Gases ~ An Overview and
Review of Concepts and Laws
J. Baumwirt, Chemistry
Granada Hills Charter High School
From a compilation of different online and textbook resources
for instructional purposes ONLY. Reproduction of this
PowerPoint is prohibited due to copyright laws.
States of Matter
• Gases are only one
form of matter
• Note the relative
distance in particles of
a gas as compared to
other states of matter
• This will be an
important factor as we
study the properties of
gases
Solid
Liquid
Gas
Plasma
Properties of Gases
• Gases are composed
of atoms or molecules
• Gas particles are far
apart and therefore are
the most compressible
state of matter
• Gases have lower
densities that solids or
liquids
• Gases will mix evenly
and completely when
confined to the same
container
Properties of Gases
• Gases expand to fill any
container
• Because atoms and
molecules are so small a
container of gas is mostly
empty space
• Gaseous particles have
relatively few attractions
or repulsions between
particles under normal
conditions
Variables that Affect the
Behavior of a Gas
•
•
•
•
Pressure
Volume
Amount of Gas
Temperature
Pressure
• Gases exert pressure on
• Pressure =
Force per unit Area any surface they contact
(Force = mass x acceleration)
P = Force
Area
• We can look at pressure
simplistically as the number
of times the particles hit the
walls of the container
( “Particle in a Box” model )
Units of Pressure
• Pressure can be designated in a variety of
different units:
• In Chemistry these
–
–
–
–
–
mmHg or torr
are the units we will
atmospheres, atm
use most commonly
Pascals, Pa
psi (pounds per square inch)
Bars
• Their equivalencies are as follows:
1 atm = 760 mmHg = 760 torr = 101,325 Pa = 14.7 psi
How Pressure is Measured
• The Barometer was
invented by
EvangelistaTorricelli
• Origin of the pressure unit
in mmHg is due to the use
of the metric system to
measure of the height of
the mercury column. This
unit is also known as a torr
• It is the atmospheric
pressure that pushes the
mercury up the inverted
glass tube
• The height (h) above the
level of the mercury in the
dish is then read in mmHg
Why was mercury used?
Manometers (mă  năh΄ mә  tŭr)
• There are two types
of manometers
• Open ended
manometers with
one end open to
the atmosphere
• Closed end
manometers where
Open ended devices require access to a barometer
to find the pressure of the atmosphere.
one end is a vacuum
• Pressure readings
require finding the
difference between
the “legs” of the
manometer
(illustrated here as h1, h2, h3)
a) Pgas= Patm - h2 b) Pgas= Patm+ h2
c) Pgas= h1
The two illustrations of an open ended manometer show:
a) a gas with a pressure less than atmospheric pressure
b) a gas with a pressure greater than atmospheric pressure
The Ideal Gas Law
• Most gases behave “ideally” under normal
conditions
• The Ideal Gas Law equates the variables
of pressure, volume, amount of gas and
temperature that affect gas behavior
together in one equation:
PV=nRT
Units of the Ideal Gas Law
PV = nRT
Pressure is in the units of atmospheres
atmospheres, torr or mmHg
Volume in Liters (L)
Amount of gas is in the unit of moles = n
Recall that
1 torr = 1 mmHg
R = gas constant 0.0821 L•atm = 62.37 L•torr
L• mmHg
mol•K
mol•K
The R constant is chosen to match the unit of Pressure used.
Temperature in Kelvin
Recall that C + 273.15 = K
A 12.25 L cylinder contains 75.5 g of neon at 24.5 oC.
Determine the pressure of the cylinder of gas.
PV = nRT
P= ?
V = 12.25 L
n=
P = nRT
V
= (3.74 mol)(62.4 L•mmHg)(297.5K)
(12.25 L) mol•K
75.5 g mol = 374 mol
20.18 g
= 5667.7 mmHg
= 5670 mmHg
What is this in atm?
R = 62.4 L• mmHg
mol • K
5667.7 mmHg 1 atm
= 7.46 atm
o
T = 24.5 C + 273 = 297.5 K
760 mmHg
Alterations of the Ideal Gas Law
• The Ideal Gas Law is used to find one
aspect about a gas:
– Pressure, Volume, number of moles or
Temperature
• Through mathematical substitutions, the
variables can be extended to
– Molecular mass and
– Density
Using Subsitutions with PV=nRT
• n in the equation = moles
• But how do we find the number of moles of
a substance?
– moles =
grams of substance
molar mass
Molar mass (or
molecular weight) = Mwt
– Substituting this back into the equation:
PV =
grams
Mwt
RT
The variables have now been
extended to include mass and
molecular weight.
More substitutions into PV=nRT
• Taking the previous substituted equation:
PV =
grams
Mwt
RT
• The equation can be rearranged to solve
for density. Density = Mass mass = grams
Volume
PV =
grams
Mwt
RT
PMwt grams
=
V
RT
The consideration here is to remember that density units are
somewhat altered now as they are in grams per Liter.
What is the density of carbon dioxide gas
at 25 oC and 725 mmHg pressure?
grams
Density = grams
PV = Mwt RT
V
grams = PMwt
RT
V
grams = 725mmHg  44.0 g/mol
V 62.4 L• mmHg/mol K  298 K
P = 725mmHg
grams = 1.72 g/L
V
V= ?
n = grams ? CO2 =
Mwt
44.0 g/mol
R = 62.4 L• mmHg
mol•K
T = 25C + 273 = 298 K
What do we
do now?
Combined Gas Law
• If there is a change in conditions such as a change in
pressure or volume of a gas, the ideal gas law can be
converted to an equality called the Combined Gas Law:
Setting the equation equal to R:
PV = R
nT
• Then since R is constant for any gas, the following can be
used to calculate a change for any of the variables.
Condition 1
P1V1
n1T1
Condition 2
=
P2V2
n2T2
• If a value is constant (does not change) it can be cancelled
out and eliminated from the equation.
A balloon contains helium gas with a volume of 2.60 L
at 25 oC and 768 mmHg. If the balloon ascends to an
altitude where the helium pressure is 590 mmHg and
the temperature is 15 oC, what is the volume of the
balloon?
What type of
problem
is this?
There are 2 sets of
conditions.
Yikes!
A balloon contains helium gas with a volume of 2.60 L at
25 oC and 768 mmHg. If the balloon ascends to an
altitude where the helium pressure is 590 mmHg and the
temperature is 15 oC, what is the volume of the balloon?
P1V1 = P2V2 n is
P1V1 = P2V2
n1T1 n1T2 constant T1
T2
Condition 1:
P1= 768 mmHg
V1= 2.60 L
T1= 25 oC + 273 = 298 K
Condition 2:
P2= 590 mmHg
V2= ?
T2= 15C + 273 = 288 K
P1V1T2 = (768 torr)(2.60 L)(288 K)
= 3.27 L
V2 =
P2T1
(590 torr) (298 K)
Avogadro’s Law and
Standard Temperature and Pressure
Avogadro’s Law states that equal volumes of any
two gases (Ideal) at the same temperature and
pressure contain the same number of molecules.
STP:
STANDARD TEMPERATURE & PRESSURE
Pressure 1 atm (760 mm Hg)
Temperature 0oC (273 K)
Standard
At STP one mole of ideal gas occupies 22.4 L
(Looks like another conversion factor to me!) 1 mol gas/22.4L