Transcript for an ideal gas

Chapter 5
Consequences of the First Law
1
There’s this unhinged physicist who is working on a device that
can be used to blow up the world. In this ground-breaking work he
needs to know  u  for a substance. He can find no
 v  T
tabulations for this quantity. Must he perform an experiment to
determine this quantity? Not necessarily. He could attempt to write
this partial in terms of other quantities that are tabulated.
Such a situation arises very often in thermodynamics and
so there is much manipulation of partial derivatives. We begin
this chapter with some useful formulae for partial derivatives.
This then adds to what we have already learned about partials in
Chapter 2.
In general results for ordinary derivatives cannot be taken over
directly when dealing with partial derivatives!
2
Mathematical Relationships:
Suppose a relationship exists among x,y and z. We can then consider
z=z(x,y) and write:
 z 
 z 
Write
Then
dz    dx    dy
 x  y
 y  x
dz  Mdx  Ndy
 M 
 N 

  

 y  x  x  y
is the condition that dz is an exact differential. (See Chapter 2)
Suppose f=f(x,y,z) and a relationship exists among x, y and z. The
function f may be regarded as a function of any two of x, y or z.
Similarly, for example, x may be regarded as a function of f and y
or z. If one considers x=x(f,y)
3
 x 
 x 
dx    df    dy
 f  y
 y  f
Using such expressions one can show
 x   y   z 
       1
 y  f  z  f  x  f
(Note!)
and also
 x 
 x 
 x   y 
         
 f  z  f  y  y  f  f  z
4
There are many relationships between thermodynamic quantities
which can be obtained by combining the definitions, laws and
rules for partial derivatives. In this course we will concentrate
on P, V, T systems, but there are similar relationships for other
systems. In such systems there are certain quantities which are
measured directly and often tabulated:
P,V , T ,  ,  , cP , (cV )
Other quantities can often be expressed in terms of the above
quantities. A quantity expressed in terms of measurable
quantities is often said to be expressed in standard form. Some
relationships will be derived in this course and others are given
as assignments.
5
The Gay-Lussac-Joule Experiment
We discussed free expansion earlier and now we consider it
in more detail. Consider u=u(T,v). An attempt was made to determine
how the internal energy changes with a change in volume if T is fixed.
 u 
( cV a known quantity)
cV  

We have
 T  V
 u   T   v 
 T   v 
 u 
 T 

 
    1 c V 
    1     c V 

 T  v  v  u  u  T
 v  u  u  T
 v  T
 v  u
 T 
We introduce the Joule coefficient    v 

u
 u 
    cV
 v  T
So how the internal energy changes with volume if the T is fixed
depends on the Joule coefficient. Let us consider this coefficient.
du=đq-đw
How can we keep u fixed?
This can be accomplished by performing a free expansion under
adiabatic conditions.
6
adiabatic walls
diaphragm
vacuum
v1  v 0
gas
v 0 T0
When the diaphragm is punctured
the gas undergoes free expansion
into the vacuum
=0
 T 
 T 
 T 
T(u, v ) dT  
 du  
 dv  
 dv   dv
 u  v
 v  u
 v  u
T1
v1
T0
v0
 dT    dv
v1
T1  T0    dv
Joule found T1  T0
v0
Much later, more refined experiments showed that T1  T0
with
 u 
K  kmol 1
 T 
so
    c V  0

  0.001
3
 v  u
m
 v  T
As mentioned earlier, we define an ideal gas to be one for which
u=u(T) only and so   0
7
Let us consider an ideal gas.
 u 
Earlier we showed that
đq  
 dT 
 T  v
đq  1  u  dT  1  u   P  dv


 

T
T  T  v
T   v  T
 u 

 v   P  dv
  T


The reason for dividing by T is that the LHS becomes an exact
differential, as will be discussed in detail in later chapters. Employing
the relationship for an exact differential:
   1  u 
   1  u   
  
         P   
  
 v  T  T  v   T  T  T  v  T
   v
1
1
   u   

    2
 
T
T
 v  T  v  
T
 u 
 1
1  P 
   u   


   P        
 v  T
 T
 T  v  T  
 v T  T  v
Interchanging the order of the partials in term on LHS results in
cancellation with a term on the RHS
8
  P 
1  u 
   P  


