# Differences in entropy between two ideal gases but under same conditions

I recently solved a problem where I had to calculate the entropy as a mole of argon gas and compare it to that of one mole of helium. The entropy was calculated during the same conditions (at room temperature and atmospheric pressure).

I'm aware of the Sackur-Tetrode equation, which relates macrostate variables such as $$U, V, N$$ to the entropy for a monoatomic ideal gas. From the Sackur-Tetrode equation, I can deduce that as the mass of the monoatomic gas increases, so does the entropy. From the derivation of the Sucker-Tetrode equation, the mass comes in when momentum is introduced into the picture.

1. This makes we wonder whether we can intuitevly understand why the entropy is greater for the monoatomic gas with greather mass, rather than having to make a case for it from the formula itself (which almost becomes like a circular argument to me).

So, here's my thought process: Since entropy is a measure of the system's non-orderliness, if the mass of the molecule increases (and I assume the internal energy is same for both systems because of the conditions previously stated), we expect the heavier molecules to move slower (on average) in space.

1. If that's the case, my intuition tells me that the rate at which something goes from ordered to non-ordered (low to high entropy) should slow down. Is this a correct way of thinking about it?

Either way, this is talking about the rate at which it's happening, and not the magnitude of the entropy itself, and I really don't really know how to further lay out my argument.

I'd be glad if anyone could help me with this question that arose as a consequence of solving that problem.

• Kinetics (your "rate at which something goes from ordered to non-ordered") is not thermodynamics. The two are separate. Commented Sep 6, 2023 at 16:21
• I am giggling at the Sucker-. The initial simple macroscopic experiments and reasoning can only give you the Sackur-Tetrode of the form that $\frac SN=\ln\left[\frac VN\left(\frac UN\right)^{\frac1{\gamma-1}}\right]+\text{const}$ and not much more, since the constant depends upon Planck's constant. The mass contribution can be deduced from aqueous and other chemical experiments that go deep into mass spectrometry, but it cannot be a simple explanation as it is, since $\frac UN\propto T$ and $p\frac VN=k_BT$ constrains the simple stuff. Commented Sep 6, 2023 at 16:31
• Entropy is an extensive property, and so it is directly proportional tot he amount of mass you have. However, the entropy per unit mass of the gas is an intensive property (like specific volume), and is independent of the amount of mass. Commented Sep 6, 2023 at 21:34
• @ChetMiller The OP refers to the dependence of entropy on the individual atoms' mass, not the total mass of the gas. Commented Sep 6, 2023 at 22:47

Sackur and Tetrode's formula for the entropy of ideal gas results from Statistical Mechanics and not classical Thermodynamics. However, the Statistical Mechanics of equilibrium is not influenced by the kinetics of the processes, i.e., the time rate of the underlying dynamics does not play any role. This fact should be evident in the ensemble theory. Once time averages over infinite times are substituted by phase space averages, more than one dynamic may lead to the same phase space density. Again, this is nicely illustrated by the possibility of evaluating average values with different dynamics, including stochastic processes.

You are asking for an intuitive reason for the mass dependence of entropy. This is a conceptually challenging question. To answer, it is helpful to start from Sackur and Tetrode's formula for the entropy of a perfect gas of $$N$$ atoms at energy $$U$$ and volume $$V$$:

$$\frac{S}{N}=k_B \ln \left[ \frac{V}{N} \left( \frac{4 \pi m}{3 h^2} \frac{U}{N}\right)^{\frac32} \right]+\frac52 k_B \tag{1}$$ where $$k_B$$ is the Boltzmann's constant, $$h$$ the Planck's constant, and $$m$$ the mass of one atom.

It is clear from the formula that a logarithmic dependence on the mass is present.

Formula $$(1)$$ does not contain any reference to the concept of order/disorder, whatever could be its definition. Then, a direct connection with the almost always ill-defined relation between entropy and disorder cannot help. What can help is realizing that the formula can be derived in the microcanonical ensemble by directly counting the microstates corresponding to fixed values of $$N$$, $$U$$, and $$V$$. However, classical microstates are a non-numerable infinity. Thus, we must introduce an elementary volume of the phase space to count them. Such elementary volume must have the physical dimension of action (energy $$\times$$ time) per each pair of position and momentum variables. Conventionally, it is indicated by $$h$$ because if we derive equation $$(1)$$ as the classical limit of a corresponding quantum formula, it turns out that $$h$$ must coincide with Planck's constant.

However, the central question is where the mass is coming from. If one follows the derivation of the formula, it directly results from the relation between momenta and energy. To evaluate the integral giving the volume of a region in phase space in the microcanonical ensemble, we should use the formula for the $$3N$$ dimensional hypersphere. $$\sum_{i=1}^{3N} p_i^2 = 2mU$$ Such an origin of the presence of the mass provides a hint for building intuition.

Indeed, we see that the number of microstates inside the volume of the phase space limited by a given energy value increases with the mass because the radius of the limiting surface is $$\sqrt{2mU}$$. At fixed $$U$$, the larger is $$m$$, the more states we count. This is the direct origin of the mass dependence of the entropy.

• Take my upvote. You spelt Tetrode wrong in the link. It is nice that you have worked out a good reason why $U$ should appear as the combination $2mU$, but then it would be difficult to explain why the rest of the thing should be $\frac12\cdot\frac{4\pi}3\cdot\frac{2mU}N\cdot\frac1{h^2}$; the division by the square of the Planck's constant is not just non-classical, but also unexpected---the $\frac{4\pi}3$ clearly invoked a 3D ball, so why a square of Planck's constant, not a cube or something? Of course, dimensional analysis fixes that it has to be square, but it is still surprising. Commented Sep 7, 2023 at 5:01
• @naturallyInconsistent Thanks for signaling my typo. Planck's constant actually appears in the log argument as $h^3$ once the simple algebra of the expression is worked out. Such a power is perfectly understandable in the context of single-particle properties in 3 dimensions. Commented Sep 7, 2023 at 5:38
• @naturallyInconsistent About the constants ($\frac12$, $\frac{4 \pi}{3}$ ), their origin can be traced back to the exact derivation of the formula, which was not the question posed by the OP, and is present in every good Statistical Mechanics textbook. Commented Sep 7, 2023 at 5:43
• That I definitely know. The exact details will always require quantum arguments since Planck's constant appeared. Am surprised that it started out as $h^3$ and became $h^2$ later. Anyway, this dead horse is beyond beat. Commented Sep 7, 2023 at 5:46
• @naturallyInconsistent The power of $h$ never changed in the formula.: $(h^2)^{\frac32}=h^3$. Commented Sep 7, 2023 at 5:49