PY2P10: Thermodynamics Dr. Graham Cross www.tcd.ie/Physics/People/Graham.Cross

PY2P10: Thermodynamics
Dr. Graham Cross
www.tcd.ie/Physics/People/Graham.Cross
Graham.Cross@tcd.ie
29.10.2014
PY2P10 Thermodynamics Lect 6
1
PY2P10: Thermodynamics – Lecture 5
1
Temperature. Zeroeth Law of Thermodynamics
Wed. Oct. 1
2
Reversible and irreversible processes. Types of work.
Wed. Oct. 8
3
Internal Energy, heat. First law of thermodynamics.
Wed. Oct. 15
4
Specific heat.
Fri. Oct. 17
5
Heat engines, Carnot cycles, Joule-Kelvin effect.
Wed. Oct. 22
6
Second law of thermodynamics. Thermodynamic (absolute)
temperature and entropy.
Wed. Oct. 29
7
More discussion of the 2nd Law of Thermodynamics. Combined
first and second laws: Central equation.
8
Thermodynamic potentials U, H, F, G and Maxwell's relations.
9
Energy equation and applications of Maxwell relations.
10
Application of thermodynamic potentials.
11
Phase changes.
12
Magnetic systems and the Third Law of Thermodynamics
13
Tutorial
14
Exam
29.10.2014
End of academic year
PY2P10 Thermodynamics Lect 6
2
Review- Carnot Cycle
A
P
Isotherms:
Adiabatics:
PV  const
PV   const
T2
Q2
W
(Ideal gas)
Q2
Engine
T1
Q1
B
T1
T2
W
Q2 T2

Q1 T1
D
C
Q1
29.10.2014
V
Ideal gas Carnot
cycle
W Q2  Q1 T2  T1
 =

Q2
Q2
T2
PY2P10 Thermodynamics Lect 6
3
PY2P10: Thermodynamics – Lecture 6
1
Temperature. Zeroeth Law of Thermodynamics
Wed. Oct. 1
2
Reversible and irreversible processes. Types of work.
Wed. Oct. 8
3
Internal Energy, heat. First law of thermodynamics.
Wed. Oct. 15
4
Specific heat.
Fri. Oct. 17
5
Heat engines, Carnot cycles, Joule-Kelvin effect.
Wed. Oct. 22
6
Second law of thermodynamics. Thermodynamic (absolute)
temperature and entropy.
Wed. Oct. 29
7
More discussion of the 2nd Law of Thermodynamics. Combined
first and second laws: Central equation.
8
Thermodynamic potentials U, H, F, G and Maxwell's relations.
9
Energy equation and applications of Maxwell relations.
10
Application of thermodynamic potentials.
11
Phase changes.
12
Magnetic systems and the Third Law of Thermodynamics
13
Tutorial
14
Exam
29.10.2014
End of academic year
PY2P10 Thermodynamics Lect 6
4
Second Law Preliminaries
•
•
First law of thermodynamics is about energy balance, and equivalence of
heat and work
The second law of thermodynamics will place limits on the efficiency of
energy conversion between heat and work
•
It requires us to establish the thermodynamic temperature scale, which
is independent of the thermometic material used
•
It also requires us to establish the concept of entropy, which is a new state
function of thermodynamic systems
29.10.2014
PY2P10 Thermodynamics Lect 6
5
Beyond the First Law of Thermodynamics
Isolated system
heat flow
Dissipative current flow
Free expansion
The reverse of these various differing processes does not occur, even though
the total energy (first law of thermodynamics) does not prevent this: we know
this intuitively.
Given this, is there some rigorous and general way we can predict which state
a system will tend towards of the various equivalent energy choices?
Ie. What direction will it go?
We want a property of the state of the system (ie. a state function) that has a
value that changes from the initial to the final state of the system.
Obviously, this cannot be energy (eg. Internal energy or enthalpy)!!
29.10.2014
PY2P10 Thermodynamics Lect 6
6
Entropy and 2nd Law of Thermodynamics
R. Clausius came up with such a function: entropy (S), specific entropy (s)
It is a state function of a system in equilibrium:
Eg.
S  S ( P,V )
For an ideal gas
Processes in which the entropy of an isolated system would decrease do not
occur. In other words, in every process taking place in an isolated system the
entropy of the system either increases or remains constant.
An isolated system must be at maximum entropy when in equilibrium
For a non-isolated system, the entropy can decrease. However, it will always be
found that the entropy of the surroundings increases by at least a similar
amount.
29.10.2014
PY2P10 Thermodynamics Lect 6
7
Thermodynamic Temperature
• The Zeroth Law of Thermodynamics led us to the concept of temperature.
• The temperature of a system is a physical property that determines whether or
not that system will change when brought into thermal contact with other systems.
• Ie. Any two systems in equilibrium with the same temperature will also have to
be in thermal equilibrium with each other.
• Kelvin suggested that one can define the ratio of any two temperatures based
on the considerations of Carnot’s cycle in a way independent of material:
• This will give us the absolute or thermodynamic temperature T. This is
significant because it allows an abstraction of material specific thermometric
properties to something more general.
• Lets look at this at some detail, considering an empirical temperature q
where X = cq , based on some thermometric property X of a system…
29.10.2014
PY2P10 Thermodynamics Lect 6
8
Reminder: Gas Pressure Thermometric
Let the thermometric parameter X be
the pressure (P) of a fixed volume of
gas in a bulb.
P  cq gas
At the limit of using small gas quantities in
the fixed volume bulb, this thermometric
parameter gives the same temperature for
any measured system, regardless of the
gas used. Eg. air, oxygen, helium, etc.
Steam Point (K)
X  cq
Liquid to
be
measured
Gas
374.0
O2
Common boiling point
373.5
Air
He
Gas becomes more ideal
373.0