T   v  T
  T  v
 u 
 P 
   T
 P
 v  T
 T  v
Since we are considering an ideal gas:
   RT  
 u 
 u 
   T

   P    0
 v  T
 v  T
 T  v   v
 u   u   v 
 v 
        0   0
 P  T  v  T  P  T
 P  T
 u 
  0
 P  T
(Not zero!)
Hence, for an ideal gas u=u(T) only.
9
The Joule-Thomson Experiment
This is a different approach from the one just discussed in
that the internal energy does not stay constant. In this experiment
gas is forced through a porous plug and is called a throttling process.
Pi v i Ti
Pf vf Tf
piston
adiabatic walls
porous plug
In this experiment work is obviously done. The pressures are kept
vf
0
constant.
w   Pi dv   Pf dv  Pf v f  Pi v i
đ q  du  đ w
vi
0
0  (uf  ui )  (Pf vf  Pi vi ) or uf  Pf vf  ui  Pi vi
From the definition of enthalpy
h f  hi
10
Hence, in a throttling process, enthalpy is conserved.
In an actual experiment, there are no pistons and there is a
continuous flow of gas. A pump is used to maintain the
pressure difference between the two sides of the porous plug.
11

Throttling Process (Joule-Thomson or Joule-Kelvin
expansion) (This process is widely used in
refrigerators.)
pump
porous plug
Pi Ti
Pf
Tf
The pump maintains the pressures Pi and Pf with Pf
smaller than Pi. In the experiment Pi, Ti and Pf are
set and Tf is measured.
12
As just discussed the enthalpy is the same on
the two sides of the porous plug i.e., hf = hi.
Consider a series of experiments in which Pi and
Ti are constant (hi constant) and the pumping
speed is changed to change Pf and hence Tf.
Since the final enthalpy does not change, we get
points of constant enthalpy. We plot Tf as a
function of Pf.
13
Tf
•
• •
•
Pf , Tf
•
•
• isenthalpic curve
• Pi, , Ti
Pf
A smooth curve is placed through the points yielding an
isenthalpic curve. {Note that this is not a graph of the
throttling process as it passes through irreversible
states.}
14
We now change Pi and Ti and obtain another
isenthalpic curve.
maximum inversion T
inversion curve
T
cooling
•d
b
•
a
•
heating
•c
ideal gas
Inversion curve
P
15
It is useful to define the Joule-Thomson
coefficient  since we are interested in the
temperature change due to the pressure
change.
T
   P h
This is the slope of an isenthalpic curve and
hence varies from point to point on the graph.
A point at which  = 0 is called an inversion
point. Connecting all of these points produces
the inversion curve.
16
If point a on the diagram ( < 0) is a starting point
and point b is the final point, then the T of the gas
will rise, i.e. we have heating.
If, on the other hand, we start at point c ( > 0)
and go to point d, then the T of the gas will drop,
i.e. we have cooling.
As higher initial starting temperatures are used,
the isenthalpic curves become flatter and more
closely horizontal. These curves are horizontal
lines for an ideal gas. As noted on the plot there
will be a maximum inversion T, the value of which
depends on the gas. For cooling to occur, the
initial T must be less than the maximum inversion
T. For such a T the optimum initial P is on the
inversion curve.
17
To make the discussion clear, we have exaggerated the slopes in the
above TP diagram. In fact, for most gases at reasonable T’s and P’s
the isenthalpic curves are approximately flat and so   0
 h 
 T 
   cP 
  cP 
 P T
 P  h
 h 
If   0 then 
  0 and so h  h(T )
 P T
u 
h 


