# Classical gases not necessarily ideal

## Main Question or Discussion Point

Systems A, B, and C are classical gases (not necessarily ideal), each with the same number of molecules N ( or same number of moles n if you prefer), where N is constant. We can measure pressures and volumes Pa,Va ; Pb,Vb ; and Pc,Vc for each system. When A and B are in thermal equilibrium, our measurements show that their pressure and volumes satisfy:
PbVb-(beta)Pb-(alpha)Vb+(alpha)(beta)-PaVa=0

When A and C are in thermal equilibrium, we find:
PcVc-PaVa-((gamma)PaVa)/Pc=0

where (alpha),(beta), and (gamma) are constants.
Find the equation relating Pb,Vb and Pc,Vc that is satisfied when system B and C are in thermal equilibrium.

Would I just set the equation for AB = AC and move the B and C's to one said and A to the other?

Bystander
Homework Helper
Gold Member
AB = AC
What does this equation have to do with the relationships you gave originally?
PbVb-(beta)Pb-(alpha)Vb+(alpha)(beta)-PaVa=0
PcVc-PaVa-((gamma)PaVa)/Pc=0

when A and B are in thermal equilibrium its gives an equation so im calling that AB then it states when A and C are in thermal equilibrium it gives a different equation so I called that AC so to find an equation that makes B and C in thermal equilibrium I was thinking to make each equation equal to each other.

is that not correct?

Bystander
Homework Helper
Gold Member
is that not correct?
"Conditionally." Please do not use a notation like
AB = AC
to indicate such an operation --- it's --- "unconventional."

how would I proceed to start to get the correct equation for B and C in thermal equilibrium?
I simply thought that is I did PbVb-(beta)Pb-(alpha)Vb+(alphabeta)-PaVa=PcVc-PaVa-(gammaPaVa)/Pc then move PaVa's to 1 side and PcVc and PbVb to the other that would be the equation I was asked to find.

Bystander
Homework Helper
Gold Member
Your first equation gives you PAVA. Plug it into the second.

oo so then B and C are in thermal equilibrium with this equation?
PcVc-PbVb+(beta)Pb+(alpha)Vb-(alphabeta)-(((gamma)PbVb-(beta)Pb-(alpha)Vb+(alphabeta))/Pc)

that may look confusing but hopefully you understand.

Chestermiller
Mentor
The gases are being kept separate from one another, and are not being mixed together; they are only being brought into thermal contact and allowed to equilibrate thermally, correct?

Chet

DrDu
Thermal equilibrium is transitive (zeroth law). I suppose you are expected to construct an empirical temperature function.

Chestermiller
Mentor
This is a pretty tricky problem. Let me articulate my understanding of the question, and see if this agrees with everyone else's.

1. You have N moles of three non-ideal gases in 3 different containers.
2. The gases are initially at different pressures, volumes, and temperatures.
3. The constant-volume containers are brought together in pairs and allowed to thermally equilibrate (i.e., reach equal temperatures).
4. From the relationship between the final equilibrated pressures of the two gases in two of the cases that have been equilibrated, you are supposed to deduce the relationship when the third pairing of the gases is carried out.

Chet

Quantum Defect
Homework Helper
Gold Member
This is a pretty tricky problem. Let me articulate my understanding of the question, and see if this agrees with everyone else's.

1. You have N moles of three non-ideal gases in 3 different containers.
2. The gases are initially at different pressures, volumes, and temperatures.
3. The constant-volume containers are brought together in pairs and allowed to thermally equilibrate (i.e., reach equal temperatures).
4. From the relationship between the final equilibrated pressures of the two gases in two of the cases that have been equilibrated, you are supposed to deduce the relationship when the third pairing of the gases is carried out.

Chet
That's what it sounds like to me.

i.e. if T_A = T_B, then the first equation is equal to zero. If T_A = T_C, the second equation is equal to zero. Derive an identity (equal to zero) for the case where T_B = T_C, using the measurables (P and V for A and B) and alpha, beta and gamma.

Quantum Defect
Homework Helper
Gold Member
That's what it sounds like to me.

i.e. if T_A = T_B, then the first equation is equal to zero. If T_A = T_C, the second equation is equal to zero. Derive an identity (equal to zero) for the case where T_B = T_C, using the measurables (P and V for A and B) and alpha, beta and gamma.
I finally figured out where I heard of "classical gas" before!

DrDu