determining an expression for an entropy equation


by Benzoate
Tags: determining, entropy, equation, expression
Benzoate
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#1
Mar11-08, 06:58 AM
P: 569
1. The problem statement, all variables and given/known data

Calculate the entropy of mixing for a system of two monatomic ideal gases, A and B ,whose relative proportion is arbitrary. Let N be the total number of molecules and let x be the fraction of these that are of species B. You should find
delta(S) mixing=-Nk[x ln x +(1-x) ln (1-x)

2. Relevant equations

delta (S(total))=delta(S(A)) + delta(S(B))=2Nk ln 2
S=Nk[ln((V/N)(((4*pi*m*U)/3Nh^2)^(3/2))+2.5]


3. The attempt at a solution

according to my thermal physics text, delta(S(A))=Nk ln 2 . The problem says that in species B , x is just a fraction of N. Then , I think I would have to conclude that delta(S(B))=x/N*(k)*ln(2).

so would my expressison be :delta(S(mixing))=delta(S(A))+delta(S(B))=Nk ln 2+ xk/N*(ln(2))
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Mapes
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#2
Mar11-08, 08:27 AM
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You can't apply [itex]\Delta S_A=Nk\ln 2[/itex] to the general problem; that's the increase in entropy for a single gas expanding into twice its original volume. If [itex]x[/itex] can vary, there's no reason to assume the volume doubles.

Also, remember that as [itex]x[/itex] increases, there are no longer [itex]N[/itex] molecules of gas A but rather [itex](1-x)N[/itex].

One common way to show your desired relation is to assume that each gas expands from its original volume into the total volume and to use the Maxwell relation


[tex]\left(\frac{\partial S}{\partial V}\right)_T=\left(\frac{\partial P}{\partial T}\right)_V=\frac{nR}{V}\quad[/tex]

[tex]dS=\frac{nR}{V}\,dV[/tex]

to calculate the change in entropy.


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