Classical gases not necessarily ideal

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Discussion Overview

The discussion revolves around the relationships between pressures and volumes of three classical gases (not necessarily ideal) in thermal equilibrium. Participants explore how to derive an equation relating the states of gases B and C based on their interactions with gas A, while considering the implications of thermal equilibrium and the constants involved.

Discussion Character

  • Mathematical reasoning
  • Debate/contested
  • Technical explanation

Main Points Raised

  • Some participants propose that to find the relationship between gases B and C in thermal equilibrium, one should equate the equations derived from the thermal equilibria of A with B and A with C.
  • Others argue that using the notation "AB = AC" is unconventional and may lead to confusion in the mathematical representation of the problem.
  • A participant suggests substituting the expression for PaVa from one equation into the other to derive the relationship for B and C.
  • Another participant articulates a structured understanding of the problem, emphasizing the need to derive an identity for the case where T_B = T_C using the measurable quantities and constants.
  • One participant mentions the existence of an empirical temperature function related to the pressures and volumes of the gases, indicating different behaviors for ideal and non-ideal gases.

Areas of Agreement / Disagreement

Participants express varying interpretations of how to approach the problem, with no consensus on the correct method for deriving the equation for gases B and C. Some agree on the need for an empirical temperature function, while others focus on the mathematical relationships without reaching a unified conclusion.

Contextual Notes

There are unresolved assumptions regarding the nature of the gases and the specific conditions under which the equations hold. The discussion also highlights the complexity of relating non-ideal gas behavior to thermal equilibrium conditions.

Logan Land
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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?
 
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Logan Land said:
AB = AC
What does this equation have to do with the relationships you gave originally?
Logan Land said:
PbVb-(beta)Pb-(alpha)Vb+(alpha)(beta)-PaVa=0
Logan Land said:
PcVc-PaVa-((gamma)PaVa)/Pc=0
 
when A and B are in thermal equilibrium its gives an equation so I am 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?
 
Logan Land said:
is that not correct?
"Conditionally." Please do not use a notation like
Logan Land said:
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.
 
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.
 
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
 
Thermal equilibrium is transitive (zeroth law). I suppose you are expected to construct an empirical temperature function.
 
  • #10
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
 
  • #11
Chestermiller said:
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.
 
  • #12
Quantum Defect said:
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!
 
  • #13
Logan Land said:
PbVb-(beta)Pb-(alpha)Vb+(alpha)(beta)-PaVa=0
PcVc-PaVa-((gamma)PaVa)/Pc=0
I get
PcVc-(1+(gamma)/Pc)(PbVb-(beta)Pb-(alpha)Vb+(alpha)(beta))=0
You can also say that there exists an empirical temperature
theta=PaVa=PbVb-(beta)Pb-(alpha)Vb+(alpha)(beta)=PcVc/(1-(gamma)/Pc)
The first expression holds for an ideal gas, the second one for a van der Waals gas and the third one is some kind of virial expansion.
 

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