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 #6
page
The Second Law
▲
First Law
∆U = q + w ,
showed the equivalence of work and heat
∫ dU = 0 for cyclic process ⇒ q = −w
Suggests engine can run in a cycle and convert heat into useful work.
▲
Second Law
•
Puts restrictions on useful conversion of q to w
•
Follows from observation of a directionality to natural
or spontaneous processes
•
Provides a set of principles for
- determining the direction of spontaneous change
- determining equilibrium state of system
Need a definition:
Heat reservoir
Definition: A very large system of uniform
T, which does not change regardless of the
amount of heat added or withdrawn.
Also called a heat bath. Real systems can come close to this
idealization.
Two classical statements of the Second Law:
Kelvin
Clausius
and a Mathematical statement
1
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 #6
page
I. Kelvin: It is impossible for any system to operate in a cycle that
takes heat from a hot reservoir and converts it to work in the
surroundings without at the same time transferring some heat to a
colder reservoir.
q> 0
w< 0
-w= q
T1 (hot)
q
T1 (hot)
q1
-w
q 1> 0
w< 0
q 2< 0
-w q 1= -w-q 2
-q 2
IMPOSSIBLE!!
T2 (cold)
OK!!
II. Clausius:
It is impossible for any system to operate in a cycle
that takes heat from a cold reservoir and transfers it to a hot
reservoir without at the same time converting some work into heat.
q 2> 0 T1 (hot)
q 1< 0
-q 1
-q 1= q 2
T1 (hot)
q2
q2
IMPOSSIBLE!! T2 (c old)
Alternative Clausius statement:
-q 1
T2 (c old)
q 2> 0
w> 0
q 1< 0
w -q 1= w+ q 2
OK!!
All spontaneous processes are
irreversible.
(e.g. heat flows from hot to cold spontaneously and irreversibly)
2
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 #6
∫
đqrev
T
đ qrev = 0
∫
Mathematical statement:
page
T
is a state function = ∫ dS
→
and
dS =
∫
đ qirrev < 0
T
đqrev
T
S ≡ ENTROPY
∫ dS
=0
→
2
∆S = S2 − S1 = ∫
1
2 qirrev
qrev
>∫
1
T
T
irrev
rev
for cycle [1] ⎯⎯⎯
→[2] ⎯⎯⎯
→[1]
2
∫1
2
∫1
1 qrev
q irrev
q
+∫
= ∫ irrev < 0
2 T
T
T
2 qirrev
qirrev
− ∆S < 0 ⇒ ∆S > ∫
1
T
T
Kelvin and Clausius statements are specialized to heat engines.
Mathematical statement is very abstract.
Let’s Link them through analytical treatment of a heat engine.
The Carnot Cycle
-
a typical heat engine
All paths are reversible
T1 (hot)
p
1
q1
isotherm (T1)
2
adiabat
w
adiabat
4
isotherm (T2)
3
q2
T2 (cold)
3
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 #6
page
1→2
isothermal expansion at T1 (hot)
2→3
adiabatic expansion (q = 0)
3→4
isothermal expansion at T2 (cold) ∆U = q2 + w 2
4→1
⇒
∴
Kelvin:
∆U = w1′
adiabatic compression (q = 0)
Efficiency =
1st Law
∆U = q1 + w 1
∆U = w 2′
work output to surroundings − (w 1 + w1′ + w 2 + w 2′ )
=
heat in at T1 (hot)
q
v∫ dU
= 0 ⇒ q1 + q2 = − (w1 + w1′ + w 2 + w 2′ )
Efficiency =
q1 + q2
q
=1+ 2
q1
q1
q2 < 0 → Efficiency ≡ ε < 1 (< 100%)
-w = q1ε = work output
Note: if the cycle were run in reverse, then q1 < 0, q2 > 0, w > 0.
It’s a refrigerator!
