Is Entropy only for Multi particle Systems?

Question

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1. What the student is asking (in plain language)

The student read that entropy is a property of a collection of particles, not of a single particle.
They then imagined a single particle that can move between

a tiny chamber (volume (V_{\text{small}})), and
a huge chamber (volume (V_{\text{large}})) that is connected by a small hole.

Because the large chamber contains many more possible places for the particle to be, after a long time the particle is “found” in the large chamber most of the time.

Questions

Can we say that the particle’s entropy has increased when it moves from the small to the large chamber?
If we can talk about the entropy of a single particle, does that contradict the statement that entropy only belongs to many‑particle systems?

2. Detailed solution (step‑by‑step)

2.1 Thermodynamic vs. statistical entropy

Thermodynamic entropy (classical)	Statistical (Boltzmann / Gibbs) entropy
Defined for macroscopic systems that contain a huge number of particles (≈ 10²³).	Defined for any probability distribution over microstates – even a single particle.
Measured experimentally via heat and temperature: (\displaystyle dS = \frac{\delta Q_{\text{rev}}}{T}).	Given by (\displaystyle S = -k_{!B}\sum_i p_i\ln p_i) (Gibbs) or (S = k_{!B}\ln \Omega) (Boltzmann), where (\Omega) = number of accessible microstates.
Extensive (proportional to the amount of matter).	Not necessarily extensive; can be zero, negative, or any value depending on the chosen distribution.

Key point: The thermodynamic notion of entropy is a special case of the statistical notion when the number of particles is astronomically large and the probability distribution is sharply peaked around equilibrium. Therefore, entropy can be defined for a single particle, but it is the statistical entropy that we are using.

2.2 The “particle in two boxes’’ thought experiment

2.2.1 What are the microstates?

The particle is a point (or a quantum wavepacket) that can occupy any position inside the total volume
[ V_{\text{tot}} = V_{\text{small}} + V_{\text{large}} . ]
If we neglect momentum (or treat momentum as already thermalised) the only degree of freedom that matters for the present question is where the particle is.
The number of position microstates is proportional to the volume available.
[ \Omega_{\text{small}} \propto V_{\text{small}},\qquad \Omega_{\text{large}} \propto V_{\text{large}} . ]

2.2.2 Entropy of the particle in each chamber

Using Boltzmann’s formula (S = k_{!B}\ln\Omega) (ignoring an overall additive constant that cancels later),

[ \begin{aligned} S_{\text{small}} &= k_{!B}\ln V_{\text{small}},\[4pt] S_{\text{large}} &= k_{!B}\ln V_{\text{large}} . \end{aligned} ]

2.2.3 Entropy change when the particle moves

[ \Delta S \;=\; S_{\text{large}} - S_{\text{small}} \;=\; k_{!B}\bigl(\ln V_{\text{large}} - \ln V_{\text{small}}\bigr) \;=\; k_{!B}\ln!\left(\frac{V_{\text{large}}}{V_{\text{small}}}\right) . ]

Because the large box is much larger than the small one, the ratio (V_{\text{large}}/V_{\text{small}}) is a huge number, so

[ \boxed{\;\Delta S > 0\;} ]

i.e. the statistical entropy of the particle increases when it goes from the small chamber to the large chamber.

2.3 Why does the particle “prefer’’ the large chamber?

The particle is not “choosing’’ anything. It wanders randomly (e.g., due to thermal agitation). The probability of finding it in a region is proportional to the number of microstates there:

[ P(\text{large}) = \frac{V_{\text{large}}}{V_{\text{tot}}},\qquad P(\text{small}) = \frac{V_{\text{small}}}{V_{\text{tot}}}. ]

Since (V_{\text{large}} \gg V_{\text{small}}), the particle will be observed in the large chamber most of the time. This statistical bias is exactly what the entropy increase (\Delta S) quantifies.

2.4 Does this contradict the “entropy only for many particles’’ statement?

No. The statement is a pedagogical shortcut used in introductory thermodynamics because:

For macroscopic systems the entropy is extensive and scales with the number of particles, making it a useful bulk property.
For a single particle the entropy is usually tiny (of order (k_{!B})) and does not affect everyday thermodynamic measurements.

Nevertheless, the formal definition of entropy (Boltzmann/Gibbs) applies to any system whose microstates can be counted or whose probability distribution is known, including a single particle. The apparent contradiction disappears once we recognise the difference between the thermodynamic and the statistical viewpoints.

2.5 What about the environment?

If the particle moves from the small to the large box without any exchange of heat with an external reservoir, the total (isolated) system is just the particle plus the two volumes. The entropy change we calculated is the total entropy change of that isolated system. No extra “environmental entropy’’ is required.

If the particle interacts with a heat bath (e.g., the walls are at temperature (T)), the same result holds because the particle’s positional degrees of freedom rapidly equilibrate with the bath. The bath’s entropy does not change appreciably because the particle’s energy exchange is negligible compared with the bath’s size.

3. Final answer

Yes, we can assign an entropy to a single particle by using the statistical definition.
When the particle moves from a small chamber of volume (V_{\text{small}}) to a much larger chamber of volume (V_{\text{large}}),

[ \boxed{\Delta S = k_{!B}\,\ln!\left(\frac{V_{\text{large}}}{V_{\text{small}}}\right) > 0 } . ]

Thus the entropy of the particle‑plus‑boxes system increases, which explains why, after a long time, the particle is overwhelmingly likely to be found in the large chamber. This does not contradict the teaching that “entropy is a property of many‑particle systems’’; that statement is merely a convenient approximation for macroscopic thermodynamics. The fundamental definition of entropy works for any number of particles, even one.

4. Common mistakes for this type of problem

Mistake	Why it’s wrong	How to avoid it
Confusing thermodynamic entropy with statistical entropy and claiming the former cannot be defined for one particle.	Thermodynamic entropy is a macroscopic limit of the statistical definition. The statistical formula (S = -k_{!B}\sum p_i\ln p_i) works for any probability distribution.	Explicitly write the statistical expression and note that the “many‑particle” claim applies only to the thermodynamic approximation.
Treating the particle’s position as deterministic (e.g., “the particle will go to the large box”).	The motion is random; the particle does not have a directed drive. The bias comes from the larger number of accessible positions.	Emphasise the probability (P \propto V) and that the particle’s trajectory is a random walk.
Using ( \Delta S = \frac{Q_{\text{rev}}}{T}) for the single‑particle case and concluding (\Delta S = 0) because no heat is exchanged.	That relation is the thermodynamic definition, which assumes a macroscopic reversible path. For a single particle we must use the statistical definition.	State both definitions, explain when each applies, and use the Boltzmann/Gibbs formula for the single‑particle scenario.
Ignoring the additive constant in (S = k_{!B}\ln V) and claiming the entropy is absolute for one particle.	Entropy is defined up to an arbitrary constant; only differences (\Delta S) have physical meaning.	Focus on the entropy change (\Delta S); the constant cancels out.
Assuming entropy must be extensive and therefore thinking a single particle must have zero entropy.	Extensivity is a property that emerges when many independent subsystems are added. A single particle can have a non‑zero (though tiny) entropy.	Remember that “extensive” refers to scaling with particle number; it does not forbid a non‑zero value for (N=1).

By keeping these points in mind, you can correctly analyse entropy changes for any system, whether it contains Avogadro’s number of particles or just one.

Original question: Is Entropy only for Multi particle Systems? on Chemistry Stack Exchange, licensed CC BY-SA.