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5.60 Thermodynamics & Kinetics
Spring 2008
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5.60 Spring 2008
Lecture #9
page 1
Entropy
•
For a reversible ideal gas Carnot cycle:
Efficiency
−w
ε=
qrev
q1 q2
+
=0 ⇒
T1 T2
⇒
•
q2rev
T
= 1 + rev = 1 − 2
q1
T1
v∫
đqrev
T
=0
The efficiency of any reversible engine has to be the same as
the Carnot cycle:
T1 (hot)
ε=
q1
w
Some
reversible
engine
w
q2
q2
T2 (cold)
Carnot
cycle
( −w )
q1
ε′ =
( −w ′ )
q1′
Assume ε ′ > ε (left
engine less efficient
than Carnot cycle)
Since the engine is reversible, we can run it backwards. Use the work
(-w’) out of the Carnot engine as work input (w) to run the left engine
backwards.
∴
Total work out = 0
(-w’ = w > 0)
But ε ' > ε ⇒
−w ′ −w
>
q1′
q1
⇒
w −w
w
>
=
q1′ q1
−q1
⇒
⇒ q1 < −q1′ since q1 < 0, q1′ > 0
− (q1′ + q1 ) > 0
This contradicts the 2nd law (Clausius). This says that we have a net
flow of heat into the hot reservoir, but no work is being done!
∴
The efficiency of any reversible engine is ε = 1 −
T2
T1
5.60 Spring 2008
Lecture #9
page 2
•
We can approach arbitrarily closely to any cyclic process using
a series of only adiabats and isotherms.
∴
For any reversible cycle
•
This defines Entropy, a function of state
dS =
Note:
•
đqrev
T
v∫
đqrev
ΔS = S2 − S1 =
⇒
∫
2
1
đqrev
T
Entropy is a state function, but to calculate ΔS requires a
reversible path.
An irreversible Carnot (or any other) cycle is less efficient
than a reversible one.
p
1
1→2
( −w )irrev < ( −w )rev ⇒ wirrev > wrev
irreversible
isotherm with pext = p2
ΔU = qirrev + wirrev = qrev + w rev
2
adiabat
4
**
=0
T
adiabat
isotherm (rev.)
∴
3
qirrev < qrev
An irreversible isothermal expansion requires less heat **
than a reversible one.
ε irrev
also
q2rev
q2rev
= 1 + irrev < 1 + rev = ε rev
q1
q1
đqirrev đqrev
T
<
T
⇒
v∫
đqirrev
T
(q2 < 0)
<0
5.60 Spring 2008
Lecture #9
• Leads to Clausius inequality
đq
page 3
≤0
v∫ T
⎧ đqrev
⎪⎪ v∫ T = 0
contains ⎨
⎪ đqirrev < 0
⎪⎩ v∫ T
The entropy of an isolated system never decreases
(A) irreversible
1
(A): The system is isolated and
irreversibly (spontaneously) changes
from [1] to [2]
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
≤0
v∫ T
⇒
∫
1
2
đqrev
T
∴
⇒
(isolated)
∫
2
1
đqirrev
=0 !
T
+∫
1
đqrev
2
T
≤0
= S1 − S2 = −ΔS ≤ 0
ΔS = S2 − S1 ≥ 0
This gives the direction of spontaneous change!
For isolated systems
1
2
But!
ΔSsurroundings
ΔS > 0
Spontaneous, irreversible process
ΔS = 0
Reversible process
ΔS < 0
Impossible
ΔS = S2 − S1
independent of path
depends on whether the process is
reversible or irreversible
5.60 Spring 2008
(a)
Lecture #9
page 4
Consider the universe as an isolated system
containing our initial system and its
surroundings.
Irreversible:
ΔSuniverse = ΔSsystem + ΔSsurroundings > 0
ΔSsurr > −ΔSsys
∴
(b)
Reversible:
′ =0
ΔSuniv = ΔSsys + ΔSsurr
′ = −ΔSsys
ΔSsurr
∴
Examples of a spontaneous process
T1
Connect two metal blocks thermally in
an isolated system (ΔU = 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
5.60 Spring 2008
Lecture #9
1 mol gas (V,T)
ΔU = 0
adiabatic
=
q=0
page 5
1 mol gas (2V,T)
w=0
Compress back isothermally and reversibly
ΔS = − ΔSbackwards
qrev ≠ 0
1 mol gas (2V,T) = 1 mol gas (V,T)
ΔSbackwards =
∴
∫
đqrev
T
= −∫
ΔS = R ln2 > 0
đw
T
=
V
∫V
2
RdV
1
= R ln
V
2
spontaneous
Note that to calculate ΔS for the irreversible process, we needed to
đq
find a reversible path so we could determine đqrev and ∫ rev .
T