Calculating Absolute 3rd Law Entropy of a Gas Solution at 25C & 1 atm

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Discussion Overview

The discussion revolves around calculating the absolute third law entropy of a gas solution containing methane and ethane at 25°C and 1 atm. Participants explore the application of the mixing entropy equation and the relevance of absolute entropies of the individual gases in this context.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant proposes using the equation dSmix = -nR(x1lnx1 + x2lnx2) to calculate the mixing entropy, questioning the necessity of the absolute entropies of methane and ethane in this calculation.
  • Another participant clarifies that the absolute entropies of the gases are not useless, as they contribute to the total entropy calculation when mixed.
  • A participant expresses confusion regarding the summation of entropies for unmixed gases and seeks clarification on the concept of extensive quantities.
  • There is a reiteration that the total entropy for unmixed gases is the sum of their individual entropies, which is a fundamental property of extensive quantities.

Areas of Agreement / Disagreement

Participants exhibit some disagreement regarding the role of absolute entropies in the mixing entropy calculation, with no consensus reached on whether they are necessary for the solution.

Contextual Notes

Participants express uncertainty about the definitions and implications of mixing entropy and absolute entropy, indicating potential limitations in understanding the relationship between these concepts.

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Homework Statement


at 25C & 1 atm, absolute entropy of methane and ethane are 186.19 J/K/mol and 229.49 J/K/mol in gas phase. Find absolute 3rd law entropy of a solution containing 1 mol of each gas in ideal behavior.


Homework Equations


I think this is the equation to use:
dSmix= -nR(x1lnx1 + x2lnx2)
where x1 and x2 i can find from 1 mole of each gas (mole fractions)

if i use this equation, then the absolute entropy of methane and ethane woud be useless in this equation. or is there another way of solving it with the absolute entropies?

Thanks in advance.
 
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What do you mean by "the absolute entropy of methane and ethane woud be useless in this equation"? (Note that the term represents an additional term to augment the sum of the unmixed entropies.)
 
Mapes said:
What do you mean by "the absolute entropy of methane and ethane woud be useless in this equation"? (Note that the term represents an additional term to augment the sum of the unmixed entropies.)

I mean the values 186.19 J/K/mol and 229.49 J/K/mol for methane and ethane given in the question. If i use the equation i said above, then these values would not be needed; so I was thinking there might be another way to do this...

"Note that the term represents an additional term to augment the sum of the unmixed entropies." - I don't get what you mean... which term are you referring to? the absolute entropies?
 
The total entropy of A and B, unmixed, is S_A+S_B.
The total entropy of A and B, mixed, is S_A+S_B+\Delta S_\mathrm{mixing}. Does this help?
 
Mapes said:
The total entropy of A and B, unmixed, is S_A+S_B.
The total entropy of A and B, mixed, is S_A+S_B+\Delta S_\mathrm{mixing}. Does this help?


i don't really get why the entropies are summed if they are ummixed.
 
Entropy is an extensive quantity. If one container has entropy S_A and the other has entropy S_B, then the total amount of entropy is S_A+S_B before mixing.
 
thanks!
 

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