BEZOUT IDENTITIES WITH INEQUALITY CONSTRAINTS

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Transcript BEZOUT IDENTITIES WITH INEQUALITY CONSTRAINTS

USSC2001 Energy
Lecture 3 Thermodynamics of Heat
Wayne M. Lawton
Department of Mathematics
National University of Singapore
2 Science Drive 2
Singapore 117543
Email [email protected]
http://www.math.nus.edu.sg/~matwml/courses/Undergraduate/USC/2007/USC2001/
Tel (65) 6516-2749
1
PRESSURE
is force per unit area and measured in Pascal’s
1 pascal(Pa) 1 N m
-2
http://www.infoplease.com/ce6/sci/A0837767.html
Pascal's law : (päskälz') [key] [for Blaise Pascal], states that pressure applied to a
confined fluid at any point is transmitted undiminished throughout the fluid in all
directions and acts upon every part of the confining vessel at right angles to its
interior surfaces and equally upon equal areas. Practical applications of the law are
seen in hydraulic machines.
Standard atmospheric pressure is 101 325 Pa
2
DEFINING TEMPERATURE
The triple point of water http://en.wikipedia.org/wiki/Triple_point
Ttriple  273.16 K
ptriple  611.73 Pa
degrees Kelvin
Pascals
We define the temperature of a gas by


p(T )

T  273.16K 
lim
gas 0,vol const p(273.16K )


It is an empirical fact that T is the same for any two
gases that are in thermal equilibrium with each other.
3
CONSTANT-VOLUME GAS THERMOMETER
The ingenious mercury thermometer shown
below can measure T at constant volume
patm  1.0110 Pa
3
3
mercury  13.6 10 kg / m
5
Questions How can constant volume be maintained at
different temperatures? How can density be measured?
p
T
Gasfilled
bulb
h
p  patm  gh
Reservoir
that can be
raised and
lowered
4
THE IDEAL GAS LAW
Amadeo Avogado 1776-1856 suggested that all
gases contained the same number of molecules
for a fixed volume, pressure and temperature
P Vol / k T
# molecules =
# moles =
P Vol / n R T
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where N A  6.02 10
32
k  1.38 10
= # molecules in a mole
J / K = the Boltzmann constant
R  8.314 joules/ ( mole  K) = the gas constant
5
COLLISION WITH A WALL
For an elastic collision between a molecule and a wall
unit
M1 M
M 2 normal
2
M1
vector V'
1
'
V1
u
n
V2
V2
M1  M 2 so the formula on page 13 of Lecture 1
'
 V1,n   V1,n  2V2,n
M1
'
V2,n  2 M 2 V1,n  V2,n
where the subscript n denotes the normal components.
 1 collision changes wall momentum by 2 M1 V1,n 6
COLLISION RATE
of an object with horizontal velocity component
on an area A wall in a length L cylinder
z
unit
vector
un
y
x
Vol  AL
is
Vx
M
V
perpendicular
to wall
| Vx | /(2L) since it travels 2 L distance between
collisions alternating between the left and right walls.
Therefore
Rcol  A | Vx | / Vol
1
2
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MOMENTUM TRANSFER RATE
Since 1 collision transfers momentum 2M | Vx | u n
un
M
V
wall
the momentum transfer rate for 1 object is
Rmom (1)  Rcol  p  ( AV M / Vol) u n
2
x
and the momentum transfer rate for all particles is
2
Rmom (all)  A  avg(Vx ) u n with   density8
PRESSURE
Since momentum transfer rate = force,
un
M
V
wall
gas is a fluid, and Pascal’s law implies that the force
of a fluid is normal to a surface, the pressure
P   avg(V )
2
x
(pressure is not a vector)
The unit of pressure is Pascal Newton /Meter2 9
EQUIPARTITION OF KINETIC ENERGY
Our discussion about pressure ignored collisions.
since the directions of the particles after collision are
very sensitive to the direction between their centers at
the time of contact, the directions are random, if x,y,z
are orthogonal coordinates with x horizontal then
avg(V )  avg(V )  avg(V )  avg(V )
2
x
2
y
2
z
1
3
2
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KINETIC THEORY OF GASES
Combining equations
  M N / Vol
where N is the number of particles, with equations
P   avg(V )
2
x
gives
and
avg(V )  avg(V )
2
x
1
3
2
2
N
2
1
P
avg ( MV )
3 Vol
2
Combining with the ideal gas law N  P Vol / k T
gives
2
2
1
k T  avg ( MV )
3
2
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i
n
s
u
l
a
t
i
o
n
lead shot
pi
W
WORK W AND HEAT Q
Vf
W   dW   pdV
Vi
pressure
pi
state
diagram
Q
thermal reservoir
pf
volume
Vi
Vf
W (and Q) depend on the thermodynamic process,
described by a path, not only on initial&final states 12
THERMODYNAMIC PROCESSES AND LAWS
Question Compute W for constant p and constant T
Vf
Wpconst   pdV  p (Vf  Vi ) WT const  V
Vf
Vi
i
nRT
Vf
dV  n RT ln
V
Vi
1st Law: There exists an internal energy function
Eint  Eint (V , p) such that during any
thermodynamic process
Eint  Q - W
2nd Law: There exists an entropy function S  S (V , p)
such that during any thermodynamic process
S  
(Vf , p f )
(Vi , p i )
Sf
dQ
 Q   T dS
Si
T
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TUTORIAL 3
1. Derive the relationship between the k and R on p 5.
2. Show that the pressure difference between heights
h1 and h2 is P2  P1    g (h2  h1 ).
3. Use this pressure difference equation to show
that a container of gas having mass m weights mg.
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TUTORIAL 3
4. Use the ideal gas law to compute the air pressure
as a function of height above the ground. Assume
that g is constant for this problem.
5. On p 13 show that if the gas expands by dV then
E_int decreases by P dV. Do this by analysing the
collisions of the molecules against the top wall of the
container – which moves by a constant speed over
some interval of time.
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