We now have 
 
  0 (Pr oblem 5.1)
 v  T  P  T
It can be shown (a problem) that
for an ideal gas.
18
If a throttling process is used to liquefy a gas,
the cooled gas is recycled through a heat
exchanger to precool the gas moving towards
the throttle. The gas continues to cool and
when a steady state is reached a certain
fraction, y, is liquefied and a fraction (1-y) is
returned by the pump.
Using the notation:
 hi = molar enthalpy of entering gas
 hf = molar enthalpy of emerging gas
 hL = molar enthalpy of emerging liquid
Since the enthalpy is constant we have
hi = y h L + (1-y)hf
Of course, as some of the gas liquefies,
additional gas must be added to the system.
19
Liquifaction of Gases:
Some gases can be liquified in a simple process. For
example, carbon dioxide can be liquified at room temperature by a
simple isothermal compression to about 60 atm.
To liquify nitrogen or air is not so simple. At room
temperature, regardless of any increase in pressure, these gases will
not undergo a phase transformation to the liquid state. A method for
these gases, using the throttling process, was invented in 1895 and is
called the Hampson-Linde Process. The basis idea is to use the gas
cooled in the throttling process to precool the gas going towards the
throttle until the T is below the maximum inversion T. Starting from
room temperature, this cycle can be used to liquify all gases except
hydrogen and helium. To liquefy H by this process, it must first be
cooled below 200K and to accomplish this liquid N at 77K is used.
To liquefy He by this process, it must first be cooled below 43K and
to accomplish this liquid H can be used. (A device called the
Collins helium liquifier is used to liquify He.)
20
It should be mentioned that Joule-Thomson liquefaction of gases
has these advantages:
• No moving parts that would be difficult to lubricate at low T.
• The lower the T , the greater the T drop for a given pressure drop.
21
Heat engines and the Carnot Cycle.
A reservoir is anything large enough so that its temperature does not
change as heat enters or leaves in a particular process.
One can easily devise a system in which work is converted to energy
into a reservoir with 100% efficiency. Can one reverse this process,
that is, use energy from some reservoir to perform work with 100%
efficiency? No!! We will study a cyclic operation, which can continue
indefinitely, with the intermediary returning to its initial state at the end
of each cycle. This intermediary is called an engine.
The next slide gives an example of an engine, the steam engine,
which led to the industrial revolution and is still used today.
An idealization of an engine is then given on the following slide.
22
Hot Reservoir
Coal, Oil, Nuclear
Boiler
Low P
Working Substance
Turbine
Steam
Pump
High P

Work
Water
Condenser
Cold Reservoir
Air, Body of water
STEAM ENGINE
23
A heat engine is any device that absorbs “heat” and converts part of
that energy to work
Q2
engine
hot reservoir
work
Q1
cold reservoir
Heat into a system is considered positive, while heat leaving a
system is negative.
24
The notation that we will use is as follows:
Q2 
heat exchange between high T reservoir and system
Q1 
heat exchange between low T reservoir and system
W 
work exchanged between system and surroundings
For a heat engine, Q2  Q1 and W
is the work done by the system
The thermal efficiency of a heat engine is defined by

W
Q2
Since there is no change in the internal energy of the system,
Q2  Q1  W
and so
 1
Q1
Q2
An efficiency of unity occurs if Q1  0 but this is not possible. Later
we will determine the maximum possible efficiency.
25
We will be considering engines. We will idealize the engines
and the idealization will be severe. Nevertheless we shall
be able to answer to the following question:
What basically limits the efficiency of the engine?
26
There are two types of practical engines:
internal combustion engine: gasoline, diesel
external combustion engine: steam, Sterling
We will consider the internal combustion engine.
The engine gases are contained in a cylinder with a moveable piston.
In part of the cycle the gases are raised to a high T and P.
The gasoline engine:
In this engine there are six processes, four of which are strokes of the
piston. An analysis of a real engine is complicated by, among other
things, the presence of friction. We obtain an upper limit to what is
possible (the efficiency) by idealizing the cycle. This idealization is
called the Otto Cycle and makes the following approximations:
(a) the working substance (air and gasoline in a real engine)
is an ideal gas with constant heat capacities
(b) no friction or turbulence
(c) no heat loss by conduction
27
(d) processes are reversible
It should be remarked that, in a gasoline engine, there is no external
high temperature reservoir. The thermal energy is produced internally
by burning fuel.
3
power stroke
P
adiabats
2
ignition
4
P0
5
compression stroke
Exhaust valve open
intake exhaust
1
V2
V1
V
In the exhaust stage what actually happens is that the piston
pushes the old mixture out through one valve and pulls in a
new mixture through another valve.
28
5 -> 1 isobaric intake (at atmospheric pressure)
1 -> 2 adiabatic compression (compression stroke)
2 -> 3 isochoric increase in T during ignition. Gas combustion is
an irreversible process, here replaced by a reversible
isochoric process in which heat is assumed to flow in from
a reservoir.
3 -> 4 adiabatic expansion (power stroke)
4 -> 1 isochoric decrease in T (exhaust valve opened)
1 -> 5 isobaric exhaust (at atmospheric pressure)
What is the efficiency? What does it depend upon?
The diagram is not to scale. In a real engine,V1 / V2 might be about 10.
1 2
T1V1 1  T2V2 1 3  4
T4V1 1  T3V2 1
T4  T1  V2 
Subtracting and dividing gives
  
