20.110J / 2.772J / 5.601J
Thermodynamics of Biomolecular Systems
Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
Lecture 16
5.60/20.110/2.772
Absolute Entropy
Third Law of thermodynamics
• Absolute Entropies
Absolute entropy of an ideal gas
Start with fundamental equation
dU = TdS − pdV
dS =
dU + pdV
T
for ideal gas:
dU = CV dT and p =
dS =
nRT
V
CV dT nR
+
dV
T
V
At constant T, dT=0
dST =
pdV
T
dST =
nRdV
V
For an ideal gas, pV = nRT
At constant T
d ( pV ) = d (nRT ) = 0
pdV = −Vdp
plugging into dST:
dST = −
nRdp
p
This allows us to know how S(p) if T held constant. Integrate!
1
20.110J / 2.772J / 5.601J
Thermodynamics of Biomolecular Systems
Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
Lecture 16
5.60/20.110/2.772
For an arbitrary pressure p,
p
S(p,T ) = S( po ,T ) − ³ p o
§ p·
nRdp
= S(p o ,T ) − nRln ¨ o ¸
p
©p ¹
where po is some reference pressure which we set at 1 bar.
Ÿ S(p,T) = So(T) – nR lnp
(p in bar)
0
S ( p, T ) = S (T ) − R ln p
S ↓ as P↑
p
S°
But to finish, we still need S o (T ) !
o
Suppose we had S ( 0 K ) (standard molar entropy at 0 Kelvin)
dH = TdS + Vdp
dH = C p dT
for ideal gas
C p dT = TdS + Vdp
dS =
2
Cp
T
dT −
V
dp
T
20.110J / 2.772J / 5.601J
Thermodynamics of Biomolecular Systems
Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
Lecture 16
5.60/20.110/2.772
Cp
§ ∂S ·
=
Then using ©
∂T ¹ p T
o
S (T ) . Integrating over dS eqn, assuming Cp constant over T range:
we should be able to get
T2
dS =
Cp
³T
T1
p2
dT −
nR
³p p dp
1
So then
§T ·
§p ·
∆S = C p ln¨¨ 2 ¸¸ − nR ln¨¨ 2 ¸¸
© T1 ¹
© p1 ¹
for p = 1bar
§T ·
= C p ln¨¨ 2 ¸¸ − nR ln p
© T1 ¹
Given Cp, T1, p1, → T2, p2, can calculate ∆S.
We will use T=0K as a reference point.
Consider the following sequence of processes for the substance A:
A(s,0K,1bar) = A(s,Tm,1bar) = A(l,Tm,1bar) = A(l,Tb,1bar) = A(g,Tb,1bar) = A(g,T,1bar)
S (T,1bar) = S o (0K ) + ³
Tm
0
So(T)
T b C p ()dT
T C p (g)dT
C p (s)dT ∆H fus
∆H vap
+
+³
+
+³
Tm
Tb
T
T
T
Tm
Tb
∆S = ³
CpdT
T
Liquid boils, ∆S =
∆H fus
Solid melts, ∆S =
T
0
3
T
∆H vap
T
20.110J / 2.772J / 5.601J
Thermodynamics of Biomolecular Systems
Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
Lecture 16
5.60/20.110/2.772
Since ∆S0 is positive for each of these processes, the entropy must have its smallest possible value
o
at 0 K. If we take S ( 0 K ) = zero for every pure substance in its crystalline solid state, then we could
calculate the entropy at any other temperature.
This leads us to the Third Law of Thermodynamics:
• THIRD LAW:
First expressed as Nernst's Heat Theorem:
Nernst (1905):
As T → 0 K , ∆S → 0 for all isothermal processes in condensed phases
More general and useful formulation by M. Planck:
Planck (1911):
As T → 0 K , S → 0 for every chemically homogeneous substance in a perfect crystalline state
Justification:
?
@
It works!
Statistical mechanics (5.62) allows us to calculate the
o
entropy and indeed predicts S ( 0 K ) = 0.
This leads to the following interesting corollary:
It is impossible to decrease the temperature of any system to T = 0 K in a finite number of steps.
How can we rationalize this statement?
Recall the fundamental equation, dU = T dS – p dV
dU = Cv dT
so
For 1 mole of ideal gas, P = RT/V
Cv dT = T dS – (RT/V) dV
dS = Cv d (ln T) + R d (ln V)
4
20.110J / 2.772J / 5.601J
Thermodynamics of Biomolecular Systems
Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
Lecture 16
T2,
5.60/20.110/2.772
For a spontaneous adiabatic process which takes the system from T1 to a lower temperature
∆S = Cv ln (T2/T1) + R ln (V2/V1) ≥ 0
but if T2 = 0, Cv ln (T2/T1) equals minus infinity !
Therefore R ln (V2/V1) must be greater than plus infinity, which is impossible. Therefore no
actual process can get you to T2 = 0 K.
But you can get very very close!
In W. Ketterle's experiments on "Bose Einstein Condensates" (recent MIT Nobel Prize in
Physics), atoms are cooled to nanoKelvin temperatures (T = 10-9 K) … but not to 0 K !
_______________
Some apparent violations of the third law (but which are not !)
Any disorder at T = 0 K gives rise to S > 0
• mixed crystals
If have an unmixed crystal, N atoms in N sites:
Ω=
N!
=1
N!
S = k ln 1 = 0
But if mixed crystal:
NA of A
NB of B
NA + NB = N
Ω=
N!
N A! N B !
S = k ln
Use Stirling’s approx:
5
N!
N A! N B !
20.110J / 2.772J / 5.601J
Thermodynamics of Biomolecular Systems
Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
Lecture 16
5.60/20.110/2.772
ln N != N ln N − N
S = k ( N ln N − N − N A ln N A + N A − N B ln N B + N B )
= − k (− N ln N + N A ln N A + N A + N B ln N B )
Using mole fractions: NA =xAN, NB =xBN
[
∆Smix = −nR X A ln X A + X B ln X B
]
> 0 Always !!! Even at T=0K
But a mixed crystal is not a pure substance, so the third law is not violated.
• Any impurity or defect in a crystal also causes S > 0 at 0 K
• Any orientational or conformational degeneracies such as in a molecular crystal causes S > 0 at 0
K, for example in a carbon monoxide crystal, two orientations are possible:
CO
CO
CO
CO
CO
CO
CO
CO
CO
CO
CO
CO
OC
CO
CO
CO
CO
OC
CO
CO
CO
CO
CO
CO
CO
CO
CO
CO
6