Chapter 15: Thermodynamics

Chapter 15: Thermodynamics
•The First Law of Thermodynamics
•Thermodynamic Processes (isobaric, isochoric, isothermal, adiabatic)
•Reversible and Irreversible Processes
•Heat Engines
•Refrigerators and Heat Pumps
•The Carnot Cycle
•Entropy (The Second Law of Thermodynamics)
•The Third Law of Thermodynamics
1
The Zeroth Law of Thermodynamics
If A is in thermal
equilibrium with C and B in
thermal equilibrium with C
then A and B have to be in
thermal equilibrium. No
heat flows!
2
Internal Energy
In thermodynamics, the internal energy of a
thermodynamic system, or a body with well-defined
boundaries, denoted by U, or sometimes E, is the total of
the kinetic energy due to the motion of molecules
(translational, rotational, vibrational) and the potential
energy associated with the vibrational and electric energy
of atoms within molecules or crystals. It includes the
energy in all the chemical bonds, and the energy of the
free, conduction electrons in metals.
3
The First Law of Thermodynamics
The first law of thermodynamics says the change in
internal energy of a system is equal to the heat flow into the
system plus the work done on the system.
ΔU = Q − W
4
First Law of Thermodynamics
The change in a systems internal energy is related to the
heat and the work.
ΔU= Uf - Ui = Q - W
Where:
Uf = internal energy of system @
start
Ui = internal energy of system @
end
Q = net thermal energy flowing
into system during process
Positive when system
gains heat
Negative when system
loses heat
W = net work done by the system
Positive when work done
by the system
Negative when work done on the
system
5
Thermodynamic Processes
A state variable describes the state of a system at time t,
but it does not reveal how the system was put into that
state. Examples of state variables: pressure, temperature,
volume, number of moles, and internal energy.
Thermal processes can change the state of a system.
We assume that thermal processes have no friction or other
dissipative forces.
In other words: All processes are reversible
(Reversible means that it is possible to return system and
surroundings to the initial states)
REALITY: irreversible
6
“Humpty Dumpty sat on a wall.
Humpty Dumpty had a great fall
All the king’s horses and all the king’s men
Couldn’t put Humpty Dumpty together again”
* Martin Schullinger-Krause (PH202 Winter 2008)
7
A PV diagram can be used to represent the state changes
of a system, provided the system is always near
equilibrium.
The area under a PV curve
gives the magnitude of the
work done on a system.
W<0 for compression and
W>0 for expansion.
8
To go from the state (Vi, Pi) by the path (a) to the state (Vf,
Pf) requires a different amount of work then by path (b). To
return to the initial point (1) requires the work to be nonzero.
The work done on a system depends on the path taken in
the PV diagram. The work done on a system during a
closed cycle can be nonzero.
9
An isothermal process
implies that both P and
V of the gas change
(PV∝T).
10
Specific Heats under constant pressure and constant volume
Specific heat
Q = m c ΔT
For a gas we use
Molar specific heat
Q = n C ΔT
Constant Volume: CV
Constant Pressure : CP
11
Thermodynamic Processes for an
Ideal Gas
No work is done on a system
when its volume remains
constant (isochoric process).
For an ideal gas (provided the
number of moles remains
constant), the change in internal
energy is
ΔU = Q − W = Q − 0
Q = ΔU = n C V ΔT
12
For a constant pressure (isobaric) process, the change in
internal energy is
ΔU = Q − W
where
W = PΔV = nRΔT
and Q = nC P ΔT .
CP is the molar
specific heat at
constant pressure.
For an ideal gas CP
= CV+R.
13
For a constant temperature (isothermal) process, ΔU = 0
and the work done on an ideal gas is
⎛ Vf ⎞
⎛ Vf ⎞
W = NkT ln⎜⎜ ⎟⎟ = nRT ln⎜⎜ ⎟⎟.
⎝ Vi ⎠
⎝ Vi ⎠
ΔU = 0 ⇒ Q = W
14
We have found for a monoatomic gas
ΔU = 3/2 n R ΔT
Constant volume:
ΔU= Q
3/2 n R ΔT = n CV ΔT
CV= 3/2 R
Constant pressure:
Q = ΔU + W
n CP ΔT = 3/2 n R ΔT + n R ΔT
CP= 5/2 R
CV – CP = R (always valid for any ideal gas)
15
Adiabatic (“not passable”) processes
(no heat is gained or lost by the system Q=0, i.e. system
perfectly isolated )
Q=0 and so ΔU= -W
P V = constant (isothermal)
P Vγ = constant (adiabatic)
γ = CP/CV
For a monoatomic gas
therefore γ = 5/3
16
Example (text problem 15.7): An ideal monatomic gas is
taken through a cycle in the PV diagram. (a) If there are
0.0200 mol of this gas, what are the temperature and
pressure at point C?
