Transcript Lecture 1: Rotation of Rigid Body

Chapter 17: Thermal Behavior of Matter
Equations of state
 State
variables and equations of state
• Variables that describes the state of the material are called state
variables: pressure, volume, temperature, amount of substance
• The volume V of a substance is usually determined by its pressure
p, temperature T, and amount of substance, described by the mass
m
• In a few cases the relationship among p, V, T and m (or n) is simple
enough to be expressed by an equation called the equation of state
• For complicated cases, we can use graphs or numerical tables.
Equations of state (cont’d)
 Ideal
gas
• An ideal gas is a collection of atoms or molecules that move
randomly and exert no long-range forces on each other. Each
particle of the ideal gas is individually point-like, occupying a
negligible volume.
• Low-density/low-pressure gases behave like ideal gases.
• Most gases at room temperature and atmospheric pressure
can be approximately treated as ideal gases.
Equations of state (cont’d)
 Definition
of a mole
• One mole (mol) of any substance is that amount of the substance
that contains as many particles (atoms, or other particles) as there
are atoms in 12 g of the isotope carbon-12 12C. This number is called
Avogadro’s number and is equal to 6.02 x 1023.
• One atomic mass unit (u) is equal to 1.66x10-24 g.
• The mass m of an Avogadro’s number of carbon-12 atoms is :
 1.661024 g 
  12.0 g
m  N A (12 u )  6.0210 (12 u )
u


23
• The mass per atom for a given element is:
molar mass
NA
4.00 g/mol
 24


6
.
64

10
g/atom
23
6.02 10 atoms/mol
matom 
mHe
1.66x10-24=1/6.02x1023
Equations of state (cont’d)
 Definition
of a mole (cont’d)
• The same number of particles is found in a mole of a substance.
• Atomic mass of hydrogen 1H is 1 u, and that of carbon 12C is 12 u.
12 g of 12C consists of exactly NA atoms of 12C. The molecular mass
of molecular hydrogen H2 is 2u, and NA molecules are in 2 g of H2 gas.
 Molar
mass of a substance
• The molar mass of a substance is defined as the mass of one mole
of that substance, usually expressed in grams per mole.
 Number
of moles
• The number of moles of a substances n is:
m
n
molar mass
m : mass of the substance
Equation of state (cont’d)
 Ideal
gas equation (Equation of state for ideal gas)
• Boyle’s law
When a gas is kept at a constant temperature, its pressure is
inversely proportional to its volume.
• Charles’s law
When the pressure of a gas is kept constant, its volume is
directly proportional to the temperature.
• Gay-Lussac’s law
When the volume of a gas is kept constant, its pressure is
directly proportional to the temperature.
p : pressure, V : volume, T : temperature in K
Ideal gas equation:
R : universal gas constant 8.31 J/(mole K)
0.0821 L atm/(mol K)
pV  nRT
1 L (litre) = 103 cm3 = 10-3 m3
The volume occupied by 1 mol of an ideal gas at atmospheric
pressure and at 0oC is 22.4 L
Equations of state (cont’d)
 The
ideal gas equation
mtot  nM
total mass = # of moles times molar mass
# of moles
Ideal-gas equation:
gas const.
mtot
pV  nRT 
RT ; R  8.314472(15) J/(mol K)
M
pM

RT
density
An ideal gas is one for which
the above equation holds
precisely for all pressure and
temperatures.
pV
 nR  constant for constant mass
T
Equations of state (cont’d)
 The
ideal gas equation (cont’d)
The condition called standard temperature and pressure (STP)
for a gas is defined to be a temperature of 0 oC=273.15 K and
a pressure of 1 atm = 1.013 x 105 Pa.
Example 18.1
If you want to keep a mole of an ideal gas in your room at STP,
how big a container do you need?
V
nRT (1 mol)(8.314 J /(mol K)(273.15 K)

