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Lecture 18 – 1/32
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Physics Topics
Phys 220
Kinematics
Forces, Torque, and Work
Conservation of Energy
Conservation of Momentum (Linear & Angular)
Density and Pressure
Harmonic Motion
Traveling Waves
Standing Waves
Heat and Temperature
Ideal Gas Law
Thermodynamics
Lecture 18 – 2/32
Molecular Picture of Gas
Phys 220
• Gas is made up of many individual molecules
• Number of moles n = N / NA
NA = Avogadro’s number = 6.0221023 mole-1
NA = number of particles (can be atoms or molecules) per
mole
1 mole = amount of substance that contains as many
elementary entities as there are atoms in exactly 12
grams of carbon-12
Lecture 18 – 3/32
Atoms, Molecules and Moles
Phys 220
• 1 mole = 6.022  1023 molecules (NA = Avogadro’s Number)
• NA = Number of atoms or molecules that make a mass equal to
the substance's atomic or molecular weight in grams.
• 1 u = 1 atomic mass unit = (mass of 12C atom)/12
▪ Approximately # of neutrons + # of protons
▪ Atomic weight A
• 1 u = 1.66  10-27 kg = 1gram/NA
• Mass of 1 mole of “stuff” in grams = molecular mass in u
▪ E.g. 1 mole of N2 has mass of 2  14 = 28 grams
Lecture 18 – 4/32
Phys 220
Question 4
Which contains more molecules ?
A) A mole of water (H2O)
B) A mole of oxygen gas (O2)
C) Same
H2O
O2
Lecture 18 – 5/32
Phys 220
Question 5
Which contains more atoms ?
A) A mole of water (H2O)
B) A mole of oxygen gas (O2)
C) Same
H2O
(3 atoms)
O2
(2 atoms)
Lecture 18 – 6/32
Phys 220
Question 6
Which weighs more ?
A) A mole of water (H2O)
B) A mole of oxygen gas (O2)
C) Same
(M=16+1+1)
H2O
O2
(M=16+16)
Lecture 18 – 7/32
Internal Energy
Phys 220
• All objects have “internal energy” (measured in Joules)
▪ random motion
of molecules
» kinetic energy
+ energy associated with
intermolecular bonds
+
potential energy
▪ collisions of molecules gives rise to pressure
• Amount of internal energy depends on
▪ temperature
» related to average kinetic energy per molecule
▪ how many molecules
» mass
▪ “specific heat”
» related to how many different ways a molecule can move
‒ Translation (the only one for monatomic gas)
‒ Rotation
‒ Vibration
» the more ways it can move, the higher the specific heat
Lecture 18 – 8/32
Ideal Gas: Experiments
• Avogadro’s Law
▪ constant P and T
▪ V is proportional to N (where N is
the number of molecules)
• Boyle’s Law
▪ constant T
▪ P V = constant
• Charles’s Law
▪ constant P
▪ change in V is proportional to a
change in T
• Gay-Lussac’s Law
▪ constant V
▪ change in P is proportional to a
change in T
Phys 220
Lecture 18 – 9/32
The Ideal Gas Law
• P V = N kB T
P = pressure in N/m2 (or Pascals)
V = volume in m3
N = number of molecules
T = absolute temperature in K
k B = Boltzmann’s constant = 1.38 x 10-23 J/K
Note: P V has units of N*m or J (energy!)
Phys 220
Lecture 18 – 10/32
The Ideal Gas Law
Phys 220
• P V = N kBT
• Another way to write this
N (number of molecules) = n (number of moles)
x NA (molecules/mole)
P V = N kB T
= (n NA) kB T
= n (NA kB) T
= n R
T
•PV=nRT
n = number of moles
R = ideal gas constant = NA kB = 8.31 J/(mol*K)
Lecture 18 – 11/32
The Ideal Gas Law
Phys 220
PV  N kB T
• N = number of molecules
• kB = Boltzmann’s constant = 1.38 x 10-23 J/K
PV  n R T
• n = number of moles
• R = ideal gas constant = 8.31 J/(mol*K)
N
n
NA
R
kB 
NA
Lecture 18 – 12/32
Phys 220
Exercise
You inflate the tires of your car so the pressure is 30 psi, when the
air inside the tires is at 20 degrees C. After driving on the highway
for a while, the air inside the tires heats up to 38 C. Which number is
closest to the new air pressure?
A) 16 psi
Pf
P0

