Question

When two ideal gases A and B mix, they don't interact. The chemical potential of each gas is independent of the other, e.g. for A:

μA=μ∘A+RT ln pA/p∘(1)

So if the two gases are independent, why is there a Gibbs energy of mixing? Shouldn't there be a requirement for an interaction energy in order to see a difference in Gibbs energy before and after mixing?

However, we know that the Gibbs energy of mixing of two ideal gases is non-zero, as it is given as:

ΔmixG=nRT(xAlnxA+xBlnxB)(2)

where x
is the molar ratio. In a related question about a problem set involving Gibbs energy of mixing, an answer uses (2)
in the first part and (1)
in the second.

In other words, how are (1)
and (2)
related to each other? Another way of asking is: if the Gibbs energy of mixing is non-zero even for ideal gases, how could you turn the process of mixing into a reversible process and show the maximal work that is done?

Answer 1

The Gibbs energy of mixing for ideal gases is non-zero due to the increase in entropy upon mixing, despite no change in enthalpy. This mixing leads to a spontaneous process, indicated by a negative change in Gibbs energy. The mixing can be thought of as a reversible process where the change in Gibbs energy represents the potential work due to entropy increase.

Step-by-step explanation:

The apparent paradox in understanding the Gibbs energy of mixing (ΔmixG) for ideal gases arises from the fact that even in the absence of interaction energy, mixing leads to an increase in entropy.

The Gibbs free energy equation (ΔG = ΔH - TΔS) factors in both enthalpy (ΔH) and entropy (ΔS), and for ideal gases, the change in enthalpy upon mixing is approximately zero (ΔHₒₕₙₐ). However, there is a significant increase in entropy due to the dispersal of gas particles, which results in a negative ΔG, signifying a spontaneous process.

Equation (1) represents the chemical potential (μ) of an ideal gas A or B, where μ = μ° + RT ln (pA/p°). Equation (2) is the equation representing the Gibbs energy of mixing for the two ideal gases, ΔmixG = nRT(xAln(xA) + xBln(xB)).

The former is a statement about the individual gas's potential at a given partial pressure, while the latter quantifies the change in Gibbs energy resulting from mixing gases in their respective molar ratios (xA and xB).

To reconcile ΔG and work, we consider the reversible process of an ideal gas expanding or mixing. In such a reversible process, the change in Gibbs energy equals the maximal work that can be extracted from the system, barring any external work.

Thus, for the mixing of two gases without external work, the non-zero ΔmixG reflects the energy that could have been harnessed as work due solely to the change in entropy.