 0.0224m3  22.4 .
5
p
1.01310 Pa
Equations of state (cont’d)
 The
van der Waals equation
The ideal gas equation can be derived from a simple molecular
model that ignores:
• the volumes of the molecules themselves
• the attractive forces between them
The van der Waals equation makes corrections for these short-comings.
an2
( p  2 )(V  nb)  nRT
V
reduces volume of gas due to
finite size of molecules
The attractive intermolecular forces reduces the pressure by gas
a : A constant that depends on the attractive intermolecular forces.
b:
A constant that represents the size of a gas molecule
Equations of state (cont’d)
 pV
diagram and phases
TA  TB  TC  TD
ideal gas
behavior
pV=const
const.temp.
• Non-ideal gasses behave differently
because the attractive forces between
molecules become comparable to or
greater than the kinetic energy of motion
• The molecules are pulled closer together
and the volume V is less than for an ideal
gas
• c is called the critical point
• For T<Tc a gas will liquify under
increased pressure
• For T>Tc a gas will not liquify under
pressure
Molecular properties of matter
 Molecules and phases of matter
• All familiar matter is made up of molecules. The smallest size of
a molecule is about of order of 10-10 m. Larger molecules may have
a size of 10-6 m.
• The force between molecules in a gas varies with distance r between
molecules. The major source of the force in this case is electromagnetic interaction between them.
• A molecule that has energy E>E0 (see below) can move around as
depicted in the figure.
• A molecule with E>0 (see below) can escape
U(r)
as in gaseous phase of matter.
• In solids molecules vibrate around fixed point
more or less.
rr
E<0
E0
• A molecule in liquid has slightly higher energy
than in solid.
Kinetic-molecular model of an ideal gas
 Assumptions
Molecules are featureless points, occupy negligible
volume.
Total number of molecules (N) is very large.
Molecules follow Newton’s laws of motion.
Molecules move independently making elastic
collisions.
No potential energy of interaction (no bonding).
Kinetic-molecular model of an ideal gas

Kinetic-molecular model
-vx
vx vx
Impulse:
Molecule of mass m moving
with speed vx in negative-x
direction in a cube of side l
collides elastically with the
wall.
p  p f  pi  mvx  (mvx )  2mvx
2l
Time between collisions
t 
vx
with the same wall:
Force exerted by molecule on the wall:
p 2mvx mvx
F


(per molecule)
t 2l / vx
l
2
Force exerted by N molecules with different speeds:
m
2
2
2
2
F  (v x1  v x2  vx3      vx N )
l
v2 y  v y
v2 x  vx
v1 y  vy
v1x   vx
Elastic collision:
y-component of
velocity does not
change before and
after the collision.
Kinetic-molecular model of an ideal gas (cont’d)
 Mean square velocity
The mean-square velocity is defined as follows:
vx1  vx2      vxN
2
 vx  
2
2
2
N
The total force exerted on the wall by N particles:
F
m
2
N  v x  (for  along x - axis)
l
Kinetic-molecular model of an ideal gas (cont’d)
 Velocities in random directions
Velocities in random directions:
v  v x  v y  vz
2
2
2
2 (true for any vector)
If the velocities have random directions:
 vx    v y    vz 
2
2
2
 v 2    vx    vx    vx   3  vx 
2
The total force on the wall:
m
 v2 
F 
N
l
3
2
2
2
Kinetic-molecular model of an ideal gas (cont’d)
 Ideal gas law and mean velocity
Pressure on the wall due to molecular impact
F F m  v2  1 N
p  2  3 N

m  v2 
A l
l
3
3V
m
 v2 
since F 
N
l
3
1
2 1
R
2
2
pV  Nm  v   N ( m  v )  NkBT  N
T  nRT
3
3 2
NA
1
3
2
m  v   k BT
2
2
Ideal Gas Law
kB = Boltzmann’s Constant
Mean translational kinetic energy is proportional to T
Kinetic-molecular model of an ideal gas (cont’d)
 The mean kinetic energy and ideal gas
1
2
1
2
2
2
pV  Nm  v   N ( m  v )  Nk BT  Ktr
3
3 2
3
Compare with ideal gas equation:
pV  nRT
kB
average translational
kinetic energy
3
Ktr 1
3nRT 3 R
2
Ktr  nRT ;
 m  v 
 ( )T
2
N 2
2N
2 NA
N A : Avogadro’s
number
N  nNA  n / N  1 / N A
1
3
3 R
2
m  v   kBT  ( )T
2
2
2 NA
6.022 1023
m olecules/ m ol
Ideal Gas Law
kB = Boltzmann’s Constant
Mean translational kinetic energy is proportional to T
Kinetic-molecular model of an ideal gas (cont’d)
 RMS speed
Molecular speed (root-mean-square speed)
1
1
3
3
2
2
N A m  v   M  v  N Ak BT  RT
2
2
2
2
vrms
3k BT
3RT
 v  

