Question

Two vessels contain the same number N of molecules of the same perfect gas. Initially the two vessels are isolated from each other, the gases being at the same temperature T but at different pressures P1 and P2. The partition separating the two gases is removed. Find the change of entropy of the system when equilibrium has been re-established, in terms of initial pressures P1 and P2. Show that this entropy change is non-negative.

Answer 1

strategy that often works is to look at the initial and the final states and then compare them. A good place to start is usually the Gibbs free energy of the system (with the intention to use ΔmixS=−∂ΔmixG∂T∣∣p,{ni}ΔmixS=−∂ΔmixG∂T|p,{ni}later): for isothermal and isobaric conditions it is

G=∑iniμi+constant=ntot∑ixi=nintotμi+constant(1)G=∑iniμi+constant=ntot∑ixi⏟=nintotμi+constant(1)

with nini, μiμi, and xi=nintotxi=nintot being the amount, the chemical potential, and the mole fraction of the iith component in the system, respectively, and ntot=∑inintot=∑ini being the total amount of gas in the system. So, now we need an expression for the chemical potential of an ideal gas. It can be shown that this is given by

μi=μ0i+RTln(piP0)(2)μi=μi0+RTln⁡(piP0)(2)

where μ0iμi0 is the chemical potential of the iith component in a given standard state, RR is the universal gas constant, TT is the temperatur, pipiis the partial pressure of the iith component, and P0P0 is the standard pressure (i.e. the pressure connected to the standard state). For ideal gases we can apply Raoult's law stating that

pi=xiPtotpi=xiPtot

where PtotPtot is the total pressure of the system. Using this with equation (2) we get

μi=μ0i+RTln(xiPtotP0)=μ0i+RTlnxi+RTln(PtotP0)(3)μi=μi0+RTln⁡(xiPtotP0)=μi0+RTln⁡xi+RTln⁡(PtotP0)(3)

Furthermore, it is known that the partial pressures in a system must add up to the total pressure, i.e.

Ptot=∑ipiPtot=∑ipi

Ok, now that the stage is set we simply have to apply equations (1) and (3) to the initial and the final state of the mixing process. At the initial state the chambers are seperated (whereby chamber ii contains only one sort of gas labeled by ii) and so these subsystems are completely independent from each other. Thus, the Gibbs free energy of the total system of chambers is simply the sum of the Gibbs free energies of each chamber, whereby the gas in each of the chambers feels the pressure Ptot=PiPtot=Pi in the system (whereby PiPi is the ambient pressure in chamber ii), since xi=1xi=1(each chamber/independent subsystem holds just one kind of gas). According to equation (1) the Gibbs free energy of the iith chamber is then given by

Gi=niμi=ni(μ0i+RTln1=0+RTln(PiP0))=ni(μ0i+RTln(PiP0))Gi=niμi=ni(μi0+RTln⁡1⏟=0+RTln⁡(PiP0))=ni(μi0+RTln⁡(PiP0))

and the Gibbs free energy of the total system at the initial state is

Gini=∑iGi=∑ini=xintot(μ0i+RTln(PtotP0))=ntot(∑ixiμ0i=G0ini+∑ixiRTln(PiP0))=ntot(G0ini+∑ixiRTln(PiP0)) .(4)Gini=∑iGi=∑ini⏟=xintot(μi0+RTln⁡(PtotP0))=ntot(∑ixiμi0⏟=Gini0+∑ixiRTln⁡(PiP0))=ntot(Gini0+∑ixiRTln⁡(PiP0)) .(4)

For a multicomponent system with more than two components I would simply use this equation. But for a system with exactly two components we can introduce some substitutions that will simplify the final equations a little. For that purpose we want to establish a relationship between the initial pressures P1P1 and P2P2 and the final pressure PfinPfin. In order to get there we use the ideal gas law and the additivity relations for the volumes and amounts: From

ntot=n1+n2 ,Vtot=V1+V2P1V1=n1RT ,P2V2=n2RT ,PfinVtot=ntotRTntot=n1+n2 ,Vtot=V1+V2P1V1=n1RT ,P2V2=n2RT ,PfinVtot=ntotRT

it follows that

P1V1+P2V2=PfinVtot .P1V1+P2V2=PfinVtot .

