Unformatted text preview: Chapter
Chapter 20
Thermodynamics III
Entropy
Entropy and the 2nd law of
thermodynamics
thermodynamics Ch 20
Reversible
Reversible and irreversible processes
Idea
Idea of entropy
The
The 2nd law of thermodynamics Opening
Opening story
The
The popcorn ‘pops’ in an oven. Can we
retrieve the original unpopped corn by put
the popcorn back into a refrigerator?
What in the world gives
direction to time? Direction of time
Why
Why time has direction?
Why oneWhy oneway processes are irreversible
From
From energy conservation?
Hot
Hot coffee cup and cold hand
Popped
Popped helium balloon Entropy
Entropy postulate
If
If an irreversible process occurs in a closed
closed
system, the entropy S of the system always
increases; it never decreases. What
What is entropy?
A state property
state
Start
Start from a Carnot engine
An
An ideal engine
Extracts
Extracts heat from environment
and
and does useful work All
All processes are reversible
Similar
Similar to frictionless in motion
Heat
Heat flow can be driven by
infinitesimal T difference Is
Is there any property
invariable in a Carnot cycle? What is entropy?
a → b & c → d : isothermal∴ ∆Eint = 0,
b
b
dV
V
QH = Qa →b = Wa →b = ∫ pdV = ∫ NkTH
= NkTH ln b
a
a
V
Va
V
QL = Qc→d = NkTL ln d
Vc b → c & d → a : adiabatic∴TV γ1 = const , γ = C p / CV
∴ TH Vb γ1 = TLVc ; TH Va
γ1 γ1 = TLVd γ1 ⇒ Vb / Va = Vc / Vd QH QL
=S
⇒
=
TH
TL Entropy! Change
Change in entropy
∆S = S f − S i = ∫ i f dQ
T (J/K) Free
Free expansion: irreversible
Can
Can be modeled by a reversible
process with same i and f state
Free
Free expansion
isothermal
isothermal
Vf
1f
Q
∆S = ∫ dQ = = nR ln
Vi
Ti
T refe For
For 1 mol He
∆S = (1 mol)(8.31 J/mol ⋅ K)ln2
= 5.76 J/K Change in entropy
What
What is ∆S of a popped popcorn?
1. Water in pericarp vaporized at 180 oC (water
1.
mass = 4 mg)
1f
Q Lm
∆S1 = ∫ dQ = = V ≈ 0.02 J/K
Ti
T
T
2. Adiabatic expansion of the vapor ∆S 2 = 0
∆S = ∆S1 + ∆S 2 ≈ 0.02 J/K
Corresponding to each audible pop
http://www.buetzer.info/fileadmin/pb/HTMLFiles/Popcorn.htm Quick
Quick quiz
Water
Water is heated on a stove. Rank the
entropy changes of the water when its
temperature
1.
2.
3. From 20 oC to 30 oC
From 30 oC to 35 oC
From 80 oC to 85 oC ∆S = S f − Si = ∫ i f dQ
Q
≈
T
Tavg Quick quiz
Is
Is the entropy change along the path to
state a larger than, smaller than, or the
same as that along the path to state b? Entropy
Entropy as a state function
For
For a reversible process
A slow series of small steps
slow dEint = dQ − dW p= dQ = pdV + nCV dT
dQ
dV
dT
= nR
+ nCV
T
V
T
∆S = nR ln ⇒∫ i Vf
Vi f nRT
V f
f
dQ
dV
dT
= ∫ nR
+ ∫ nCV
i
i
T
V
T + nCV Tf
Ti Only depends on initial and final states, not pathdependent Entropy and 2nd law of dQ
= dS
thermodynamics
T
In
In a reversible process, there is no gain or
loss of Q/T
Entropy
Entropy remains constant For
For irreversible processes
The
The entropy of the world is increased
No
No conservation of entropy ∆S ≥ 0 !!
refe Example
Example of irreversibility
Work
Work done on an object by friction
Q=W
Entropy
Entropy of the world increase by W/T Hot
Hot stone in cold water
∆S of stone?
–Q/T1 ∆S of water?
Q/T
Q/T2 ∆S of the world?
Q/T
Q/T2–Q/T1>0
refe Entropy of reversible processes
In
In a Carnot engine
∆S = 0 for the reversible cycle
How
How about a single reversible process such
as
as c
d?
d?
∆S = QL/T < 0??
Need
Need to consider the whole
system, including the heat
reservoir.
∆S = QL/T + QL/T = 0 Entropy
Entropy and time
Entropy
Entropy is the arrow of time
For
For irreversible processes, the entropy always
increases
the
the forward direction of time
To
To return in time requires an entropy
decrease
decrease
Not
Not allowed by thermodynamics Efficiency of an engine
W
energy we get
=
ε=
energy we pay for QH For
For a Carnot engine
εC = QH − QL
QH = 1− QL
QH TL
= 1−
TH ...
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 Fall '11
 K.Y.Liao
 Thermodynamics, Entropy, Heat

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