SME
Thermal & Statistical Physics · Problem type
A two-state paramagnet: spins up and down
Use this for “A paramagnet of 10 spins has 7 pointing up. How many microstates have 7 up, and what is the entropy?”
Diagram
Row 12: 1, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1; they add to 2¹² = 4,096.
Entry 6 of row 12: C(12, 6) = 924.
The two above add to it: C(11, 5) + C(11, 6) = 462 + 462 = 924.
This is an example. Type your own values to start a new problem.
Ω
=
C(
N
,
n
)
=
σ
Entropy over kB
worked out (not typed)
S
Entropy
worked out (not typed) · J/K
Formulas
Ω
=
C(
N
,
n
)
924 = C(12, 6)
σ
=
ln(
Ω
)
6.8287 = ln(924)
S
=
1.381 ×
10
-23
×
σ
9.4305 ×
10
-23
=
1.381 ×
10
-23
×
6.8287
Assumptions
•
Each spin points up or down, the spins can be told apart, and the macrostate is how many point up.
•
Ω counts the ways to choose which n of the N spins are up; half up gives the most.
•
Every microstate is equally likely, so S = kB ln Ω, with kB = 1.381 × 10⁻²³ J/K; all up is one microstate, S = 0.
Step-by-step
WE KNOW
N = 12, n = 6
FIND
multiplicity (Ω), entropy over kB (σ), entropy (S)
Step 1 · Find multiplicity (Ω)
Use Ω = C(N, n)
Choose
which
n
of
the
N
spins
point
up:
entry
n
of
row
N
of
Pascal’s
triangle.
Ω
=
C(12,
6)
Ω
=
924
Step 2 · Find entropy over kB (σ)
Use σ = ln(Ω)
Boltzmann’s
entropy
in
units
of
kB,
σ
=
S
kB
:
the
log
of
the
number
of
microstates.
σ
=
ln(924)
σ
=
6.828712
Step 3 · Find entropy (S)
Use S = 1.381 × 10⁻²³ × σ
S
=
kBσ:
multiply
by
kB
for
joules
per
kelvin.
S
=
1.381 ×
10
-23
×
6.828712
S
=
9.4305 ×
10
-23
J/K
Check
Put the numbers back in. Both sides should match.
924 = C(12, 6) ✓
6.828712 = ln(924) ✓
9.4305 ×
10
-23
=
1.381 ×
10
-23
×
6.828712
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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