Transcript Chapter 15: Thermodynamics

Fisica Generale - Alan Giambattista, Betty McCarty Richardson
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
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§15.1 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.
U  Q  W
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§15.2 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.
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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.
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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.
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An isothermal process
implies that both P and
V of the gas change
(PVT).
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§15.3 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
Q  U  nCV T .
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For a constant pressure (isobaric) process, the change in
internal energy is
U  Q  W
where W  PV  nRT and Q  nCP T .
CP is the molar specific heat at constant
pressure. For an ideal gas CP = CV+R.
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For a constant temperature (isothermal) process, U = 0
and the work done on an ideal gas is
 Vi
W  nRT ln 
 Vf

.

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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
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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 U = Q.
 3  PBVB PAVA 
U  Q  nCV T  n R 


nR 
 2  nR
3
 PBVB  PAVA 
2
3
 V PB  PA   200 J
2
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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=½VP = 66 J.
(d) What is the total change in internal energy of this
gas in one cycle?
 3  Pf Vf PiVi 
U  nCV T  n R 


nR 
 2  nR
3
The cycle ends where
 Pf Vf  PiVi   0
it began (T = 0).
2
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Example (text problem 15.11): 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 U = Q + W = 0 (for
an ideal gas) and W = -Q = -5.00 kJ. Heat flows from
the gas to the reservoir.
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§15.4 Reversible and Irreversible
Processes
A process is reversible if it does not violate any law of
physics when it is run backwards in time. For example 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.
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A collision between two billiard balls is reversible.
Momentum is conserved if time is run forward; momentum
is still conserved if time runs backwards.
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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.
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§15.5 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 (U= 0).
W net  Qnet
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The efficiency of an engine is defined as
Wnet
net work done by the engine
e

.
heat input
Qin
Note: Qnet = Qin - Qout
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§15.6 Refrigerators and Heat
Pumps
In a heat engine, heat flows from hot to cold, with work as
the output. In a heat pump, heat flows from cold to hot, with
work as the input.
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The efficiency of a heat engine can be rewritten as
net work output Wnet
e

heat input
QH

QH  QC
QH
 1
QC
QH
.
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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
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§15.7 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).
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From the second law of thermodynamics, no engine can
have an efficiency greater than that of an ideal reversible
engine that uses the same two reservoirs. The efficiency of
this ideal reversible engine is
er  1 
TC
.
TH
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§15.8 Details of the Carnot Cycle
The ideal engine of the previous section is known as a
Carnot engine.
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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.
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The Carnot cycle
illustrated
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Example (text problem 15.28): 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.
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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
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§15.9 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.
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If an amount of heat Q flows into a system at constant
temperature, then the change in entropy is
Q
S  .
T
Every irreversible process increases the total entropy of the
universe. Reversible processes do not increase the total
entropy of the universe.
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The second law of thermodynamics (Entropy Statement):
The entropy of the universe never decreases.
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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
S  
 1.22 J/K.
T
273 K
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§15.10 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.
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probabilit y of a macrostate 
number of microstate s correspond ing to the macrostate
total number of microstate s for all possible macrostate s
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The number of microstates for a given macrostate is related
to the entropy.
S  k ln 
where  is the number
of microstates.
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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?
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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)
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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   k ln 6  1.79k.
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§15.11 The Third Law of
Thermodynamics
It is impossible to cool a system to absolute zero.
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Summary
•The Three Laws of Thermodynamics
•Thermodynamic Processes
•Reversible and Irreversible Processes
•Heat Engines and Heat Pumps
•Efficiency of an Engine
•Entropy
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