m
M
2
At a given temperature, gas molecules of different mass m have the
SAME average kinetic energy but DIFFERENT rms speeds.
Kinetic-molecular model of an ideal gas (cont’d)
 Collision between molecules
• Consider N spherical molecules
with radius r, and suppose only
one molecule is moving.
r
• When it collides with another molecule
the distance between centers is 2r.
• The moving molecule collides with any
other molecule whose center inside a
cylinder of radius 2r.
r
2r
v
r
r
vdt
• In a short time dt a molecule with speed v travels a distance vdt,
during which time it collides with any molecule that is in the cylindrical
volume of radius 2r and length vdt.
• The volume of the cylinder is 4pr2vdt.
• There are N/V molecules per unit volume.
Kinetic-molecular model of an ideal gas (cont’d)
 Collisions between molecules (cont’d)
• The number of molecules dN with centers
in the cylinder:
dN  4pr 2vdtN / V
• The number of collisions per unit time: r
dN 4pr 2vN
dt

r
2r
v
r
r
V
• If there are more than one molecule moving,
it can be shown that:
dN 4p 2r vN

dt
V
2
vdt
• The average time between collision (the mean free time):
V
tmean 
4p 2r 2vN
• The average distance traveled between collision (the mean free path):
V
k BT
  vtmean 

( pV  Nk BT )
2
2
4p 2r N 4p 2r p
Heat capacities
 Heat capacities of ideal gases
(~ monatomic gasses)
Change of translational kinetic energy due to change in
temperature:
3
3
Ktr  nRT  dK tr  nRdT
2
2
Heat input needed for a change in temperature:
dQ  nCV dT; CV :
heat capacity (specific heat) at const.volume
From dK=dQ,
nCV dT 
3
3
nRdT  CV  R  12.47 J/(mol  K)
2
2
For an ideal gas
Heat capacities (cont’d)
 Theorem of equipartition of energy
Each quadratic term in the expression of the average total energy
of a particle in thermal equilibrium with its surrounding contributes
on the average (1/2)kT to the total energy.
OR
Each degree of freedom contributes an average energy of
(1/2)kT.
• Translational kinetic energy comprises 3 terms (degree of freedom):
1
1
1
1
m  v x2 , m  v 2y , m  vz2  3( kT )
2
2
2
2
• Rotational energy of diatomic molecule has 2 degree of freedom:
1
1
1
2
2
I1  1 , I 2  2  2( kT )
2
2
2
Heat capacities (cont’d)
 Heat capacities of gases in general
Heat capacities (cont’d)
 Heat capacities of gases in general (cont’d)
However, which degree of freedom is available depends on the
temperature. Furthermore, in case of diatomic molecules, for
example, two more degrees of freedom are possible from two
possible modes of vibration.
diatomic molecule
Heat capacities (cont’d)
 Heat capacities of solids
• Consider a crystalline solid
consisting of N identical atoms
(monatomic solid).
• Each atom is bound to an equilibrium position by interatomic forces.
• Each atom has three degrees of
freedom, corresponding to its three
components of velocity.
• In addition each atom acts as 3D
harmonic oscillator because of the
potential created by interatomic
forces – three more degrees of freedom.
6 degrees of freedom
Etot  3Nk BT  3nRT
CV  3R  24.9 J /(mol K )
Phases of matter
 Phase equilibrium
A transition from one phase to another ordinarily takes place
under conditions of phase equilibrium between two phases, and
for a given pressure this occurs at only one specific temperature.
 Phase diagram
Phase Diagram for Water
All three phases exist in
equilibrium at the triple
point. (273.16 K and 610
Pa for water)
Phases of matter (cont’d)
 Phase diagram (cont’d)
Water has an Unusual Property
• Most substances contract
when transforming from a
liquid to a solid (e.g. carbon
dioxide).
• Water is unusual in that it
expands upon freezing (solidliquid interface curve has a
negative slope).
water
• Ice floats on liquid water.
Carbon dioxide
Phases of matter (cont’d)
 Phase diagram (cont’d)
Exercises
Problem 1
A hot-air balloon stays aloft because hot air at atmospheric pressure
is less dense than cooler air at the same pressure. If the volume of the
balloon is 500 m3 and the surrounding air is at 15.0oC, what must the
temperature of the air in the balloon be for it to lift a total load 290 kg
(in addition to the mass of the hot air)? The density of air at 15.0oC and
atmospheric pressure is 1.23 kg/m3.
Solution
The density  ' of the hot air must be  '    m / V where  is the density
of the ambient air and m is the load. The density is inversely proportional
to the temperature, so


