Q_{s_0})(\# S)$$ which is clearly extensive.

Answer 4

Extensive means a physical quantity whose magnitude is additive for sub-systems.

The state of any system is defined physically by four parameters, $p$ pressure, $T$ temperature, $V$ volume, and $n$ amount (moles -- could be number of particles or mass). A physical equation of state exists for any system, so only three of the four physical parameters are independent.

A state function (or state property) is the same for any system at the same values of $p, T, V$.

Define $P_s$ as a state function (property) for a system at a given set of $p, T, V$. Since it is a function (or property) for a specific system, we must determine whether it is either extensive (defined as above) or intensive to the system. Intensive means that $P_s$ is a physical quantity whose magnitude is independent of the extent of the system. Assume that $P_s$ is defined as not extensive. Let's prove that this means it is intensive. Take two systems with the same substance at the same state $p, T, V$. They must have the same $P_s$ by definition. Combine those two systems. Since $P_s$ is defined to be not extensive, the total $P_s$ is not the sum of the two values of $P_s$. Since the combined system is at the same $p, T$ as its two initial sub-systems, the combination must be at the same intensive $P_s$ as the two sub-systems. Therefore $P_s$ is intensive by definition.

Since $P_s$ is intensive, we can correspondingly define an extensive state function or state property $P'_s = nP_s$. The state function $P'_s$ will depend on the extent (volume) of the system, so it will not be intensive. The state function $P'_s$ will be additive for sub-systems, so it will be extensive.

For any state function $U, S, H, G, A$, we can choose to consider it in the intensive form $P_s$ or in the extensive form $P'_s$. We have no need to prove anything specific to any one of the properties/functions themselves. Reading between the lines to your question, see here next when you intended instead to ask how to prove that entropy is a state function using classic thermodynamics.