n2a 
 P  2 V  nb   nRT
V 

29.10.2014
PY2P10 Thermodynamics Lect 6
N2
H2
0
500
1000
Gas pressure PTP at
water triple point (torr)
9
Thermodynamic Temperature
Carnot process for PVq system:
ie. a reversible cycle ABCDA
q
A
q2
Q2
B
First law:
U  W  Q U  0
W  Q2  Q1
(That’s all 1st Law has to say!)
q1
D
Q1
(Kelvin) asserted that for any material:
C
Q2
(Shape of reversible adiabatics will
depend on physical nature of system)
q is some thermometric
temperature, eg.
29.10.2014
q gas  P c
V
Q1
 f (q 2 ,q1 )
Where f may depend on thermometric
property chosen, but not on the physical
nature (eg. kind of gas, fluid, etc.) of the
cycled system
PY2P10 Thermodynamics Lect 6
10
Thermodynamic Temperature
Consider an intermediate process ABEFA
liberating heat Qi at temperature qi on the
EF isothermal leg.
q
A
q2
qi
Q2
B
Q2
E
Qi
Qi
F
q1
D
 f (q 2 ,qi )
Similarly for FECDF:
Q1
Qi
C
Q1
V
f (q 2 ,qi )  f (qi ,q1 ) 
Implies f can be decomposed in the form
 f (qi ,q1 )
Q2
Qi

Qi
Q1

Q2
Q1
f (q2 ,q1 )  f (q2 ,qi )  f (qi ,q1 )
29.10.2014
PY2P10 Thermodynamics Lect 6
11
Thermodynamic Temperature
f (q2 ,q1 )  f (q2 ,qi )  f (qi ,q1 )
q
A
q2
qi
Q2
B
This is possible if and only if f can
be decomposed into the ratio of
some function f:
E
f (q 2 )
f (q 2 ,qi ) 
f (qi )
Qi
F
q1
D
Q1
f (qi )
f (qi ,q1 ) 
f (q1 )
Ex. Show if only part
C
f (q 2 )
 f (q 2 ,q1 ) 
Q1
f (q1 )
Q2
V
Since the ratio f (q2 ) f (q1 ) is independent of substance, Kelvin suggested
defining an absolute temperature:
Q2 T2
where T  Af (q )

