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5.60 Thermodynamics & Kinetics
Spring 2008
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5.60 Spring 2008
Lecture #10
page 1
Entropy and Disorder
• 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
= V
A +
VB
To calculate ΔSmix , we need to find a reversible path between
the two states.
constant T
A+B
piston
permeable
to A only
piston
permeable
to B only
⇒
ΔU = 0
⇒
B
back to initial state
ΔSdemix = −ΔSmix
For demixing process
A
function of state
qrev = −w rev = pAdVA + pB dVB
work of compression of each gas
∴
ΔSdemix =
∫
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 =
VB
V
XA =
nA
n
XB =
nB
n
5.60 Spring 2008
Lecture #10
∴
⇒
page 2
ΔSdemix = nR [XA ln X A +XB ln XB ]
ΔSmix = −nR [X A ln XA +XB ln XB ]
Since XA , XB < 1
⇒
ΔSmix > 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
∴
For an isolated system (or the universe)
ΔS > 0
ΔS = 0
ΔS < 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.
Examples of ΔS 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)
ΔSmix = −nR [X A ln XA +XB ln XB ]
5.60 Spring 2008
Lecture #10
page 3
(b) Heating (or cooling) at constant V
A ( T1 , V) = A ( T2 , V ) ΔS =
∫
đqrev
T
=
T2
∫T
1
CV dT
T
if CV is
=
T -independent
T
CV ln 2
T1
[Note ΔS > 0 if T2 >T1 ]
(c)
Reversible phase change at constant T and p
e.g. H
2O (l, 100°C, 1 bar) = H2O (g, 100°C, 1 bar)
q p = ΔHvap
q pvap ΔH vap
ΔSvap (100°C) =
=
Tb
Tb
(Tb = boiling Temp at 1 bar)
(d) Irreversible phase change at constant T and p
e.g. H
2O (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 ΔS.
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)
= −ΔHfus
5.60 Spring 2008
Lecture #10
page 4
ΔS = ΔSheating + ΔSfus + ΔScooling
=
Tfus
∫T
1
∴
ΔS =
−ΔHfus
T
C p ( A ) dT −ΔHfus
T C p ( s ) dT
+
+∫
Tfus
T
Tfus
T
ΔS =
1
−ΔHfus
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