Introduction to
Gas Quantities
© Mr. D. Scott; Celina High School
Everyone has
seen hot air
balloons floating
in the sky.
How do they
work?
By examining how a
hot air balloon works,
we can learn most of
the basics about gases
and how they work.
The Main
Parts:
Obviously the “hot” in “hot air
balloon” is a key part of how
they work. The temperature
of the air inside must be
higher than the outside air.
Consequently, the balloons
must have a heat source.
Propane burners are used for
this purpose.
Gas variables:
•Temperature (T)
•Volume (V)
•Pressure (P)
Tinside>Toutside
Pinside=Poutside
•Number of gas
particles (n)
Hot air balloons are open
at the bottom
Air particles are free
to move in and out of
the balloon in order to
maintain and equal
pressure with the
surrounding air.
Gas Particles in motion
When heated, the
gas particles move
faster and bounce
farther apart.
With greater
distances between
the particles, some
of them must exit
the fixed volume of
the balloon.
Number of
particles
inside
per liter
Number of
particles
outside
per liter
The number of particles relates to
the mass of the gas inside the
balloon.
Density = mass
volume
So, the density
inside is LESS
than the density
outside.
Buoyancy
Air
Density is
LOWER
Air
Density is
HIGHER
The upward
force that
keeps things
afloat.
The net
upward
buoyancy force
is equal to the
magnitude of
the weight of
air displaced
by the balloon.
Denser, heavier air pushes the
balloon up and out of its way.
Examining the Gas Variables
(T) Temperature = a measure that is proportional to the average
kinetic energy of the particles in a sample of matter.
KE  12 mv2
m  mass
v  velocity
Units: Celsius & Kelvin
Kelvin=Celsius + 273.15
Celsius = Kelvin – 273.15
For a given particle, higher T means it is faster moving.
But, when comparing two different particles, it may not. Let’s see how.
Consider two particles at the same temperature:
KE
=
KE
If they are at the same T, then their KE is equal.
1 mv2
2
=
1 mv2
2
For two particles that have the same mass, equal changes in their
temperature will result in equal changes in their velocity.
Now, consider two particles at the
same temperature that have different masses.
=
KE
KE
If they are at the same T, then their KE is still equal.
1 mv2
2
=
1 mv2
2
With different masses however, equal changes in their temperature will
result in un-equal changes in their velocity.
1
2
m
v2
=
v
1m
2
2
The smaller particle must move faster to have the same KE
as the larger one.
Examining the Gas Variables
(P) Pressure = the force exerted by the particles as they collide with
the walls of their container.
P F
A
The Mercury Barometer
F  force
A  area
Vacuum space
Height
measured in
millimeters
Glass tube
Liquid Hg
Evangelista
Torricelli is
credited with
inventing the
barometer in 1643
Units: atmospheres, Pascals, bars, Torr, mmHg, psi
1.00 atm = 101325 Pa = 1.01325 bar = 760 Torr = 760 mmHg = 14.696 psi
Examining the Gas Variables
(V) Volume = the space available for the moving gas particles inside
their container. Gas volume is measured in Liters.
(n) Number of gas particles = the total number of individual gas
molecules, atoms, or ions expressed in moles.
Putting it all together
For any given gas trapped in a container:
PV =constant
nT
All four of these variables can be changing whenever chemical changes
are happening. In order to keep things simple, we will first look at gas
relationships between only (n) and (V).
(n) And (V) can be examined for their relationship alone
only when (T) and (P) are held constant.
STP
Standard Temperature and Pressure
Standard Temperature =
0.00°C = 273K
Standard Pressure =
1.00 atm = 760 mmHg (Torr)
This now allows us to discover Avogadro’s Principle.
Equal volumes of gas at the same temperature and pressure have equal
numbers of particles regardless of the type of gases.
In other words, at STP, one mole of gas will occupy a definite
predictable volume. By setting our standard quantity at 1.00 mole of gas
particles, the volume they occupy at STP turns out to be 22.4 Liters.
Standard Molar Volume
MOLE
22.4 liters
1 mol
X
mol 22.4L  L
1mol
1mol  mol
X L 22.4
L
Gas Volume
at STP
(L)
Let’s do some conversions.
MOLE
X
mol 22.4L  L
1mol
1mol  mol
X L 22.4
L
What volume would 1.75 moles of
hydrogen gas occupy at STP?
1.75mol H2
Gas Volume
22.4 L
1 mol
= 39.2 L H2
at STP
(L)
15.0L of oxygen gas at STP
consists of what number of moles?
15.0 L O2
1 mol
22.4 L
= 0.670mol O2
Gas Density at STP
Density = mass
volume
Gas density can
change by
changing the
volume of the
container while
maintaining the
same number of
gas particles.
Gas Density at STP
Density = mass
volume
Gas density can
also change by
changing the
number of
particles while
maintaining the
same volume of
the container.
Gas Density at STP
Density = mass
volume
Gas density is normally measured using
grams for mass and liters for volume.
Gas Density = grams = g
L
liters
When the mass and
volume are each
known, the density can
be solved directly by
simply dividing.
Example: What is the density of 2.00 g
of CO2 that occupy 1.05 L?
2.00g CO2
D m
 1.90 g L
V
1.05L CO2
Gas Density at STP
Knowing the identity of a gas at STP also allows us to find its density.
Here’s how:
Example: What is the density of nitrogen gas at STP?
From STP we know:
22.4 L
1 mole
From molar mass we know:
And, we know that g
L
density must be
Putting them together:








 
 






g 1mol  g
1mol 22.4L L
1 mole N2
28.0134 g N2
So we have:














28.0134 g N2 1 mole 

  1.25 g
L
1 mole N2  22.4 L 


Gas Density at STP
Using the density of a gas at STP, we can determine a molar mass.
Here’s how:
Example: A certain noble gas has a density of 3.74 g/L
at STP. Which one is it?
Since we are at STP we
know:
22.4 L
Having the density given we
also know:
3.74 g
1 Liter
1 mole
Making the units
cancel to give us
molar mass we have:


















3.74 g 22.4 L = g
1 Liter 1 mol 1 mol
Carrying out the arithmetic we
have:
 83.8 g
The noble gas is:
mol
Krypton