Q1
29.10.2014
T1
PY2P10 Thermodynamics Lect 6
12
Thermodynamic Temperature
T  Af (q )
f (q )
may not be a known function
Now, ratio of Carnot cycle heats for any system is given:
(ie. Not just ideal gas as shown on slide 15 of Lecture 5)
Q2
T2

Q1 T1
In particular, let one Carnot reservoir be at the water triple point T3
Q
T

Q3 T3
T  T3
Q
Q3
NB. T ≥ 0 by definition
For T3 assigned value 273.16, T assumes units of Kelvin
29.10.2014
PY2P10 Thermodynamics Lect 6
13
Entropy State Function
Arbitrary reversible cyclic process
T
Approximate this process by large
number of small Carnot cycles
traversed in same direction
Q2
T2

Q1 T1
T2
Q2

T1
Q1
Adiabatics: Eg. Ideal:
29.10.2014
 1
Tv
v
 Const
Each Carnot:
PY2P10 Thermodynamics Lect 6
For each small
Carnot cycle
(from previous slides)
Restore sign
convention
Q1 Q2

0
T1 T2
14
Entropy State Function
Each Carnot:
T
Q1 Q2

0
T1 T2
Sum over all Carnot cycle terms:
Qi
T 0
i
(Reversible heat
exchange Qi)
Take limit of small Carnot cycles:
dQr
 T 0
v
While
29.10.2014
PY2P10 Thermodynamics Lect 6
(Reversible heat
exchange)
đQr is not exact,
đQr
T
is!
15
Entropy State Function
This allows us to define a property of the system independent of process:
đQ
dS  r
T
FOR ALL THESE: MUST BE A
REVERSIBLE PROCESS FOR
DEFINITION TO STAND!!
 dS  0
Cyclic relation for exact differential

b
a
dS  Sb  Sa
Path independence for exact differential
S is Clausius’ entropy, a new system state function which we have defined,
so far, to within an arbitrary constant (similar to situation for energy.)
0
We can define the entropy S 298K of a substance at ambient conditions, the
integral of dS from zero absolute temperature to room temperature.
29.10.2014
PY2P10 Thermodynamics Lect 6
16
Entropy in Reversible Processes
• For reversible adiabatic processes:
dS  0
đQr  đQ  0
Isentropic process
• For reversible isothermal processes:
b
b
a
a
S  Sb  Sa   dS  
đQr 1 b
Qr
  đQr 
T
T a
T
Put equilibrated system in contact with heat bath slightly above or below
its temperature:
• For higher temperature bath: Qr is postive and entropy increases.
• For lower temperature bath: Qr is negative and entropy decreases.
Great isothermal example is constant pressure phase change (eg.
melting with latent heat l):
S  l T
29.10.2014
PY2P10 Thermodynamics Lect 6
17
Entropy in a Carnot Cycle
f ( P,V , T )  0
T
T2
Q2
A
U  U PV ( P,V )
U  UTV (T ,V )
B
Q  Q2  Q1
T1
U  U PT ( P, T )
S  S PV ( P,V )
S  STV (T ,V )
S  SPT ( P, T )
P  PUT (U , T )
T  TSV (S ,V )
Integrate on this diagram:
D
C
Q1
S1
S2
B
đQ2
A T (S )dS  A T T A đQ2
 Q2
B
S
B
 TdS  Q
2
29.10.2014
PY2P10 Thermodynamics Lect 6
 Q1  Q
18
One way processes
Isolated system
heat flow
Dissipative current flow
Free expansion
• The reverse of these various differing processes does not occur, even though
the total energy (first law of thermodynamics) does not prevent this.
How does our new entropy state function S capture this behaviour??
29.10.2014
PY2P10 Thermodynamics Lect 6
19
Entropy in Irreversible Processes
• Our defining entropy formula is true only for reversible processes:
dS 
đQr
T
(reversible)
• We will use a standard approach (trick?) to calculate state variable change for
irreversible processes:
• Since entropy is a state function, changes to it depend on only the initial
and final state.
• For a given irreversible process that brings our system from state to
another (after settling back into equilibrium), we will construct a reversible
process that does the same thing.
• We can then go ahead and calculate the entropy change.
29.10.2014
PY2P10 Thermodynamics Lect 6
20
Entropy in Irreversible Processes
Heat flows from a large reservoir at T2 into a small system
initially at T1 under condition of finite temperature difference:
T1
T2
or
Isolated system
heat flow
T1 > T2
Process is irreversible
T1 < T2
(An infinitesimal change in temperature will not reverse process)
• Consider the final state for system 1 (small) which will end up at temperature T2.
Let’s get there by a reversible isobaric process instead:
đQP
dT
đQP  CP dT
Q   đQr   CP dT  CP (T2  T1 )
T1
T1
T2
T2 đQ
T2 C dT
r
P
S1  ST2  ST1   dS  

