Determining an expression for an entropy equation

[Benzoate]

Homework Statement

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)

Homework 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]

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))

 

[Mapes]

You can't apply \Delta S_A=Nk\ln 2 to the general problem; that's the increase in entropy for a single gas expanding into twice its original volume. If x can vary, there's no reason to assume the volume doubles.

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

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


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

dS=\frac{nR}{V}\,dV

to calculate the change in entropy.