How Is PT Related to PA, PB, and PC in a Rigid Gas Mixture?

[ksle82]

A rigid container is filled with a mixture of three gases: A, B and C. The pressure gage reads PT. The container is evacuated and filled with an amount of gas A equal to that in the original mixture. The gage now reads PA. This process is repeated for gases B and C with the pressure gage reading PB and PC respectively. How is PT related to PA, PB, and PC? What assumptions are made?

Assumptions: 1) constant volume, 2) no heat loss, 3)PT>TA,TB,TC
-these are the assumptions i can think of as of right now. are there any other relevant ones?

As for the question How is PT related to PA, PB, and PC?
Answer: PT = PA + PB + PC
- is this the right aswer or it's more complicated than that?

 

[LeonhardEuler]

A rigid container is filled with a mixture of three gases: A, B and C. The pressure gage reads PT. The container is evacuated and filled with an amount of gas A equal to that in the original mixture. The gage now reads PA. This process is repeated for gases B and C with the pressure gage reading PB and PC respectively. How is PT related to PA, PB, and PC? What assumptions are made?

Assumptions: 1) constant volume, 2) no heat loss, 3)PT>TA,TB,TC
-these are the assumptions i can think of as of right now. are there any other relevant ones?

As for the question How is PT related to PA, PB, and PC?
Answer: PT = PA + PB + PC
- is this the right aswer or it's more complicated than that?

You're final answer would be right with one more assumption: what kind of gas does it need to be?

Also, I think assumption #2 is worded a little wierd. Since the mixed gas is removed and a new source of gas is used to fill the container with A, I don't think it makes sense to say "no heat loss". I think a better way to phrase it would be "constant temperature".

 

[ksle82]

Also, I think assumption #2 is worded a little wierd. Since the mixed gas is removed and a new source of gas is used to fill the container with A, I don't think it makes sense to say "no heat loss". I think a better way to phrase it would be "constant temperature".

You're right. The "no heat loss" assumption was kind of vague. I should have said that the container is fully insulated so no heat can escape once the it is fully closed.

I don't think i can asume constant temperature since no info in the question giving hint that i can make that assumption.

 

[Miro]

I think u know that this depends on gases,
if it's the case of perfect gases, just use PV=nRT
so cos it's a constant volume & temperature, u'll get:
V/(RT)=nA/PA=nB/PB=nC/PC=(nA+nB+nC)/PT

I wish I was helpful,

 

[Andrew Mason]

A rigid container is filled with a mixture of three gases: A, B and C. The pressure gage reads PT. The container is evacuated and filled with an amount of gas A equal to that in the original mixture. The gage now reads PA. This process is repeated for gases B and C with the pressure gage reading PB and PC respectively. How is PT related to PA, PB, and PC? What assumptions are made?

Assumptions: 1) constant volume, 2) no heat loss, 3)PT>TA,TB,TC
-these are the assumptions i can think of as of right now. are there any other relevant ones?

As for the question How is PT related to PA, PB, and PC?
Answer: PT = PA + PB + PC
- is this the right aswer or it's more complicated than that?

I would say that PT=PA+PB+PC but it assumes the individual gases in the container are kept at the same temperature as the original mixture.

In the mixture, A occupied only a portion of the volume, as did B and C. So when it is all by itself, A must occupy a larger volume. If A is injected at the same temperature as the original mixture, it will cool as it expands to the larger volume. I am assuming it is warmed to the original temperature of the mixture before PA is read. Ditto for the other gases.

AM

 

[LeonhardEuler]

Don't forget that Dalton's law of partial pressures does not apply to all gases. It applies to ideal gases. In the case of a non-ideal gas the partial pressure does not have the interpretation that it is the pressure that the gas would exert if it alone occupied the entire volume of the container.