B) 32 psi
C) 57 psi
Tf
T0
 Tf
 Pf  P0 
 T0



Careful, you need to use the
temperature in K!
 38  273 
Pf  (30 psi)
  32 psi
 20  273 
Lecture 18 – 13/32
Phys 220
Question
What happens to the pressure of the air inside a
hot-air balloon when the air is heated?
(Assume V is constant)
A) Increases
B) Same
C) Decreases
Balloon is still open to atmospheric
pressure, so it stays at 1 atm
Lecture 18 – 14/32
Phys 220
Question
What happens to the buoyant force on the balloon
when the air is heated? (Assume V remains
constant)
A) Increases B) Same
FB = r V g
r is density of outside air!
C) Decreases
Lecture 18 – 15/32
Phys 220
Question
What happens to the number of air molecules
inside the balloon when the air is heated? (Assume
V remains constant)
A) Increases B) Same
C) Decreases
PV = NkBT
P and V are constant. If T increases N decreases.
Lecture 18 – 16/32
Balloon
Phys 220
• In terms of the ideal gas law, explain how a hot air
balloon works.
▪ Air in the balloon gets hot
» Net weight of the balloon plus hot air is less than the same
volume of cold air
» This results in the balloon rising
▪ Temperature increases, then volume of the gas increases
» Density of gas decreases since mass decreases and volume is
constant
» Density of the hot air in the balloon is less than the density of
the surrounding air
Note: This is not a pressure effect. It is a density effect.
As T increases, density decreases. The balloon then floats due
to Archimedes principle. The pressure remains constant.
Lecture 18 – 17/32
Ideal Gas
• Model of particles in an ideal gas
Phys 220
Lecture 18 – 18/32
Kinetic Theory
Phys 220
• Helps to explain gas laws by applying Newton’s
laws to the microscopic molecular motions
p=2mv (momentum transfer for one collision)
Lecture 18 – 19/32
Maxwell-Boltzmann Distribution
Phys 220
• How many molecules have speeds in a certain
range?
Lecture 18 – 20/32
Phys 220
Kinetic Theory
• The relationship between
energy and temperature
(for a monatomic ideal gas)
Force is caused by collisions
and is given by:
p x  2 m vx
p 2 m vx
F

t
t
Total Force for N molecules
depends on the number
reaching the +x wall:
N d traveled N vx t
N  x, reaching wall 

6
L
6 L
2 m vx  N vx t  N m vx2
 Ftotal 


t  6 L 
3L
L
Pressure is given by:
F N m vx2 N m vx2
P 

A
3A L
3V
N m vx2
 PV 
3
Remember:
PV  N kB T
N m vx2

 N kB T
3
Kinetic Energy:
1
3
2
KE  N m vx  N k B T
2
2
Lecture 18 – 21/32
Phys 220
Kinetic Theory
• For N molecules:
• For one molecule:
1
3
2
KE  N m vx  N k B T
2
2
1
3
2
KE  m vx  k B T
2
2
• Solving for the rms velocity
of one molecule in a monatomic gas:
vrms 
3 kB T
m
• Internal Energy (U)
▪ The kinetic energy is held “internally” by the gas molecules
3
3
U  N kB T  n R T
2
2
Only for a monatomic gas!
Lecture 18 – 22/32
Internal Energy
Phys 220
• Energy of all molecules including
▪ Translational and rotational kinetic energy of
Only
molecules due to their individual random motions.
translational
▪ Vibrational energy (both kinetic and potential) of
kinetic energy
molecules and of atoms within molecules due to
considered for a
random vibrations about their equilibrium points. monatomic gas!
▪ Potential energy due to interactions between
the atoms and molecules of the systems.
▪ Chemical and nuclear energy (kinetic and potential
energy associated with the binding of atoms to form
molecules, the binding of electrons to nuclei to form
atoms, and the binding of protons and neutrons to form
nuclei).
• DOES NOT INCLUDE
▪ Macroscopic motion of object
▪ Potential energy due to interactions with other objects
Lecture 18 – 23/32
Example
Phys 220
• What is the rms speed of a nitrogen N2 molecule in this
classroom?
vrms 
3 kB T
m
This equation is for the velocity of one
molecule in a gas!
m is the mass of one molecule of N2 gas:
2 (14.01 g)
 4.65 1023 g  4.65 1026 kg
NA
1.66 10-27 kg
m  2 (14.01 u )
 4.65 1026 kg
1u
m
vrms
3 kB T
3 (1.38 1023 J/K ) (293 K)


