Transcript first law of thermodynamics.

Chapter 3.
The First Law of Thermodynamics
1
In this chapter we will introduce the concepts of work and
energy, which will lead to the 1st Law of Thermodynamics.
2
Configuration work: We consider some reversible process
đ W= (intensive variable ) (change in extensive variable)
Let y=intensive variable
X= extensive variable
dW 
 y dX
i
i
i
đ W is not an exact differential.
The variables X i determine the configuration of the system.
3
Consider a system and its surroundings.
surroundings
system
When the system as a whole
does work on the surroundings or the
surroundings do work on the system as a whole, we
call this external work.
When one part of the system does work on another
part of the system, it is called internal work. Internal
work cannot be discussed within the framework of
macroscopic thermodynamics.
When we use the term work, we mean external work.
4
Convention:
 Work done on system by surroundings is negative
 Work done by system on surroundings is positive
[NOTE: Be careful. Often the opposite convention is used.]
We will be considering quasi-static processes.
This is an ideal situation in which an infinitesimal
unbalanced force results in an infinitesimally slow
transition from an initial state to a final state through a
succession of states near thermodynamic equilibrium.
At each instant an equation of state is valid for the whole
system. At present we will ignore dissipative processes
such as friction.
5

Configuration work in changing a volume
adiabatic walls
F
A
P
PISTON (adiabatic)
The gas expands by exerting a pressure
infinitesimally greater than the pressure on the
gas exerted by the force acting on the piston.
F=PA (good approx.)
The work done by the gas is
đW = Fdx = PAdx
6
đW because this is an inexact differential.
W is not a function of the system variables
W  W(P,V,T).
We cannot write
dW   WT P dT   WP T dP
Volume increases and Adx = dV
đ W = PdV
The symbol đW indicates that a small amount of work is
done and the bar indicates that it is an inexact differential.
(The work done will depend on the path.)
7
If dV is negative, then đW is negative and
so work is done on the system.
If dV is positive then đW is positive and
work is done by the system.
8

PV Diagrams (Indicator Diagrams)
Wi f   Pd V
Vf
P
i
is positive
Vi
f
V
Work done by system.
Wf i   Pd V
Vi
i
Vf
P
f
V
i
P
f
V
is negative
Work done on system.
A cycle
(not zero!)
W   PdV
Net work is positive.
Net work done by system
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Work is not a state function.
Work is an activity or process that results in a
change of the state of a system.
In going from an initial thermodynamic state to a
final state through intermediate states, the work
done will depend on the path taken.
i
W 
I
P
II
V
f

Vf
Vi
Pd V
W is area under the curve
and obviously depends
upon the path taken.
10
Calculation of work for quasi-static processes
Isothermal expansion of an ideal gas:
i
W 
P
f

Vf
Vi
Pd V
P V  nRT
V
T constant so
W  nRT ln
W 
  dV
Vf
Vi
nRT
V
 
Vf
Vi
W >0 work done by gas
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Isothermal increase of the pressure on a solid
We defined:

1
V
 
V
T P
   V1   VP T
Volume expansivity
Isothermal compressibility
V T, P
d V    dT    dP
đ W  P   dP  P V   dP (since dT=0)
V
T P
V
P T
1
V
V
P T
V
P T
đ W   P V dP
For a solid Vconstant and over a limited range constant.
W   V 
Pf
Pi
PdP
W    V P  P
1
2
2
f
2
j

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Dissipative work
Dissipative work is work done in an irreversible process. It
is always done on a system. (We will discuss irreversible
processes later in the course.) Dissipative effects include friction,
viscosity, inelasticity, electrical resistance and magnetic hysteresis.
If there are no dissipative effects, all the work done by the
system in one direction can be returned to the system during the
reverse process.
Reversible processes: The dissipative work must be zero.
An example of dissipative work is work done on a liquid by
stirring (see textbook). Regardless of the direction of rotation of
the stirrer shaft, the external torque is always in the same direction
as the angular displacement of the shaft and the work done due to
the external torque is always negative, regardless of the direction
of rotation. (Work is done on the system.)
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Free expansion
vacuum
diaphragm
gas
If the diaphragm is punctured, the gas expands
and fills all of the container. In this process, no
external work is done.
In a free expansion no work is done!
14
Adiabatic work.
Consider a system going from some initial state i to a final
state f. Many different processes are possible and the work done
will depend on the path taken.
Now we limit ourselves to adiabatic paths. (System surrounded
by an adiabatic boundary and so the temperature of the system is
independent of that of the surroundings. No energy transfer.)
Consider the processes shown on the PV diagram below:
adiabatic
i
free expansions
expansion
a
adiabatic expansion
b
P
f
work
V
15
First, we have a free expansion (no work) from i to a, then an adiabatic
expansion from a to f. The work done by the system is the area under the af
curve.
Now we consider another path. First, an adiabatic expansion is made from i
to b, such that a free expansion results in the same final state f. The work
done by the system is the area under the ib curve.
Notice that the free expansions are denoted by dotted lines because they do
not proceed quasi-statically through a series of equilibrium states. The area
under these dotted lines is not the work done. (The work done is, of course,
zero.)
It is an experimental fact that the work done during these two different
processes is the same.
The total work is the same in all adiabatic processes between any two
equilibrium states having the same kinetic and potential energy!!! This is
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called the restricted statement of the first law of thermodynamics.
Because of this independence of path (for these adiabatic processes)
there exists a function of the thermodynamic coordinates of the
system such that
Wi f (adiabatic)  (U f  U i )
U is the internal energy function
U  U (, ,) where , ,
represent state variables such
as T, P, …..
In differential form, dU= -đ W If the system performs work, it
comes at the expense of the internal energy.
The internal energy function:
U f  U i  increase in the internal energy
Wi f (adiabatic)  (U f  U i )
(a) conservation of energy
then expresses:
(b) state function U exists
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For example, if we consider U=U(T,V)
 U 
 U 
 dT  
 dV
dU  
 T V
 V T
Suppose a system undergoes a process from an initial state i to a
final state f, by a diathermic process. In this case
Wi f (diathermic)  (U f  Ui )
By conservation of energy, energy must have entered (or left) the
system by some non-mechanical process.
The energy transferred between the system and surroundings
by a non-mechanical process or electrical process is called heat (Q).
We define Q  (U f  Ui )  Wi f (diathermic) or
U f  Ui  Q  W
Mathematical formulation of the
first law of thermodynamics
Units: U, Q, W are all energies and must all have units of joules.
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This law includes the concepts:
(a) a state function U exists
(b) energy is conserved
(c) energy is transferred due to a temperature difference.
Heat is that energy which is transferred due to a temperature difference
between a system and its surroundings. Heat is transferred from a
higher temperature region to a lower temperature region.
Heat and work are not state functions.

