# Question about chemical potential of a multicomponent ideal gas

Consider a pure ideal gas, i.e. only having one component. The chemical potential and free energy of this gas are as follows:

$$\mu_1=\mu^\circ(T)+RT\ln\left(P\right)$$

$$G_1=n\mu_1$$

Where n is the total number of gas moles.

Now, just assume that we divide the molecules in gas by two, and treat each of these tow parts as a separate component. This is just an imaginary division, and none of P, T, composition, number of molecules are changed. So, we expect the free energy to remain the same as well.

Now, the chemical potential for each of these two component is identical and reads as

$$\mu_2=\mu^\circ(T)+RT\ln\left(P\frac{1}{2}\right)$$

and the free energy is

$$G_2=\mu_2\left(n\frac{1}{2}+n\frac{1}{2}\right)=n\mu_2$$

Now my question is that why is not G_1 equal to G_2?

Edit: To illustrate this I have envisaged two cases for making the gas divided in two parts, please see that attached image. In case 1, the gas is divided into to compartments with half of the volume of the initial compartment. In case 2, the gas is divided into to compartments with the same volume as the initial compartment.

I got two different Delta G values for these two cases. In the first case Delta G is zero and chemical potential does not change, but for case 2 chemical potential changes as described above.

• Why did you write P/2 when pressure is supposed to remain same? Commented Mar 25 at 2:59
• The partial pressure of each part will be p/2. See Eq. 12.24 here bcp.fu-berlin.de/en/chemie/chemie/forschung/PhysTheoChem/… Commented Mar 25 at 8:43
• Thanks for your comment! No, there is no change in the gas, no dimerization or change in molecular interactions. It is just an imaginary division of the gas molecules to two identical parts. Commented Mar 25 at 9:16
• @bloodybutunbowed0 are the volumes of our imaginary compartments equal or not? Commented Mar 25 at 11:30
• @NatashaJ yes it is equal along with all other properties such as pressure and temperature. Commented Mar 25 at 12:12

If you first divide the space into two with a wall, the chemical potential will be the same, and the sum of the Gibbs energies will be the same as the original Gibbs energy.

Then, you mark the molecules in one space but not in the other. There is no change in Gibbs energy. However, if you now remove the barrier, you get an irreversible process of mixing, lowering the Gibbs energy. To unmix (or separate) the two flavors of particles, you would need to do work on the system.

For a visual of these ideas, see Why would there be a non-zero Gibbs energy of mixing for ideal gases?.