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Gas Laws
AP Physics B
The Periodic Table
Atomic # - # of protons OR electrons
All of the elements on the
periodic table are
referred to in terms of
their atomic mass. The
symbol u is denoted as
an atomic mass unit.
1 amu = 1.66 x 10-27 kg
Atomic Mass - # of protons + Neutrons
The Mole
The word “mole” is derived from Latin to mean “heap”
or “pile”. It is basically a chemical counting unit.
Below, what we have is called the Mole Road Map
and it summarizes what a mole equals.
Atomic Mass from P.T
Avogadro’s # = 6.02 x 1023 Particles
1 MOLE
STP = Standard Temperature
and Pressure
•ST = 0 degrees Celsius or
273 Kelvin
22.4 Liters gas @ STP
•SP = atmospheric pressure
Example
A flexible container of Oxygen(O2) has a volume of 10.0
m3 at STP. Find the # moles and molecules that exist
in the container
Factors that effect a GAS
1.
2.
3.
4.
The quantity of a gas, n, in moles
The temperature of a gas, T, in Kelvin
(Celsius degrees + 273)
The pressure of a gas, P, in pascals
The volume of a gas, V, in cubic meters
Gas Law #1 – Boyles’ Law
(complete TREE MAP)
“The pressure of a gas is
inverse related to the
volume”
 Moles and Temperature
are constant
1
k
P  P 
V
V
k  constant of proportion ality
PoVo  k
PV  k
Gas Law #2 – Charles’ Law
“The volume of a gas is
directly related to the
temperature”
 Pressure and Moles are
constant
Vo To  Vo  kTo
Vo
k
To
V
k
T
Gas Law #3 – Gay-Lussac’s Law
“The pressure of a gas is
directly related to the
temperature”
 Moles and Volume are
constant
Po To  Po  kTo
Po
k
To
P
k
T
Gas Law #4 – Avogadro’s Law
“The volume of a gas is
directly related to the #
of moles of a gas”
 Pressure and
Temperature are
constant
Vo no  Vo  kno
Vo
k
no
V
k
n
Gas Law #5 – The Combined Gas Law
You basically take
Boyle’s, Charles’ and
Gay-Lussac’s Law and
combine them together.
 Moles are constant
PoVo To  PoVo  kTo
PoVo
k
To
PV
k
T
Example
Pure helium gas is admitted into a leak proof cylinder containing a
movable piston. The initial volume, pressure, and temperature
of the gas are 15 L, 2.0 atm, and 300 K. If the volume is
decreased to 12 L and the pressure increased to 3.5 atm, find
the final temperature of the gas.
Gas Law #6 – The IDEAL Gas Law
All factors contribute! In the previous examples, the constant, k,
represented a specific factor(s) that were constant. That is
NOT the case here, so we need a NEW constant. This is
called, R, the universal gas constant.
PV  nT
R  constant of proportion ality
J
R  Universal Gas Constant  8.31
mol  K
Example
A helium party balloon, assumed to be a perfect sphere, has a
radius of 18.0 cm. At room temperature, (20 C), its internal
pressure is 1.05 atm. Find the number of moles of helium in
the balloon and the mass of helium needed to inflate the
balloon to these values.
So we have:
The absolute pressure P of an ideal gas is directly proportional to the
Kelvin temperature T and the number of moles n of the gas and is
inversely proportional to the volume V of the gas:
Kinetic Theory of Gases
1.
2.
3.
The container holds a very large number N of
identical molecules. Each molecule has a mass m,
and behaves as a point particle.
The molecules move about the container in a
random manner. They obey Newton’s laws of
motion at all times.
When the molecules hit the walls of the container
or collide with one another, they bounce elastically.
Other than these collisions, the molecules have no
interactions.
For conditions of low gas density, the distribution of speeds of the
particles within a large collection of molecules at constant
temperature was calculated by Maxwell.
See the figure below for Maxwellian distribution curve.
To find the force exerted by the molecule impacts
on the container wall consider:



ideal gas of N identical particles in a cubical container whose sides have
length L.
Except for elastic collisions (where on average there is no gain or loss of
translational K) the particles do not interact
focus on one particle of mass m as it strikes a wall perpendicularly and
rebounds elastically.
 while approaching the wall, it has a velocity +v and linear momentum
+mv
 the particle rebounds with a velocity –v and linear momentum –mv ,
travels to the opposite wall, and rebounds again
 the time t between collisions with the first wall is the round-trip distance
2L divided by the speed of the particle: t = 2L / v
 Impulse-momentum theorem says the force exerted on the particle by
the wall is equal to the change in momentum of the particle per unit time:
final momentum  initial momentum
Average Force 
time between successive collisions
(mv)  ( mv)  mv2


2L / v
L
Cont.



Newton’s 3rd law tells us that this force (of wall-on-particle) is equal and
opposite to the force of the particle on the wall. Thus, F= + mv2/L .
the total force exerted on the wall is equal to the number of particles
times the average force for each particle.
Since N particles move randomly in 3-D, about 1/3 of them will strike the
wall, that is N/3. Therefore the total force is:
 N  mv
F  
3 L

2
Notice v2 has been replaced with , the average squared speed.
(remember Maxwellian distribution). This quantity is called the rootmean-square speed, or rms speed; . Thus:
2
mv
N
 
F    rms
3 L
Cont.

Pressure is F/A, so P acting on a wall of area L2 is

Volume of the box is V = L3, so:

Of course, is the average translational kinetic energy :

This is similar to PV= N k T. set these equal (PV = PV):

This can be rearranged:

Thus, . Rearranging yields:
Example
The atmosphere is composed primarily of nitrogen N2
(78%) and oxygen O2 (21%). (a) Is the rms speed of N2
(28.0 g/mol) greater than, less than, or the same as the
rms speed of O2 (32.0 g/mol)? (b) Find the rms speed of
N2 and O2 at 293 K.
The Internal Energy of a Monatomic Ideal
Gas

Internal Energy – sum of the various kinds of energy that the atoms
or molecules of the substance possess.
The only energy contributing to internal energy of a monatomic ideal
gas is translational kinetic energy. Thus, the total internal energy, U
is the number of particles times the kinetic energy of each particle:
U = N (). Using the equation derived above, this becomes:

Remember that k = R / NA and N / NA = n