P
i
f
Q, W depend on the path. They are
path functions, not state functions.
However (Q-W) is independent of
path and is a state function.
V
19
Heat is a process, not a quantity of anything. It is not correct to say
“heat in a body”. After a process is completed, there is no further
use for Q.
The differential form of the first law is :
dU=đQ-đW
We have described how to calculate a reversible đW in a hydrostatic
(i.e. P,V,T) system: đWrev=PdV and so dU= đQ-PdV.
However, as yet, we have no way of writing đQ in terms of
system coordinates. Later we will see how to do this.
[Great care must be taken in the use of “heat” or “to heat”. Some
authors insist that heat should never be used as a noun or even a
verb! However avoiding such usage results in awkward
circumlocutions and so most of us are somewhat careless in this
regard. In teaching an elementary course one should be careful!]
20
Now that we have an equation relating energy to work,
how do we use it to make calculations for particular
systems?
We know how to calculate mechanical work. How can we
calculate either Q or U? If we could calculate one of them,
the equation would permit us to obtain the other. Clearly
we have more work to do, but first let us do some
examples.
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EXAMPLE: Chapter 3
Compute the work done against atmospheric pressure when 10kg
of water is converted to steam occupying 16.7 m3
For water
kg
  10 3
m
3
m  V
10 kg
V 


3 kg
10 3
m
m
V  10 2 m3
Vf
W 
 PdV  P(V
f
 Vi )
Vi
W  (1.01  105 Pa)(16.7  102 )m3
W  1.69  106 J
The system, initially water, does work in pushing aside the atmosphere.
22
EXAMPLE: Chapter 3
Steam at a constant pressure of 30atm is admitted to the cylinder of a
steam engine. The length of the stroke is 0.5m and the radius of
the cylinder is 0.2m. How much work is done by the steam per stroke?
W   PdV
V  Ax
dV  Adx
xf
W  PA  dx  PA( x f  xi )
xi
W  (30  1.01  10 Pa) (0.2m) (0.5m)
5
2
W  1.90  105 J
The system (steam) does work on the piston.
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EXAMPLE: Chapter 3
A volume of 10 m3 contains 8 kg of oxygen at 300K. Find the
work necessary to decrease the volume to 5 m3
(a) At constant pressure
(b) At a constant temperature
(c) What is the temperature at the end of process (a)?
(d) What is the pressure at the end of process (b)?
(e) Show both processes on an indicator (PV) diagram.
We assume an ideal gas law equation of state.
O2
2(16)amu  32 amu
n
8kg
kg
32
kmole
n  0.25 kmole
(a) Constant P
Pi 
nRTi

Vi
0.25kmole(8.314  10 3
10 m3
J
)300 K
kmole  K
Pi  6.24  10 4 Pa
24
Vf
W 
4
3
3
PdV

P
(
V

V
)

6
.
24

10
Pa
(
5
m

10
m
)
i
f
i

Vi
(b) Constant T
W  3.12  105 J
nRT
P
V
Vf
W   PdV  nRT ln
Vi
 Vi
Vf
1
J
W  0.25kmole(8.314  10
)(300 K ) ln  
kmole  K
2




3
W  4.32  105 J
Ti T f
(c) For constant P V  V
i
f
(d) Constant T Pi Vi  Pf Vf
Vf
T f  Ti 
 Vi

 
  300 K  1 

2

 Vi
Pf  Pi 
 Vf
T f  150 K

  (6.24  104 Pa)( 2)