m 1
T' T 
 T (1 
)
 '   (m / V )
V
 (288.15K )(1 
(290kg )
1
)
 545 K  272C.
3
3
(1.23kg / m )(500m )
Exercises
Problem 2
The vapor pressure is the pressure of the vapor phase of a substance
when it is in equilibrium with the solid or liquid phase of the substance.
The relative humidity is the partial pressure of water vapor in the air
divided by the vapor pressure of water at that same temperature,
expressed as a percentage. The air is saturated when the humidity is
100%. a) The vapor pressure of water at 20.0oC is 2.34 x 103 Pa. If the
air temperature is 20.0oC and the relative humidity is 60%, what is the
partial pressure of water vapor in the atmosphere ( the pressure due to
water vapor alone)? b) Under the conditions of part (a), what is the mass
of water in 1.00 m3 of air? (The molar mass of water is 18g/mol.)
Solution
(a)
(b)
(0.60)(2.34 103 Pa)  1.40103 Pa
MpV (18.0  103 kg / mol)(1.40  103 Pa)(1.00m3 )
m

 10 g
RT
(8.3145J /(mol K ))(293.15K )
Problem 3
Exercises
Modern vacuum pumps make it easy to attain pressures of order 10-13 atm
in the laboratory. At a pressure of 9.00 x 10-14 atm and an ordinary
temperature (say T=300 K), how many molecules are present in a volume
of 1.00 cm3?
Solution
pV
(9.119 109 Pa)(1.00  106 m3 )
N  nNA 
NA 
(6.023 1023 m olecules/ m ol)
RT
(8.3145J /(m ol K ))(300K )
 2.20  106 m olecules
Exercises
Problem 4
A balloon whose volume is 750 m3 is to be filled with hydrogen at
atmospheric pressure (1.01 x 105 Pa). a) If the hydrogen is stored
in cylinders with volumes of 1.90 m3 at a gauge pressure of 1.20 x
106 Pa, how many cylinders are required? b) What is the total weight
(in addition to the weight of gas) that can be supported by the balloon
if the gas in the balloon and the surrounding air are both at 15.0oC?
The molar mass of hydrogen (H2) is 2.02 g/mol. The density of air at
15.0oC and atmospheric pressure is 1.23 kg/m3.
Solution
(a) The absolute pressure of the gas in a cylinder is:
(1.20106  1.013105 ) Pa  1.30106 Pa.
At atmospheric pressure, the volume of hydrogen will increase by a
factor of 1.30  106 / 1.01 105 so the number of cylinders is:
750m3 /[(1.90m3 )(1.30 106 / 1.01105 )]  31.
(b) The difference between the weight of the air displaced and the weight
of hydrogen is:
(  air   H 2 )Vg  (  air 
pM H 2
)Vg
RT
(1.01 105 Pa)(2.02  103 kg / m ol)
3
 [(1.23kg / m ) 
](9.80m / s 2 )(750m3 )
(8.3145J /(m ol K ))(288.15K )
 8.42  103 N .