m
4.65 1026 kg
Temperature must be
in Kelvin.
(Room temperature is
about 20 °C)
v = 510 m/s
= 1150 mph!
Lecture 18 – 24/32
Phys 220
Question
Suppose you want the rms (root-mean-square) speed of
molecules in a sample of gas to double. By what factor
should you increase the temperature of the gas?
A) 2
B) √2
C) 4
1
3
2
KE  m vx  k B T
2
2
vrms 
3 kB T
m
Lecture 18 – 25/32
Phys 220
Diffusion
2


r
t
D
• Diffusion is the motion of a particle in a random
zig-zag path as it collides with other particles in a gas
• D is the Diffusion Constant and D depends on
▪ v (typical particle speed)
▪ l (mean free path)
» Mean free path is the average distance between collisions
» depends on ρ (density)
Lecture 18 – 26/32
Ideal Gas Law
PV  N kB T
NA = 6.02 x 1023 mol-1
N = number of molecules
kB = Boltzmann’s constant = 1.38 x 10-23 J/K
PV  n R T
n = number of moles
R = ideal gas constant = 8.31 J/(mol*K)
Pressure in Pa
Volume in m3
Temperature in K
Phys 220
Room
Temperature is
20°C = 293 K
N
n
NA
R
kB 
NA
9
TF  TC  32
5
TK  TC  273.15
Lecture 18 – 27/32
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Physics Topics
Phys 220
Kinematics
Forces, Torque, and Work
Conservation of Energy
Conservation of Momentum (Linear & Angular)
Density and Pressure
Harmonic Motion
Traveling Waves
Standing Waves
Heat and Temperature
Ideal Gas Law
Thermodynamics
Lecture 18 – 28/32
First Law of Thermodynamics
Phys 220
• The change in the internal energy of a system (∆U) is
equal to the heat flow into the system (Q) minus the
work done by the system (W)
∆U = Q – W
Increase in internal
energy of system
Heat flow
into system
Work done
by system
U  Q  W
Lecture 18 – 29/32
First Law of Thermodynamics
• Energy Conservation
U  Q  W
Q  U  W
Phys 220
Lecture 18 – 30/32
Signs Example
Phys 220
You are heating some soup in a pan on the stove.
To keep it from burning, you also stir the soup.
Apply the 1st law of thermodynamics to the soup.
What is the sign of
A) Q
Positive, heat flows into soup
B) W Negative, stirring does work on soup
C) U Positive, soup gets warmer
Lecture 18 – 31/32
Work Done by a System
Phys 220
M
M
• W > 0 if V > 0
y
W = F d cos q= P A d =
= P A y = P V
▪ expanding system does positive work
• W < 0 if V < 0
▪ contracting system does negative work
• W = 0 if V = 0
▪ system with constant volume does no work
W  P V
Lecture 18 – 32/32
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Work on P-V Diagram
Phys 220
Pressure vs Volume = PV Diagram
The initial state is a point on the diagram
The process is the line or curve
W  P V
The final state is another point
Work = P ∆V which is the area under the curve
Positive Work
Negative Work