T3  T2  V1 
We define the compression ratio rc  V1 / V2
 1
which gives
29
T4  T1
1
  1 (1)
T3  T2 rc
T1 T2

Dividing the two equations for the adiabats gives
so
T4 T3
T4  T1
T4  T1
T1(T4  T1 ) T1



T3  T2 T4T2
T2 (T4  T1 ) T2
 T2
T1
From the definition of specific heat at constant volume, and
assuming that the specific heat is constant:
Q 2  CV (T3  T2 )
Q1  CV (T4  T1 ) giving
T4  T1
T1
Using equation (1) gives
 1
or   1 
T3  T2
T2
1 A higher compression ratio increases the efficiency!
  1   1
For rc  8,   0.56 if   1.4
rc
For a real gasoline engine
  0.2 to 0.3
30
Hence, for a gasoline engine, the efficiency can be increased by
reducing friction and making other improvements. However the
maximum attainable efficiency is limited by the compression
ratio.
31
The Diesel Engine:
To increase the efficiency of gasoline engines one could try to
increase the compression ratio. Unfortunately there are severe
preignition problems associated with high compression ratios. The
diesel engine avoids preignition problems by compressing only air,
then spraying fuel into the cylinder after the air is hot enough to
ignite the fuel. The spraying-ignition is done as the piston begins to
move outward, at a rate adjusted to maintain approximately constant
pressure. Diesels typically have compression ratios of about 20.
Again we make an idealization, neglecting such problems as friction,
all of which will reduce the efficiency.
The idealized diesel cycle is shown on the next slide.
32
injection-ignition Q 2
P
3
2
power-stroke: T drops
adiabats
4
air
compressed
P0
Q1
5
1
intake
V2
exhaust
V3
V1
V
4 -> 1 constant volume so Q1  CV (T4  T1 )
2 -> 3 constant pressure so Q2  C P (T3  T2 )
(T4  T1 )
Using the definition of gamma   1   (T  T )
3
2
33
Introduce the expansion ratio: rE  V1
V3
1  2 T1V1 1  T2V2 1
so
T2  T1rc 1 (1)
V2 V3

2 -> 3 ideal gas and P=constant, so
T2 T3
rc
V3
V3 V1
T3  T2
 T2
T3  T2
V2
V1 V2
rE
Using (1)
3 -> 4
rc
T3  T1  (2)
rE
 1
T3V3
 1
 T4V1
so

and using (2)
T3
T4   1
rE
 rc 
T4  T1   (3)
 rE 
Using (1), (2), and (3) in the expression for the efficiency
34
  r 