Carnot cycle for an ideal gas
1→2
2→3
⎛V ⎞
2
∆U = 0; q1 = −w1 = ∫1 pdV = RT1 ln ⎜ 2 ⎟
V
⎝
q = 0; w1′ = CV (T2 −T1 )
Rev. adiabat
⎛T2 ⎞ ⎛V2 ⎞
⎜ ⎟=⎜ ⎟
⎝T1 ⎠ ⎝V3 ⎠
⇒
⎛V4 ⎞
⎟
⎝V3 ⎠
4
∆U = 0; q2 = −w2 = ∫3 pdV = RT2 ln ⎜
4→1
q = 0; w2′ = CV (T1 −T2 )
⇒
⎠
γ −1
3→4
Rev. adiabat
1
⎛T1 ⎞ ⎛V4 ⎞
⎜ ⎟=⎜ ⎟
⎝T2 ⎠ ⎝ V1 ⎠
γ −1
4
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 #6
page
q2 T2 ln (V4 V3 )
=
q1 T1 ln (V2 V1 )
γ −1
γ −1
⎛ V1 ⎞
⎜ ⎟
⎝V4 ⎠
⎛T ⎞ ⎛V ⎞
=⎜ 2⎟=⎜ 2⎟
⎝T1 ⎠ ⎝V3 ⎠
or
q1 q2
+
=0 ⇒
T1 T2
⇒
v∫
⎛V4 ⎞ ⎛V1 ⎞
⎜ ⎟=⎜ ⎟
⎝V3 ⎠ ⎝V2 ⎠
đqrev
T
⇒
q2
T
=− 2
q1
T1
=0
this illustrates the link between heat engines to the
mathematical statement of the second law
Efficiency
ε =1+
q2
T
=1− 2
q1
T1
For a heat engine (Kelvin):
→ 100% as T2 → 0 K
q1 > 0, w < 0, T2 < T1
⎛T −T2 ⎞
⎟ q1
⎠
Total work out = −w = ε q1 = ⎜ 1
⎝ T1
⇒
For a refrigerator (Clausius):
q2 > 0, w > 0, T2 < T1
( −w ) < q1
Note: In the limit T2 → 0 K, (-w) → q1, and ε → 100% conversion of
heat into work. 3rd law will state that we can’t reach this limit!
Total work in
But
q1
q
=− 2
T1
T2
⎛T −T ⎞
= w = ⎜ 2 1 ⎟ q1
⎝ T1 ⎠
⎛T −T ⎞
⇒ w = ⎜ 1 2 ⎟ q2
⎝ T2 ⎠
Note: In the limit T2 → 0 K, w → ∞. This means it takes an infinite
amount of work to extract heat from a reservoir at 0 K ⇒ 0 K
cannot be reached (3rd law).
5
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 #6
page
6
The efficiency of any reversible engine has to be the same as the
Carnot cycle, this can be shown by running the reversible engine
as a refrigerator, using the work output of a Carnot engine to
drive it so that the total work out is zero, and showing that, if
the efficiency of the reversible engine is higher, then the second
law is broken.
Additionally:
•
We can approach arbitrarily closely to any cyclic process using a
series of only adiabats and isotherms.
∴
So, for any reversible cycle
•
This is consistent with the mathematical statement of the second
law, which defines Entropy, a function of state, with
dS =
Note:
•
đqrev
T
v∫
∆S = S2 − S1 =
⇒
∫
2
1
đqrev
T
=0
đqrev
T
Entropy is a state function, but to calculate ∆S from q
requires a reversible path.
An irreversible Carnot (or any other) cycle is less efficient than a
reversible one.
p
1
irreversible
isotherm with p ext = p 2
2
adiabat
4
adiabat
isotherm (rev.)
3
1→2
( −w )irrev < ( −w )rev ⇒ wirrev > w rev
∆U = qirrev + wirrev = qrev + w rev
∴
qirrev < qrev
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 #6
page
An irreversible isothermal expansion requires less heat **
than a reversible one.
ε irrev = 1 +
also
q2rev
q2rev
<
1
+
= ε rev
q1irrev
q1rev
đqirrev đqrev
T
<
T
⇒
v∫
(q2 < 0)
đqirrev
<0
T
• This leads to the Clausius inequality
đq
v∫ T
≤0
⎧ đqrev
⎪⎪ v∫ T = 0
contains ⎨
⎪ đqirrev < 0
⎪⎩ v∫ T
• Important corollary: The entropy of an isolated system never
decreases
(A): The system is isolated and
irreversibly (spontaneously) changes
from [1] to [2]
(A) irreversible
1
2
(B) reversible
(B): The system is brought into contact with a heat
reservoir and reversibly brought back from [2] to [1]
qirrev = 0
Path (A):
Clausius
đq
v∫ T
≤0
⇒
(isolated)
∫
2
1
đqirrev
T
=0 !
+∫
1
2
đqrev
T
≤0
7
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 #6
⇒
∫
1
2
đqrev
T
∴
page
8
= S1 − S2 = −∆S ≤ 0
∆S = S2 − S1 ≥ 0
This gives the direction of spontaneous change!
For isolated systems
1
2
But!
∆Ssurroundings
∆S > 0
Spontaneous, irreversible process
∆S = 0
Reversible process
∆S < 0
Impossible
∆S = S2 − S1
independent of path
depends on whether the process is
reversible or irreversible
(a)
Irreversible:
Consider the universe as an isolated system
containing our initial system and its
surroundings.
∆Suniverse = ∆Ssystem + ∆Ssurroundings > 0
∴
(b)
∆Ssurr > −∆Ssys
Reversible:
′ =0
∆Suniv = ∆Ssys + ∆Ssurr
∴
′ = −∆Ssys
∆Ssurr
∆Suniverse ≥ 0 for any change in state (> 0 if irreversible, = 0 if reversible)