Here, we’re interested in

calculating the entropy change when you mix two ideal gases. An important constraint

is we’re doing this at constant pressure and

constant temperature. So the situation we’re looking at is, we start with two gases for example, in separate containers, separated say, by a diaphragm here. So gas A, gas B, nA is number of moles of A, pressure one, temperature one, important the same pressure, same temperature. And the idea is, if we were to break that diaphragm, and so we end up with one container, and we have A and B together. So we’d like to know the entropy change and the way we’re going to do that, is to treat each of the gases separately. So, elsewhere we’ve derived an equation. I’m just going to write it down here for entropy change for an ideal gas. So this entropy change

due to temperature change, and then due to a pressure change, P2 is the final, P1 is the initial. Of course, for our situation, this term is zero, so we’re only looking at

the change due to pressure. But if we treat each gas independently, then we start out at pressure P1, and end up with a final

pressure that’s lower, and this is the reason

the entropy increases, we’ve lowered the pressure

of each of the gases. So let’s look at the entropy change of A, delta S is per mole, so minus R log initial pressure P1, our final pressure is the

mole fraction of A times P1, final partial pressure, so

that the entropy change for A is just minus R log

the mole fraction of A, well, the completely

equivalent term for B. Entropy change of B. Just a reminder, this is per mole of B, in that case and the first term, per mole of A. So then the total entropy change for the system we’re looking at, this is total, meaning not per mole, but for the system we have. So we have two terms, we have the number of moles of A then R log mole fraction of A and the number of moles of B, and we call this the

entropy change of mixing. We’re gonna divide by the number of moles, so entropy change per

mole of our final mixture. So I’m gonna divide nA by

the total number of moles. All right, I’ll divide the right side, and then likewise, nB. Well, I can simplify this. This is the mole fraction of A and this is the mole fraction of B, entropy change of mixing per mole. I can be arranged slightly, minus R[yA logyA] plus yB logyB, we’re using natural log of course. So entropy change in mixing for

a binary mixture in general, then delta S of mixing would be minus R, the

summation of mole fractions. So we could have three or

four or five components if we’re mixing at

constant total pressure. Here’s the entropy change of mixing. So mixing, and this of course,

is only good for ideal gases, constant temperature. Temperatures should not

change with mixed ideal gases because there’s no interaction,

but constant pressure, meaning the initial pressure of the gases and final total pressure is the same.

What if nA = nB but pA different than pB. And the process is adiabatic

Is there any entropy change for the mixing of two identical gas samples, instead of two different gases?

Doesn't your final equation, (delta) S mix = -R (sum) yi ln ( yi) need to have the total number of moles in the front? ie) (delta) S mix = -R Ntotal (sum) yi ln ( yi)