Class Note 14
3
n A A atoms n B B atoms September 18, 2001 Reading: Chapter Four Homework: 4.1,4.2,4.3,4.4 Boltzmann Equation: Consider adding a small amount of heat to a system of fixed V and T, So, δw = 0. First law: q w q dU δ δ δ = − = T dU T q dS = = δ B B k dS T k dU d = = Ω) (ln So, ) (ln Ω = d k dS B ) (ln Ω = B k S --- Boltzmann equation. Configurational Entropy: Consider two crystalline materials that are completely miscible in each other in physical contact, such as Au(A) and Ag(B). Suppose there are n A number of A atoms and n B number of B atoms. Initially, two crystals are completely separately. Number of ways of distributing n A atoms on n A sites is 1 ! ! = = Ω A A A n n ; similarly, number of ways of distributing n B atoms on n B sites is 1 ! ! = = Ω B B B n n . Thus, 1 = Ω Ω = Ω B A initial Now, let two crystals completely mixing up, i.e. distributing n A A atoms and n B B atoms on (n A +n B ) sites. The number of ways in doing so is ! ! )! ( b A B A final n n n n + = Ω
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Transcript of Class Note 14
nA A atoms
nB B atoms
September 18, 2001 Reading: Chapter Four Homework: 4.1,4.2,4.3,4.4 Boltzmann Equation: Consider adding a small amount of heat to a system of fixed V and T, So, δw = 0. First law: qwqdU δδδ =−=
T
dUTqdS == δ
BB k
dSTk
dUd ==Ω)(ln
So, )(ln Ω= dkdS B )(ln Ω= BkS --- Boltzmann equation. Configurational Entropy: Consider two crystalline materials that are completely miscible in each other in physical contact, such as Au(A) and Ag(B). Suppose there are nA number of A atoms and nB number of B atoms. Initially, two crystals are completely separately. Number of ways of distributing nA atoms on nA
sites is 1!!
==ΩA
AA n
n; similarly, number of ways of distributing nB atoms on nB
sites is 1!!
==ΩB
BB n
n.
Thus, 1=ΩΩ=Ω BAinitial Now, let two crystals completely mixing up, i.e. distributing nA A atoms and nB B atoms on (nA+nB) sites. The number of ways in doing so is
!!)!(
bA
BAfinal nn
nn +=Ω
Thus, the entropy of mixing (configurational entropy), i.e., the increase of entropy due to mixing is
ffBiBfBif SkkkSSS =Ω=Ω−Ω=−=∆ lnlnln
!!)!(
lnlnbA
BABfB nn
nnkkS +=Ω=
Using Stirling’s relation:
]lnln)()ln()[( BBBAAABABABAB nnnnnnnnnnnnkS +−+−+−++= ]ln)ln(ln)ln([ BBBABAABAAB nnnnnnnnnnk −++−+=
BA
BBB
BA
AAB nn
nnknn
nnk+
−+
−= lnln
So, )lnln( BBAAB NkS χχχχ +−=
Where BA
BB
BA
AA nn
nnn
n+
=+
= χχ , , is the mole fraction, i.e., concentration.
BA nnN += = total number of particles.
For 1 mole, 0NN = = Avogadro’s number
)lnln( BBAARS χχχχ +−= This is known as configurational entropy. It corresponds to maximum entropy of mixing with uniform composition. Since U and V are fixed (no heat, no work), the criterion for equilibrium is to maximize entropy. Example: Prove that the configurational entropy is indeed the maximum in a two-component system if the composition is uniform. Solution: Starting with the uniform composition, suppose we move a few atoms from one side to another to create a compositional inhomogeneity. So, let’s say we shift n0 A atoms from left to right, and n0 B atoms from right to left.
χAnA - n0 A atoms χBnA + n0B atoms
χAnB + n0 A atoms χBnB - n0 B atoms
nA sites nB sites
The new Ω can be calculated as
)!()!(!
)!()!(!
0000 nnnnn
nnnnn
BBBA
B
ABAA
A
−+⋅
+−=Ω
χχχχ
Thus,
)ln()()ln()(ln)ln()()ln()(lnln
0000
0000
nnnnnnnnnnnnnnnnnnnn
BBBBBABABB
ABABAAAAAA
−−−++−+++−−−−=Ω
χχχχχχχχ
The objective is to determine n0 for which Ωln is a maximum.
1)ln(1)ln(1)ln(1)ln()(ln0 00000
+−+−+−−+−+−=Ω≡ nnnnnnnndn
dBBBAABAA χχχχ
)ln()ln()ln()ln( 0000 nnnnnnnn ABBABBAA +++=−+− χχχχ
))(())(( 0000 nnnnnnnn ABBABBAA ++=−− χχχχ
)()( 0
200
20 ABBABABABBAABABA nnnnnnnnnnnn χχχχχχχχ +++=+−+
0)(0 =+++ ABBABBAA nnnnn χχχχ
00 =+ BA nnn
Since 0,0 0 =≠+ nnn BA Furthermore,
01111
1111)ln(
00000020
2
00
<−−−−=
−−
+−
+−
−−=Ω
==
BBBAABAA
nBBBAABAAn
nnnn
nnnnnnnndnd
χχχχ
χχχχ
Thus, Ωln is a maximum when n0=0, i.e., when the composition is uniform.
