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

[higherme]

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.

 

[Mapes]

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.)

 

[higherme]

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?

 

[Mapes]

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?

 

[higherme]

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.

 

[Mapes]

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.

 

[higherme]

thanks!