Pf  1.25  105 Pa
25
OXYGEN
12.5
300K
4
P(10 Pa)
150K
6.24
0
5
10
V (m3 )
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EXAMPLE: Chapter 3
An ideal gas originally at temperature and pressure T1, P1 is
compressed reversibly against a piston to a volume equal to
one-half its original volume. The temperature is varied during the
compression so that, at each instant, the relation P  KV is satisfied.
(a) Draw an indicator diagram for the process.
(b) Find the work done in terms of n,R and the initial temperature.
isotherms
P=KV
P
P1
T1
V2
V2  21 V1
V1
V
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V1 / 2
(b)
V1 / 2
K
W   PdV  K  VdV 
2
V1
V1
1 2
2
 4 V1  V1 


3
W   KV12
8
But
P1V1  nRT1
P1  KV1
KV12  nRT1
so
3
W   nRT1
8
Work is done on the gas.
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EXAMPLE: Chapter 3
An ideal gas, and a block of Cu, have equal volumes of 0.500 m3
at 300K and 1 atmosphere. The pressure of each is increased
reversibly and isothermally to 5 atmospheres.
(a) Explain, with the aid of an indicator diagram, why the work is not
the same in the two processes.
(b) In which process is the work greater?
6
1


0
.
7

10
atm
.
(c) Find the work done on each. For Cu,
(d) Find the change in volume for each case.
gas
Cu
5
isotherms
W=area under isotherm
P(atm)
Cu compressed much less
1
V0
0
V (m3 )
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(b) Area under curve for gas > area under curve for Cu
W(on gas)>W(on Cu)
 Pi
V f  Vi 
P
 f
Vf
(c) Gas
 PdV
W 
PiVi  Pf V f
Vi
PiVi
nRT
P

V
V
PiVi  nRT
W  PiVi
 1
  Vi
 5

Vi / 5

Vi

dV
V

1 
1
N 
W  PiVi ln    1.01  105 2  0.5m3 ln  
m 
5 
5
0
Vf
Cu W   PdV
Vi
1  V 
    
V  P T
W ( gas)  8.13  10 4 J
 V 
 V 
dV  
 dT  
 dP
 T  P
 P  T
Pf
dV  VdP
W    VPdP
Pi
30
, V  constant


1
W   V Pf2  Pi2
2

1 0.7  10 6
W 
0.5m3 25  1atm2
2
atm
1
5
(d) Ideal gas V f  Vi
V  V f  Vi 

W (Cu)  0.424 J
1
4
Vi  Vi   Vi
5
5
V ( gas)  0.4m3
For Cu
dV  VdP


0.7  10 6
V  VP  
0.5m3 4atm
atm
V (Cu)  1.4  106 m3
More elegantly,
Vf
Pf
dV
Vi V   Pi  dP
 Vf
ln
 Vi

    P

 Vi  V 
    P
ln
 Vi 
31
But, for x<<1
ln(1  x)  x
 Vi  V 

V  V
  ln 1 
 
   P
Hence ln
Vi  Vi
 Vi 

V    V i P as before
32
EXAMPLE: Chapter 3
(a) Derive the general expression for the work per kilomole of a
van der Waals gas in expanding reversibly and at a constant T from
a specific volume v 1 to a specific volume v 2
(b) Find the work done when 2 kmoles of steam expand from a volume
of 30m3 to a volume of 60m3 at a temperature of 100  C
For steam
3
J

m
a  580  10 3
kmole2
m3
b  0.0319
kmole
(c) Find the work of an ideal gas in the same expansion.
a 
RT
a

(a)  P  2 ( v  b)  RT P 
 2
vb v
v 

v2
a 
 RT

W  
 2 dv   RT ln( v  b ) 
vb v 

v1 
v
a 2
v  v1
 v b  1
 v b  1
1 
1 
  a
  a 

W  RT ln 2
  W   RT ln 1
 v1  b   v 2 v1 
 v 2  b   v1 v 2 
33
3
(b) 2 kmoles of steam V1  30 m
V2  60 m3
T  373 K


m3 
 30m 3

0
.
0319


3
3
 8.314  10 J
 2kmole 2kmole  
3 J m
2
kmole
kmole


W  2kmole
( 373K ) ln
 580  10


3
3
2 
3
3 
60
m
m
kmole

K
kmole
30
m
60
m






 0.0319


kmole 
 2kmole


W  4.267  10 J
6
V2
(c) ideal gas
V2
nRT
W   PdV  
dV
V
V1
V1
 V2 
W  nRT ln 
 V1 


J
3
(373 K ) ln(2)
W  2kmole 8.314  10
kmole  K 

W  4.299  106 J
Not a great deal of difference. What is initial pressure?
34
Now we have added new concepts and ideas:
Work has been introduced:
đW=PdV. It is not an exact differential
W is not a state function, but rather a process!
We also introduced the internal energy of a system. If the
system is not isolated then, since energy must be conserved,
energy must enter or leave the system by non-mechanical
means and this energy is represented by Q.
đQ=?. It is not an exact differential
Q is not a state function, but rather a process!
1st Law of Thermodynamics: dU=đQ-đW
35