Question

15.4 Two systems are identical in all respects except that in one the molecules are distinguishable and in the other they are indistinguishable.

Calculate the difference between their molar entropies.

Physical Chemistry (4th edition), Laidler, Meise, Sanctuary

Answer 1

Solution: Difference between their molar entropies

Two-level systems, that is systems with essentially only two energy levels are important kind of systems, as, at low enough temperatures, only the two lowest energy levels will be involved. Especially important are solids where each atom has two levels with different energies depending on whether the electron of the atom has spin up or down.

We consider a set of N distinguishable “atoms” each with two energy levels. The atoms in a solid are of course identical but we can distinguish them, as they are located in fixed places in the crystal lattice. The energy of these two levels is ε0 and ε1. It is easy to write down the partition function for an atom

Z = e ^{−ε0 / kB T} + e ^{−ε1 / kBT} = e ^{−ε 0 / k BT} (1+ e ^{−ε / kB T}) = Z0 ⋅ Z term

Where ε is the energy difference between the two levels. We have written the partition sum as a product of a zero-point factor and a “thermal” factor. This is handy as in most physical connections we will have the logarithm of the partition sum and we will then get a sum of two terms: one giving the zero point contribution, the other giving the thermal contribution.

At thermal dynamical equilibrium we then have the occupation numbers in the two levels

n₀ = N/ Z e^{−ε 0/kBT} = N/ 1+ e ^{−ε/k BT}

n₁ = N/ Z e ^{−ε 1 /k BT} = Ne ^{−ε/k BT} /1 + e^{−ε /k BT}

We see that at very low temperatures almost all the particles are in the ground state while at high temperatures there is essentially the same number of particles in the two levels. The transition between these two extreme situations occurs very roughly when kBT ≈ε or T ≈ θ = ε/kB, the so-called scale temperature θ that is an important quantity.

In this case we can directly write down the internal energy

E = n₀ε₀ + n₁ε₁ = N ε₀ e ^{−ε 0/kBT} + ε₁e ^−ε1/kBT /e ^{−ε 0/kBT} + e ^−ε1/kBT = N ε₀ + Nεe ^−θ/T /1 + e ^−θ/T

The internal energy is a monotonous increasing function of temperature that starts from E(0) = Nε 0 and asymptotically approaches E(0) + Nε /2 at high temperatures.

One way of calculating ΔS for a reaction is to use tabulated values of the standard molar entropy (S°), which is the entropy of 1 mol of a substance at a standard temperature of 298 K; the units of S° are J/(mol·K). Unlike enthalpy or internal energy, it is possible to obtain absolute entropy values by measuring the entropy change that occurs between the reference point of 0 K [corresponding to S = 0 J/(mol·K)] and 298 K.

From Table "Standard Molar Entropy Values of Selected Substances at 25°C", for substances with approximately the same molar mass and number of atoms, S° values fall in the order S°(gas) > S°(liquid) > S°(solid). For instance, S° for liquid water is 70.0 J/(mol·K), whereas S° for water vapor is 188.8 J/(mol·K). Likewise, S° is 260.7 J/(mol·K) for gaseous I2 and 116.1 J/(mol·K) for solid I2. This order makes qualitative sense based on the kinds and extents of motion available to atoms and molecules in the three phases.