From the graph:
Pc = 98.0 kPa
Using the ideal gas law
PcVc
Tc =
= 1180 K.
nR
17
Example continued:
(b) What is the change in internal energy of the gas as
it is taken from point A to B?
This is an isochoric process so W = 0 and ΔU = Q.
⎛ 3 ⎞⎛ PBVB PAVA ⎞
−
ΔU = Q = nCV ΔT = n⎜ R ⎟⎜
⎟
nR ⎠
⎝ 2 ⎠⎝ nR
3
= (PBVB − PAVA )
ΔU = 3/2 n R ΔT
2
3
= V (PB − PA ) = −200 J
2
18
Example continued:
(c) How much work is done by this gas per cycle?
The work done per cycle is the area between the
curves on the PV diagram. Here W=½ΔVΔP = 66 J.
(d) What is the total change in internal energy of this
gas in one cycle?
⎛ 3 ⎞⎛ Pf Vf PiVi ⎞
−
ΔU = nCV ΔT = n⎜ R ⎟⎜
⎟
nR ⎠
⎝ 2 ⎠⎝ nR
3
The cycle ends where
= (Pf Vf − PiVi ) = 0
it began (ΔT = 0).
2
19
Example: An ideal gas is in contact with a heat reservoir so
that it remains at constant temperature of 300.0 K. The gas
is compressed from a volume of 24.0 L to a volume of 14.0 L.
During the process, the mechanical device pushing the
piston to compress the gas is found to expend 5.00 kJ of
energy. How much heat flows between the heat reservoir
and the gas, and in what direction does the heat flow occur?
This is an isothermal process, so ΔU = Q - W = 0 (for
an ideal gas) and W = Q = 5.00 kJ. Heat flows from
the gas to the reservoir.
Vf
W = nRT ln( ) → n =
Vi
W
Vf
RT ln( )
Vi
− 5000 J
=
= 3.7 mol
8.31 ⋅ ln(14 / 24)
20
An ice cube placed on a countertop in a warm room will
melt. The reverse process cannot occur: an ice cube will
not form out of the puddle of water on the countertop in a
warm room.
21
Any process that involves dissipation of energy is not
reversible.
Any process that involves heat transfer from a hotter object
to a colder object is not reversible.
The second law of thermodynamics
(Clausius Statement): Heat never flows
spontaneously from a colder body to a hotter
body.
22
Heat Engines
A heat engine is a
device designed to
convert disordered
energy into ordered
energy. The net work
done by an engine
during one cycle is
equal to the net heat
flow into the engine
during the cycle (ΔU= 0).
W net = Qnet
23
The efficiency of an engine is defined as
net work done by the engine Wnet
e=
=
.
heat input
Qin
(e.g. a efficiency of e=0.8 means 80% of the heat is converted
to mechanical work)
Note: Qnet = Qin - Qout
net work output Wnet
e=
=
QH
heat input
QH − QC
QC
=
= 1−
.
QH
QH
24
Refrigerators and Heat Pumps
Here, heat flows from
cold to hot but with
work as the input.
Pump
Refrigerator
25
26
Example (text problem 15.15): (a) How much heat does an
engine with efficiency of 33.3 % absorb in order to deliver
1.00 kJ of work?
QH =
Wnet
e
1.00 kJ
=
= 3.00 kJ
0.333
(b) How much heat is exhausted by the engine?
e = 1−
QC
QH
QC = (1 − e ) QH = 2.00 kJ
27
Reversible Engines and Heat
Pumps
A reversible engine can be
used as an engine (heat
input from a hot reservoir
and exhausted to a cold
reservoir) or as a heat
pump (heat is taken from
cold reservoir and
exhausted to a hot
reservoir).
28
From the second law of thermodynamics, no engine can
have an efficiency greater than that of an ideal reversible
engine “Carnot engine” that uses the same two reservoirs.
The efficiency of this ideal reversible engine is
TC
er = 1 − .