T1
T1 T
T1
T
T2 dT
T2
 CP 
 CP ln
T1 T
T1
T2
T2
CP 
(Assume CP is relatively constant over temperature range!)
29.10.2014
PY2P10 Thermodynamics Lect 6
21
Entropy in Irreversible Processes
T1
T2
Isolated system
heat flow
Heat flows from a large reservoir at T2 into a small system
initially at T1 under condition of finite temperature difference:
or
T1 > T2
Process is irreversible
T1 < T2
(An infinitesimal change in temperature will not reverse process)
• Consider the final state for system 2 (large). The heat flow in the actual
irreversible process from reservoir to small system, was Q  CP (T2  T1 )
• To calculate S 2 , consider a reversible isothermal process for the
reservoir (at its temperature of T2) that involves this much heat:
Q
S2 
T2
29.10.2014
T2  T1
 CP
T2
PY2P10 Thermodynamics Lect 6
System 1
previous slide
22
Entropy in Irreversible Processes
• The sum of these two processes is easily expressed
T1
S  S1  S2  CP ln
T2
 T2 T2  T1 
 CP  ln 

T2 
 T1
Isolated system
heat flow
S
CP
1
S
CP
lnT2 T1
T2  T1
T2
0
-1
(reversible case)
0
29.10.2014
1
T2
T T
 CP 2 1
T1
T2
ln
T2 T2  T1

T1
T2
S  0
Entropy of universe (both systems)
changes when heat flow across
finite temperature difference
T2 T1
PY2P10 Thermodynamics Lect 6
23
Next Time
More Second Law, Fundamental Equation of Thermodynamics
Practice problems - Chapter 4 of Finn:
Problem 4-1: Simple heat engine calculation
Problem 4-3: Intersection of adiabatics
Problem 4-5: Efficiency of Carnot cycle
Problem 4-8: Comparison of efficiencies
Problem 4-9: Carnot engine in outer space
Problem 4-10: Efficiency of non-Carnot engine
Problem 4-11: The diesel engine cycle
Practice problems - Chapter 5 of Finn on Entropy:
Problem 5-4: Entropy change of resistive heating
Problem 5-5: Entropy change of electrical dissipative heating
Problem 5-6: Entropy change of ideal gas given specific heat
Problem 5-7: Entropy change of a sack of sand hitting ground
Problem 5-8: Entropy change of ideal gas free expansion
Problem 5-9: Entropy change during mixing of water at different temp.
Problem 5-10: Final temperature of system for different processes
29.10.2014
PY2P10 Thermodynamics Lect 6
24