# Can entropy in a closed system remain constant when you heat a liquid?

This is in regards to an exam question that I did not understand and got wrong.

The question was: When you heat acetone from 200 K to 400 K, what happens to the entropy? Answers: Increases, decreases, more info needed

I put more info needed, but this was incorrect. My logic was that if you have a sealed container completely filled with acetone (there is no gas or empty space in the container) and you heat it, there are no new states that the liquid particles can occupy so there is no change in entropy. Is this reasoning incorrect?

• Technically, that's right: there are no new states. But the particles distribute more uniformly among the existing states. – Ivan Neretin May 14 at 7:14
• Exactly, as the higher energy states of electrons, of bond vibrations and molecule rotations are not occupied at low temperatures, or are occupied unevenly. – Poutnik May 14 at 11:24

Poutnik's answer gives an explanation for the increase in Entropy from a classical thermodynamic point of view. I would like to add the statistical mechanical perspective, which is what you based you logic on.

So entropy in statistical mechanics is interpreted as the randomness in the distribution of particles in various microstates.

$$S = k_B \ln \Omega$$

where $$\Omega$$ is the number of microstates.

When the temperature of the system is raised, the energy of the system increases. This leads to more availability of higher energy microstates, and thus the particles are more randomly distributed (since there are more microstates available to them).

I would like to know where your assumption "there are no new states that the liquid particles can occupy" came from.

PS: I think the question in the exam was intended to make you think about entropy in thermodynamic terms, but there is also a way to look at it statistically.

The quantitative differential classical thermodynamic definition of entropy is $$\mathrm{d}S=\mathrm{d}Q/T$$

Therefore providing heat $$Q$$ to the acetone increases its entropy as $$\Delta S=\int_{T_\mathrm{1}}^{T_\mathrm{2}} \frac{C(T)}{T}\cdot \mathrm{d}T$$

where $$C$$ is the heat capacity of the system.

The statistical thermodynamics definition has an alternative definition to $$S=k_\mathrm{b} \cdot \ln \Omega$$ as $$S=-k_\mathrm{b} \cdot \sum_\mathrm{i} p_\mathrm{i} \cdot \ln p_\mathrm{i}$$ as the sum over all states.

The higher energy states of electrons, of bond vibrations and molecule rotations are not occupied at low temperatures, or are occupied unevenly. In such a case, even for the same number of states, more evenly distributed probabilities at higher temperatures lead to higher entropy.