c
    1
1   rE 

 1  

 rc
 1

 rc 
 rE



(T4  T1 )
 1
 (T3  T2 )
Divide all terms by
EXAMPLE: If

c
r
1  rE  rc 
  1   1
1 
  rE  rc 
rE  10.0 rc  20   1.4
then   0.65
Real diesel engines have efficiencies of about 0.4
Both the gasoline and diesel engines take in the working substance at
the beginning of the cycle and eject it at the end of the cycle. These are
not closed systems and hence not appropriate systems for the
thermodynamic principles that we have used. However they can, in
fact, be replaced by the closed systems that we have been discussing. 35
Since the efficiency of a heat engine cannot be 1.0, the question
arises as to the upper limit for the efficiency and what factors
influence the efficiency. These questions were investigated by a
brilliant French engineer, Sadi Carnot. He considered an engine
operating cyclically and reversibly between two heat reservoirs.
A Carnot cycle is a reversible series of adiabatic, isothermal,
adiabatic, isothermal processes which returns the thermodynamic
system to its original state. The “working substance” can be any
thermodynamic material, but here we will consider an ideal gas.
The cycle is shown on the next slide.
NOTE: You may wonder what the relevance of this cycle
is to the cycles that we have discussed. The answer will be
discussed in the next chapter.
36
P
2
isotherm
Carnot Cycle
adiabat
T2
1
Q2
{Not to scale!}
3
isotherm
adiabat
T1
Q1
4
V
1 -> 2 No heat enters or leaves. As the compression slowly continues
work is done on the system and the internal energy increases. Thus
the temperature also increases.
2 -> 3 The T is constant and hence the internal energy does not change.
As the volume slowly increases, the system does work on its
surroundings and so heat must enter from the reservoir to keep the 37
T constant.
3 -> 4 The system is thermally isolated so that no heat enters or
leaves. As the slow expansion continues the system does
work on its surroundings. The internal energy decreases and
so T decreases.
4 -> 1 The T is constant and therefore the internal energy does not
change. As the volume is slowly decreased, the surroundings
do work on the system, and so heat must go from the system
to the low-temperature reservoir to maintain constant internal
energy.
A Carnot refrigerator is a Carnot cycle operating in the opposite
direction. The magnitudes of the work and heat transfers remain the
same.
38
In the Carnot cycle heat transfer occurs during the isothermal
processes. For these portions of the cycle there is no change in the
internal energy.
Q W
and
W   PdV
Since we are considering an ideal gas we obtain:
 V3 
Q2  nRT2 ln 
 V2 
 V4 
Q1  nRT1 ln 
 V1 
For the adiabatic processes: T2V2 1  T1V1 1
T2V3 1  T1V4 1
Dividing these last two equations gives (V2 / V3 )  (V1 / V4 )
Hence ln(V3 / V2 )  ln(V4 / V1 )
Q1
T1

Therefore
T2
Q2
and so
T1
 1
T2
(What a simple and
beautiful result!)
We will make use of this equation in the next chapter.
39
Real Refrigerator (or air conditioner)
We are concentrating on the Carnot cycle, but it should be noted
that, while we talk of a Carnot Refrigerator, the refrigerator in
our kitchen does not use the Carnot cycle, but uses a modification
of what is called the Rankine Cycle. The COP is more difficult
to work out and we will not do so.
A schematic diagram of an actual refrigerator is shown on
slide 42. The working substance changes back and forth from
a gas to a liquid. The most common substance was Freon-12
(CCl2F2). However escaping freon destroys the Earth’s ozone
layer and so it has been replaced by other substances such as
C2H2F4.
Notice that the throttling process is used. On the diagram
it is represented by a dashed line (green) because it is an
irreversible process. Some liquid is vaporized and the
liquid-vapor system cools. (T4<T3)
40
The evaporator consists of a series of pipes connected to
the interior of the refrigerator. After the throttling process the
working substance is cooler than the interior of the refrigerator, so
it absorbs heat from the interior, cooling the interior. (T1=T4). A
change of phase takes place.
The gas is then compressed so that its pressure and temperature
increase and then reaches the condenser. (T2>T1)
The condenser consists of a series of pipes connected to
the room. (You can see these underneath the refrigerator and it
might be necessary to clean this system periodically.) The gas
is now at a higher temperature than the outside. In the
condenser the gas condenses, giving up heat to the outside. A
change of phase again takes place.
41
1  2 Gas compressed: P,T raised
2  3 Liquefies, giving up heat to
Real Refrigerator
room
Q2
41
W
compressor
condenser
Throttle
gas
liquid
2
3
P
throttle
1
4
evaporator
Q1
refrigerator
34
room via network of pipes
Through throttle, reducing
P and T
Absorbs heat and becomes
a gas via a network of pipes
liquid + gas
V
42
EXAMPLE: A Carnot cycle operates between two heat reservoirs at
temperatures of 400K and 300K.
(a) If the engine receives 1200J from the high-T reservoir in each cycle
how much energy does it reject to the low temperature reservoir?
(b) If the engine is operated as a refrigerator and receives 1200J from
the low-T reservoir, how much energy does it deliver to the high-T
reservoir?
(c) How much work is done by the engine in each case?
T1
300 K
1
  0.250
(a)   1 
400K
T2
400 K
-1600J
1200J
Q1
Q1
 1
 0.75 Q1  0.75 Q2
Q2
Q2
300J
-400J
Q1  900 J
-900J
1200J
W  Q2  Q1  300 J W  300 J
300K
43
(b) Carnot refrigerator
Again
Q1
Q2
 0.75
4
Q2  Q1
3
Q2  1600 J
W  Q2  Q1  (1600  1200 )J
W  400 J
44
EXAMPLE:
(a) Show that for Carnot engines operating between the same high-T
reservoirs but different low-T reservoirs, the engine operating over
the largest T difference has the greatest efficiency.
(b) Is the more effective way to increase the efficiency of a Carnot
engine is to increase the temperature of the high-T reservoir keeping
the low-T reservoir fixed, or vice-versa?
T1 T2  T1
T
(a)
 1 

