Thermodynamics
•The First Law of Thermodynamics
•Thermodynamic
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
th
thermal
l equilibrium
ilib i
with
ith C
then A and B have to be in
thermal equilibrium.
q
No
heat flows!
2
Internal Energy
From http://en.wikipedia.org/wiki/Internal_energy.........
In thermodynamics,
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.
free,
metals
3
The First Law of Thermodynamics
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
work.
ΔU= Uf ‐ Ui = Q ‐ W
Where:
Uf = internal energy of system @ end
Ui = internal energy of system @ start
i
l
f
@
Q = net thermal energy flowing into y
gp
system during process
Positive when system gains heat
Negative when system loses heat
W = net work done by
W
net work done by the system
the system
Positive when work done by the system
Negative when work done on the system
5
Thermodynamic Processes
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
wall.
Humpty Dumpty had a great fall
All the king’s horses and all the king’s men
Couldn’tt put Humpty Dumpty together again”
Couldn
again
* Martin Schullinger-Krause (PH202 Winter 2008)
7
A PV diagram
g
can be used to represent
p
the state changes
g
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
p
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
nonzero.
9
An isothermal process
implies
p
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
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
ΔU constant),
= Q −W = Q − 0
the change in internal
Q = ΔU = n C ΔT
energy is
V
12
For
o a co
constant
sa p
pressure
essu e ((isobaric)
soba c) p
process,
ocess, the
e cchange
a ge 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
Q=0
0, ii.e.
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: An ideal gas is in contact with a heat reservoir so
that it remains at constant temperature of 300
300.0
0K
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
i t tto compress the
th gas is
i ffound
d tto expend
d5
5.00
00 kJ off
energy. How many moles of the ideal gas are in the system?
How much heat flows between the heat reservoir and the
gas, and in what direction does the heat flow occur?
Vf
W = nRT
RT ln( ) → n =
Vi
W
Vf
RT ln( )
Vi
=
− 5000 J
= 3.7 moll
8.31 ⋅ ln(14 / 24)
This is an isothermal process, so ΔU = Q - W = 0 (for
an ideal gas) and W = Q = - 5.00
5 00 kJ
kJ. Heat flows from
the gas to the reservoir.
20
An ice cube placed on a countertop in a warm room will
melt The reverse process cannot occur: an ice cube will
melt.
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
Heat Engines
A heat engine is a
device designed to
convert disordered
energy into
i
ordered
d d
energy. The net work
g
done byy 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
QH
(e.g. a efficiency of e=0.8
means 80% of the heat is
converted to mechanical
work)
Note: Qnet = Qin - Qout
Wnet
net
e work
wo ou
output
pu
e=
=
QH
heat input
QH − QC
QC
=
= 1−
.
QH
QH
24
Refrigerators and Heat Pumps
Refrigerators and Heat Pumps
Here, heat flows from
cold to hot but with
work as the input.
Pump
Refrigerator
K = Coefficient of performance
25
26
Reversible Engines and Heat Pumps
Reversible Engines and Heat Pumps
A reversible engine can be
used as an engine
g
((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
h
t d tto a h
hott
reservoir).
28
From the second law of thermodynamics, no engine can
have an efficiencyy g
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
Details of the Carnot Cycle
The ideal engine of the previous section is known as a
Carnot engine.
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
3.
I h
l compression:
i
H
Heat QC is
i exhausted;
h
d gas
temperature remains at TC.
4 Adiabatic compression: raises the temperature back to
4.
TH.
30
The Carnot
engine model was
graphically
expanded upon
by Benoit Paul
Émile Clapeyron
p y
in 1834 and
mathematically
elaborated upon
by Rudolf
Clausius in the
18 0 and
1850s
d 60s
60
from which the
concept
p of
entropy emerged
The Carnot cycle
illustrated
31
The Otto cycle
Its power cycle consists of adiabatic
compression heat addition at constant
compression,
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
Entropy Entropy
Heat flows from objects of high temperature to objects at
low temperature because this process increases the
disorder off the system. Entropy is a measure off 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
E
i
ibl process iincreases th
the ttotal
t l entropy
t
off th
the
universe. Reversible processes do not increase the total
entropy of the universe.
36
The second
Th
d law
l
off thermodynamics
th
d
i
(Entropy Statement): The entropy of the
universe never decreases.
37
Example: 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
333.7
7 J of energy
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.
h t =
T
300 K
Q + 300 J
ΔS cold = =
= 60 J/K.
T
5K
300K
Q
5K
htt //
http://www.youtube.com/watch?v=Xa6Pctf23tQ
t b
/ t h? X 6P tf23tQ
39
Statistical Interpretation of Entropy
A microstate specifies the state of each constituent particle
in a thermodynamic
y
system.
y
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 g
given macrostate is related
to the entropy.
S = k ln Ω
where Ω is the number
of microstates.
42
The Third Law of Thermodynamics
The Third Law of Thermodynamics
It is impossible to cool a system to absolute zero.
46