20.110J / 2.772J / 5.601J
Thermodynamics of Biomolecular Systems
Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
20.110/5.60 Fall 2005
Lecture #7
page 1
Entropy, Reversible and Irreversible Processes and
Disorder
Examples of spontaneous processes
T1
Connect two metal blocks thermally in
an isolated system (∆U = 0)
T2
Initially
T1 ≠T2
dS = dS1 + dS2 =
đq1
T1
−
đq2
T2
= đq1
(T2 −T1 )
TT
1 2
( đq1 = −đq2 )
dS > 0 for spontaneous process
⇒
if T2 >T1
T2 <T1
gas
V
vac.
V
⇒
⇒
đq1 > 0
đq1 < 0
in both cases heat flows
from hot to cold as expected
Joule expansion with an ideal gas
1 mol gas (V,T)
∆U = 0
adiabatic
=
q=0
1 mol gas (2V,T)
w=0
Need a reversible path to compute ∆S from q! Close the cycle and go
back to the initial state reversibly and isothermally
∆S = −∆Sbackwards
20.110J / 2.772J / 5.601J
Thermodynamics of Biomolecular Systems
Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
20.110/5.60 Fall 2005
Lecture #7
page 2
qrev ≠ 0
1 mol gas (2V,T) = 1 mol gas (V,T)
∆Sbackwards =
∴
∫
đqrev
= −∫
T
đw
T
=
V
∫V
2
RdV
1
= R ln
V
2
spontaneous
∆S = R ln2 > 0
IMPORTANT!! To calculate ∆S for the irreversible process, we
needed to find a reversible path so we could determine
đqrev and
∫
đqrev
T
.
• Mixing of ideal gases at constant T and p
nA A (g, VA, T) + nB A (g, VB, T) = n (A + B) (g, V, T)
nA
VA
nB
VB
spontaneous
mixing
n = nA + nB
V = VA + VB
To calculate ∆Smix , we need to find a reversible path between
the two states.
constant T
A
B
A+ B
piston
permeable
to A only
piston
permeable
to B only
∆Sdemix = −∆Smix
back to initial state
function of state
20.110J / 2.772J / 5.601J
Thermodynamics of Biomolecular Systems
Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
20.110/5.60 Fall 2005
Lecture #7
page 3
For demixing process
⇒
∆U = 0
⇒
qrev = −w rev = pAdVA + pB dVB
work of compression of each gas
∴
∆Sdemix =
∫
VA p dV
VB p dV
V
dqrev
V
= ∫ A A + ∫ B B = nAR ln A + nB R ln B
V
V
T
T
T
V
V
Put in terms of mole fractions
Ideal gas ⇒
∴
⇒
XA =
VA
V
XB =
XA =
nA
n
XB =
nB
n
VB
V
∆Sdemix = nR [XA ln X A +XB ln XB ]
∆Smix = −nR [XA ln XA +XB ln XB ]
Since XA , XB < 1
⇒
∆Smix > 0
mixing is always spontaneous
The mixed state is more “disordered” or “random” than the
demixed state.
Smixed > Sdemixed
This is a general result:
Entropy is a measure of the disorder of a system
20.110J / 2.772J / 5.601J
Thermodynamics of Biomolecular Systems
Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
20.110/5.60 Fall 2005
\
Lecture #7
page 4
For an isolated system (or the universe)
∆S > 0
∆S = 0
∆S < 0
Spontaneous, increased randomness
Reversible, no change in disorder
Impossible, order cannot “happen” in isolation
There is an inexorable drive for the universe to go to a
maximally disordered state.
Microscopic understanding: Boltzmann Equation of Entropy:
S = k ln Ω
Where k is Boltzmann’s constant (k=R/NA).
And Ω is the number of equally probable microscopic
arrangements for the system.
This can also be used to calculate ∆S
In the case of the Joule expansion of an ideal gas in volume V
expanding to a volume 2V (as on the first page of these notes):
if we divide the initial volume V into m small cubes, each with
volume v, so that mv=V, the number of ways of placing the N
molecules of ideal gas into these small cubes initially is mN.
After the expansion the number of ways of placing the n
molecules of ideal gas into the now 2m small cubes is (2m)N.
The number of probably microscopic arrangements initially is:
N
N
Ω ∝ ( m ) , or Ω = C ( m ) (C is a constant)
20.110J / 2.772J / 5.601J
Thermodynamics of Biomolecular Systems
Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
20.110/5.60 Fall 2005
Lecture #7
page 5
After the expansion it is:
N
N
Ω ∝ (2m ) , or Ω = C (2m )
So using Boltzmann’s equation to calculate ∆S for the
expansion:
N
∆S = k ln ⎡(2m ) ⎤ − k ln ⎡⎣m N ⎤⎦ = kN ln 2 = R ln 2
⎣
⎦
As we had found above!
More examples of ∆S calculations
In all cases, we must find a reversible path to calculate
(a)
∫
đqrev
T
Mixing of ideal gases at constant T and p
nA A (g, VA, T) + nB A (g, VB, T) = n (A + B) (g, V = VA + VB, T)
∆Smix = −nR [X A ln X A +XB ln XB ]
(b) Heating (or cooling) at constant V
A ( T1 , V) = A ( T2 , V )
∆S =
∫
đqrev
T
=
T2
∫T
1
CV dT
T
if CV is
=
T -independent
T
CV ln 2
T1
[Note ∆S > 0 if T2 >T1 ]
(c) Reversible phase change at constant T and p
e.g. H2O (l, 100∞C, 1 bar) = H2O (g, 100∞C, 1 bar)
20.110J / 2.772J / 5.601J
Thermodynamics of Biomolecular Systems
Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
20.110/5.60 Fall 2005
Lecture #7
page 6
q p = ∆Hvap
∆Svap (100°C ) =
(d)
q pvap
Tb
=
∆H vap
(Tb = boiling Temp at 1 bar)
Tb
Irreversible phase change at constant T and p
e.g. H2O (l, -10∞C, 1 bar) = H2O (s, -10∞C, 1 bar)
This is spontaneous and irreversible.
\
We need to find a reversible path between the two states
to calculate ∆S.
irreversible
H2O (l, -10∞C, 1 bar)
=
H2O (s, -10∞C, 1 bar)
đqrev = C p ( A )dT
H2O (l, 0∞C, 1 bar)
đqrev = C p ( s ) dT
reversible
q
=
rev
p
H2O (s, 0∞C, 1 bar)
= −∆Hfus
Note: ∆Hfus is for the process going from the solid state to the
liquid state, the opposite of what we have above, same for ∆Sfus.
∆S = ∆Sheating − ∆Sfus + ∆Scooling
=
Tfus
∫T
1
∴
∆S =
−∆Hfus
T
C p ( A ) dT −∆Hfus
T C p ( s ) dT
+
+∫
Tfus
T
Tfus
T
∆S =
1
−∆Hfus
T
Tfus
+∫
T1
Tfus
T1
+ ⎡⎣C p ( A ) − C p ( s ) ⎤⎦ ln
dT
⎡⎣C p ( A ) − C p ( s ) ⎤⎦
T
if Cp values are T-independent