Transcript Physics 106P: Lecture 1 Notes

Okay, so I was watching this video of some U.S. warplane breaking the sound barrier and
there was this, sort of, vertical halo around the entire mid-section of the plane----a ring of
smoke, so to speak. It looked really cool. What the hell was that?!
UB, Phy101: Chapter 14, Pg 1
Physics 101: Chapter 14
Ideal Gas Law & Kinetic Theory

Today’s lecture will cover Textbook Chapter 14.1-14.2
UB, Phy101: Chapter 14, Pg 2
Aside: The Periodic Table
UB, Phy101: Chapter 14, Pg 3
The Periodic Table Explained ?
Look carefully
proton
neutron
electron
UB, Phy101: Chapter 14, Pg 4
Energy vs Mass (more in Physics 102)
He (m=4.0026 u)
O (M=15.9995 u)
4 x He = 16.01 u
Mass difference = 0.01 u
= binding energy
So energy is the same as mass somehow ??
E = mc2
UB, Phy101: Chapter 14, Pg 5
Atoms, Molecules, and Moles



1 mole = 6.022 x 1023 molecule (NA = Avogadro’s
Number)
1  = 1 atomic mass unit = (mass of 12C atom)/12
» approximately #neutrons + #protons
» atomic weight W
Mass of 1 mole of “stuff” in grams = molecular mass in 
e.g., 1 mole of N2 has mass of 2x14=28 grams
UB, Phy101: Chapter 14, Pg 6
Back to Physics 101: Chapter 14, Preflight
Which contains the most molecules ?
1. A mole of water (H2O)
2. A mole of oxygen gas (O2)
3. Same
correct
H2O
O2
UB, Phy101: Chapter 14, Pg 7
Chapter 14, Preflight
Which contains the most atoms ?
correct
1. A mole of water (H2O)
2. A mole of oxygen gas (O2)
3. Same
H2O (3 atoms)
O2 (2 atoms)
UB, Phy101: Chapter 14, Pg 8
Chapter 14, Preflight
Which weighs the most ?
1. A mole of water (H2O)
2. A mole of oxygen gas (O2)
3. Same
H2O (M = 16 + 1 + 1)
O2 (M = 16 + 16)
UB, Phy101: Chapter 14, Pg 9
The Ideal Gas Law

pV = NkBT
p = pressure in N/m2 (or Pascals)
V = volume in m3
N = number of molecules
T = absolute temperature (K)
k B = Boltzmann’s constant
» kB = 1.38 x 10-23 J/K
» note: pV has units of N-m or J (energy!)
UB, Phy101: Chapter 14, Pg 10
UB, Phy101: Chapter 14, Pg 11
The Ideal Gas Law



pV = NkBT
Alternate way to write this
N = number of moles (n) x NA molecules/mole
pV= NkBT
» nNAkB T
» n(NAkB)T
» nRT
pV = nRT
R = ideal gas constant = NAkB = 8.31 J/(mol-K)
UB, Phy101: Chapter 14, Pg 12
The Ideal Gas Law

pV = nRT
 R = ideal gas constant = NAkB = 8.31 J/mol/K

Other units:
 R = .0823 l-atm/mol/K
 R = 1.99 cal/mol/K
UB, Phy101: Chapter 14, Pg 13
Ideal Gas Law: Applications
pV = nRT



When T is constant, pV is constant (Boyle’s Law)
Isothermal process demo
When p is constant, V is proportional to T
Isobaric process demo
Hot air balloon, helium and oxygen in LN2
When V is constant, p is proportional to T
Isohoric process demo
Open tube manometer
UB, Phy101: Chapter 14, Pg 14
Chapter 14, Preflight
In terms of the ideal gas law, explain briefly how a hot air balloon works.
Hot air, being less dense than cold air, can lift the balloon.
now we now that hot air rises. so therefore the bolloon collects all the hot air
until it over comes the weight of the basket below and you have lift off
Charles Law. As a gas is heated, it expands. When the air in the balloon expands,
its density goes down. Thus, it is less dense than surrounding air and "floats"
For a gas with a constant molecular weight, the pressure is proportional to the
product of the density and the temperature. Less molecules of hot air are needed
compared to cold air to achieve the same pressure. The density of hot air is less
than that of cold air, which allows the ballon to float.
Note! this is not a pressure effect, it is a volume effect. As T increases,
volume increase, so that density decreases the balloon then floats due to
Archimedes principle. The pressure remains constant!
UB, Phy101: Chapter 14, Pg 15
UB, Phy101: Chapter 14, Pg 16
Kinetic Theory:
The relationship between energy and temperature
(for monatomic ideal gas)
Per molecule
1
3
2
KE  m v  k BT
2
2
v rms 
v
2
DEMO
3k BT

m
3
3
internal energy  U  N KE  N k BT  nRT
2
2
UB, Phy101: Chapter 14, Pg 17
Chapter 14, Preflight
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?
1. 2
2. 2
correct
3. 4
1
3
2
KE  m v  k BT
2
2
• If v doubles, v2 quadruples
• Therefore, T quadruples
UB, Phy101: Chapter 14, Pg 18