TH
29
Details of the Carnot Cycle
The ideal engine of the previous section is known as a
Carnot engine.
The Carnot cycle has four steps:
1. Isothermal expansion: takes in heat from hot reservoir;
keeping the gas temperature at TH.
2. Adiabatic expansion: the gas does work without heat
flow into the gas; gas temperature decreases to TC.
3. Isothermal compression: Heat QC is exhausted; gas
temperature remains at TC.
4. Adiabatic compression: raises the temperature back to
TH.
30
The Carnot
engine model was
graphically
expanded upon
by Benoit Paul
Émile Clapeyron
in 1834 and
mathematically
elaborated upon
by Rudolf
Clausius in the
1850s and 60s
from which the
concept of
entropy emerged
The Carnot cycle
illustrated
31
The Otto cycle
Its power cycle consists of adiabatic
compression, heat addition at constant
volume, adiabatic expansion and
rejection of heat at constant volume and
characterized by four strokes, or
reciprocating movements of a piston in a
cylinder:
intake/induction stroke
compression stroke
power stroke
exhaust stroke
32
Example: An engine operates between temperatures 650
K and 350 K at 65.0% of its maximum efficiency. (a)
What is the efficiency of this engine?
The maximum possible efficiency is
TC
350 K
er = 1 −
= 1−
= 0.462.
TH
650K
The engine operates at e = 0.65er = 0.30 or 30% efficiency.
33
Example continued:
(b) If 6.3×103 J is exhausted to the low temperature reservoir,
how much work does the engine do?
The heats exchanged at the reservoirs are related to each
other through
QC = (1 − e ) QH .
Wnet = QH − QC
⎛ e ⎞
=
− QC = ⎜
⎟ QC = 2.7 kJ
(1 − e )
⎝1− e ⎠
QC
34
Entropy
Heat flows from objects of high temperature to objects at
low temperature because this process increases the
disorder of the system. Entropy is a measure of a system’s
disorder. Entropy is a state variable.
35
If an amount of heat Q flows into a system at constant
temperature, then the change in entropy is
Q
ΔS = .
T
Every irreversible process increases the total entropy of the
universe. Reversible processes do not increase the total
entropy of the universe.
36
The second law of thermodynamics
(Entropy Statement): The entropy of the
universe never decreases.
37
Example (text problem 15.50): An ice cube at 0.0 °C is
slowly melting. What is the change in the ice cube’s entropy
for each 1.00 g of ice that melts?
To melt ice requires Q = mLf joules of heat. To melt one
gram of ice requires 333.7 J of energy.
The entropy change is
Q 333.7 J
ΔS = =
= 1.22 J/K.
T
273 K
38
Q − 300 J
ΔS hot = =
= −1 J/K.
T
300 K
Q + 300 J
ΔS cold = =
= 60 J/K.
T
5K
300K
Q
5K
http://www.youtube.com/watch?v=Xa6Pctf23tQ
39
Statistical Interpretation of Entropy
A microstate specifies the state of each constituent particle
in a thermodynamic system. A macrostate is determined
by the values of the thermodynamic state variables.
40
probability of a macrostate =
number of microstates corresponding to the macrostate
total number of microstates for all possible macrostates
41
The number of microstates for a given macrostate is related
to the entropy.
S = k ln Ω
where Ω is the number
of microstates.
42
Example (text problem 15.61): For a system composed of
two identical dice, let the macrostate be defined as the sum
of the numbers showing on the top faces. What is the
maximum entropy of this system in units of Boltzmann’s
constant?
43
Example continued:
Sum
Possible microstates
2
(1,1)
3
(1,2); (2,1)
4
(1,3); (2,2); (3,1)
5
(1,4); (2,3); (3,2); (4,1)
6
(1,5); (2,4), (3,3); (4,2); (5,1)
7
(1,6); (2,5); (3,4), (4,3); (5,2); (6,1)
8
(2,6); (3,5); (4,4) (5,3); (6,2)
9
(3,6); (4,5); (5,4) (6,3)
10
(4,6); (5,5); (6,4)
11
(5,6); (6,5)
12
(6,6)
44
Example continued:
The maximum entropy corresponds to a sum of 7 on
the dice. For this macrostate, Ω = 6 with an entropy of
S = k ln Ω = k ln 6 = 1.79k .
45
The Third Law of Thermodynamics
It is impossible to cool a system to absolute zero.
46