Question

What would be the change in entropy (if the change is zero, explain why, if not then calculate) when one mixes the following gases, at a temperature of 1000K and room pressure, if the gases are:

a) A mole of H with another mole of H.

b) A mole of gaseous C₁₂ with a mole of gaseous C₁₃.

Answer 1

The effect is the same as allowing each gas to expand to twice its volume; the thermal energy in each is now spread over a larger volume.

This question is imprecisely stated. I assume that for case (a), your teacher is implicitly assuming that the isotopic composition of both samples of hydrogen (which would normally exist as H2 molecules, not as individual H atoms) have identical isotopic compositions. (Remember that natural hydrogen is composed of two isotopes, protium, or H-1, and deuterium, H-2. Tritium, H-3, has a vanishingly small natural abundance on Earth).

I am also assuming that this mixing occurs at a constant volume (as well as pressure and temperature).

With the above assumption, the entropy change for case (a) is zero, because the particles in "parcel 1) of the hydrogen are indistinguishable from the particles in "parcel 2" of the hydrogen. Another way of thinking of this is that the average density of protium and deuterium atoms in the system does not change when the two parcels of gas are mixed. This is a version of the Gibbs' paradox. The key here is that the particles that are mixing in the two parcels are indistinguishable from one another.

In case (b), we clearly have distinguishable particles. The entropy of mixing is most easily calculated from the standard formula for the entropy of mixing:

ΔS_mix = -n*R*[x1*ln(x1) + x2*ln(x2)]

where x1 and x2 are the mole fractions of components 1 and 2 in the mixture, n is the total number of moles involved, and R is the universal gas constant. For a 2-component mixture (as we have here) x2 = 1 - x1. Specifically, in this case, x1 = x2 = 0.5, and n = 1 mole

The entropy of mixing in this case is therefore:

ΔS_mix = -1mol*R*[0.5*ln(0.5) + 0.5*ln(0.5)] = R*ln(2)*mol = 5.763 J/K