T fixed   T
T2
(b)
T2
  
T  1
T

  12   1 
 T2 T1 T2
 T2  T2
  
1

  
T2
 T1 T2
T2
2
  
T  1
1

   1  
 T2 T1  T2  T2 T2
  
1

 
T2
 T1 T2
  
  

  

 T1 T2
 T2 T1
Hence decreasing the T of the low-T reservoir causes the largest
increase in efficiency.
45
EXAMPLE: The “sadly Carnot” cycle.
This is called the “sadly Carnot” cycle
P
1
because, with a cursory analysis, the
T,U max
efficiency appears to be unity.
H

Q reverses We are used to cycles involving

isobaric, isothermal, isochoric or
M
adiabatic processes. We calculate the
adiabat
net work done and use the end points
2
to calculate the change in internal
Ideal gas
energy. Such a calculation will lead to
an efficiency of unity.
V
The error comes about by assuming that the heat transfer is
unidirectional along the linear portion of the cycle. For the linear
portion P=aV+b. Along the linear part 1 -> 2, heat both leaves and
enters the system. There is a transition point M at which the direction
of heat transfer changes. At this point dQ(V )  0 and d 2Q(V )  0
dV
dV 2
46
There is also a point H at which the temperature, and hence U, is a
maximum. At this point dU (V )  0
dV
One can solve for the points H and M in a straightforward manner.
47
EXAMPLE:
One kilomole of an ideal gas is the working substance of a Carnot
engine. During the isothermal expansion the volume doubles. The
ratio of the final volume to the initial volume in the adiabatic
expansion is 5.7. The work output of the engine in each cycle is
Calculate the temperatures of the reservoirs between
9  105 J
which the engine operates. Take CV  (3 / 2)R
A
A -> B isothermal expansion
isotherm
U  0 0  Q2  WAB Q2  WAB
P
T2
VB
 VB 
B
WAB   PdV  PAVA ln    PAVA ln(2)
 VA 
VA
adiabat
WAB  RT2 ln(2) Q2  RT2 ln(2)
T1
C
V
But

W
Q2
1
T1 T2  T1

T2
T2
so
48
9  105 J T2  T1

RT2 ln(2)
T2
T2  T1 
9  105 J

J
ln(2) 8.314  103 
K

T2  T1  156 K (1)
B -> C adiabatic expansion. From
PV   constant and PV  nRT so TV  1  constant
 1
 1
T2VB
 1
 T1VC
(2) into (1)
T2  VC 
  
T1  VB 
 (5.7)
2/3
T2
 3.19 (2)
T1
3.19T1  T1  156 K
T1  71.2K
T2  71.2K  156 K
T2  227 K
49
In this chapter we examined the throttling process, which is
isenthalpic. This process is used in refrigerators and to liquify
real gases.
The concept of a heat engine was introduced and the efficiency
of the gasoline and diesel engines studied.
The most important concept introduced, for this course, was
that of a Carnot Cycle.
50