Now, we can express P2P2 in terms of PfinPfin by introducing the deviation factor αα which is defined by P2=(1−α)PfinP2=(1−α)Pfin and use this in the previous equation

P1V1+P2V2P1V1+(1−α)PfinV2P1V1P1=PfinVtot=PfinVtot=Pfin(Vtot−V2=V1+αV2)=(1+αV2V1)PfinP1V1+P2V2=PfinVtotP1V1+(1−α)PfinV2=PfinVtotP1V1=Pfin(Vtot−V2⏟=V1+αV2)P1=(1+αV2V1)Pfin

which gives us the relation between P1P1 and PfinPfin. Introducing these equations into equation (4) yields

Gini=ntot(G0ini+x1RTln(P1P0)+x2RTln(P2P0))=ntot(G0ini+x1RTln(1+αV2V1)PfinP0+x2RTln(1−α)PfinP0)=ntot(G0ini+x1RTln(1+αV2V1)+x2RTln(1−α)+(x1+x2)=1RTln(PfinP0))=ntot(G0ini+x1RTln(1+αV2V1)+x2RTln(1−α)+RTln(PfinP0)) .Gini=ntot(Gini0+x1RTln⁡(P1P0)+x2RTln⁡(P2P0))=ntot(Gini0+x1RTln⁡(1+αV2V1)PfinP0+x2RTln⁡(1−α)PfinP0)=ntot(Gini0+x1RTln⁡(1+αV2V1)+x2RTln⁡(1−α)+(x1+x2)⏟=1RTln⁡(PfinP0))=ntot(Gini0+x1RTln⁡(1+αV2V1)+x2RTln⁡(1−α)+RTln⁡(PfinP0)) .

At the final state the chambers are connected and the gases mix, so the chambers are not independent subsystems anymore and the Gibbs free energy cannot be broken down into independent parts like before. Also, the ambient pressure in the system changes to Ptot=PfinPtot=Pfin. Now, the complete system of connected chambers must be treated via equation (1)

Gfinal=∑ini=xintot(μ0i+RTlnxi+RTln(PfinP0))=ntot(∑ixiμ0i+∑iRTxilnxi+∑ixiRTln(PfinP0)=constant)=ntot(∑ixiμ0i=G0ini+RT∑ixilnxi+RTln(PfinP0)∑ixi=1)=ntot(G0ini+RT∑ixilnxi+RTln(PfinP0))=ntot(G0ini+RT(x1lnx1+x2lnx2)+RTln(PfinP0)) .Gfinal=∑ini⏟=xintot(μi0+RTln⁡xi+RTln⁡(PfinP0))=ntot(∑ixiμi0+∑iRTxiln⁡xi+∑ixiRTln⁡(PfinP0)⏟=constant)=ntot(∑ixiμi0⏟=Gini0+RT∑ixiln⁡xi+RTln⁡(PfinP0)∑ixi⏟=1)=ntot(Gini0+RT∑ixiln⁡xi+RTln⁡(PfinP0))=ntot(Gini0+RT(x1ln⁡x1+x2ln⁡x2)+RTln⁡(PfinP0)) .

Now, that we have the Gibbs free energies of the initial and final state we simply need to subtract the former from the latter one in order to obtain the change in Gibbs free energy during the mixing process:

ΔmixG=Gfinal−Gini=ntot(G0ini+RT(x1lnx1+x2lnx2)+RTln(PfinP0))−ntot(G0ini+x1RTln(1+αV2V1)+x2RTln(1−α)+RTln(PfinP0))=ntotRT(x1lnx1+x2lnx2)−ntotRT(x1ln(1+αV2V1)+x2ln(1−α))(5)ΔmixG=Gfinal−Gini=ntot(Gini0+RT(x1ln⁡x1+x2ln⁡x2)+RTln⁡(PfinP0))−ntot(Gini0+x1RTln⁡(1+αV2V1)+x2RTln⁡(1−α)+RTln⁡(PfinP0))=ntotRT(x1ln⁡x1+x2ln⁡x2)−ntotRT(x1ln⁡(1+αV2V1)+x2ln⁡(1−α))(5)

Now you see why we went through those substitutions for P1P1 and P2P2. This way the ntotRTln(PfinP0)ntotRTln⁡(PfinP0) terms cancelled in equation (5).

Differentiating equation (5) with respect to the temperature we get the entropy of mixing since ∂G∂T∣∣p,{ni}=−S∂G∂T|p,{ni}=−S

ΔmixS=−∂ΔmixG∂T∣∣∣p,{ni}=−ntotR(x1lnx1+x2lnx2)+ntotR(x1ln(1+αV2V1)+x2ln(1−α))ΔmixS=−∂ΔmixG∂T|p,{ni}=−ntotR(x1ln⁡x1+x2ln⁡x2)+ntotR(x1ln⁡(1+αV2V1)+x2ln⁡(1−α))

The first term −ntotR(x1lnx1+x2lnx2)−ntotR(x1ln⁡x1+x2ln⁡x2)is just the usual entropy of mixing. The second term ntotR(x1ln(1+αV2V1)+x2ln(1−α))ntotR(x1ln⁡(1+αV2V1)+x2ln⁡(1−α))describes the contribution of the pressure equilibration to the entropy.