• There is a lot of “free” space
in a gas.
• Gases can be expanded
infinitely.
• Gases fill containers
uniformly and completely.
• Gases diffuse and mix rapidly.
So you see there is no such thing as still
air. The air molecules are constantly
moving at an average of 1,000 miles per
hour.
Gas properties can be modeled using math.
Model depends on—
• V = volume of the gas (L)
• T = temperature (K)
– ALL temperatures in MUST be in Kelvin!!!
No Exceptions!
• n = amount (moles)
• P = pressure
(atmospheres)
9 hits
9 sec
1 hit
sec
=
0
Seconds
9
Hits
18
½ volume
Pressure comes from the
gas molecules hitting the
side of the container. Let’s
count them out loud.
18 hits
9 sec
=
2 hits
sec
00
01
02
03
04
05
06
07
So we saw that as
volume decreases the
pressure increases.
V=0.5
P=2
V=6 ,,P=5
V=0.1,
P=10
V=3, P=10
They are inversely
proportional.
BOYLES LAW
We can also show this by
having them multiply by
each other.
27 ºC = 300 K
9 hits
9 sec
=
1 hit
sec
0
Seconds
9
18
27 ºC = 300 K
We saw that we can increase pressure by reducing the
volume, but we can also do it by increasing the
temperature and therefore the speed of the gas
molecules. At room temperature the hits are 1 hit/sec
00
01
02
03
04
05
006
K
07
327 ºC = 600 K
9 hits
4.5 sec
=
2 hits
sec
327 ºC
= 600 K
0
Seconds
9
18
27 ºC = 300 K
We are going from room temperature 27
ºC = 300 K to double that temperature,
which is 600 Kelvin. Let’s count the
number of collisions at this higher speed.
We get twice the number of collisions
and therefore twice the pressure.
00
01
02
03
04
05
006
K
07
15 psi, 300 K
30 psi 600 K
3 psi
So we just saw that when temperature goes up,
so does the pressure. This makes sense
because higher temperature means the gas
molecules are going faster, colliding more
often, and hitting harder.
Gay-Lussac’s Law
60 K
Pressure is proportional
to the number of gas
molecules, which we
count in moles.
Another way to increase pressure is to
increase the number of gas molecules.
This is the approach the steam engine
used by heating water.
This is also a safety problem. Any closed
container that has liquid in and gets
heated will likely increase pressure
dramatically until the container bursts.
Let’s review what we learned. If the
volume decreases the pressure will
increase. Then the reverse happens if
the volume increases. The pressure
drops as gas molecules are farther apart.
As we also learned, we can
increase pressure by
introducing more molecules of
the gas into the volume.
doubles
We also learned that if temperature
doubles, the pressure doubles if
volume is fixed. Or if the container
is flexible, the volume will double
with pressure staying constant. Or
both can increase such that the
product of the two doubles.
• P is pressure measured in atmospheres.
• V is volume measured in Liters
• n is moles of gas present.
• R is a constant that converts the units. It's value is 0.0821
atm•L/mol•K
• T is temperature measured in Kelvin.
• Simple algebra can be used to solve for any of these values.
• P = nRT V = nRT n = PV
T = PV R = nT
•
V
P
RT
nR
PV
To make these quantities equal, we need a conversion
constant. We call it R (the Universal Gas Constant)
•
•
•
•
•
Pressure=1 atmosphere
Volume=1 Liter
n = 1 mole
R=0.0821 L atm mol-1 K-1
What is the temperature?
Let’s find what temperature the gas must be if we have the following
readings for these other properties.
Normally 1 mole of a gas at 1 atmosphere pressure takes up 22.4
liters. So it must be very cold to only have a volume of 1 liter.
Pressure of air is
measured with a
BAROMETER
(developed by Torricelli
in 1643)
Hg rises in tube until force of Hg (down)
balances the force of atmosphere
(pushing up). (Just like a straw in a
soft drink)
P of Hg pushing down related to
• Hg density
• column height
Evangelista Torricelli
1608 –1647
Pressure of air is
measured with a
BAROMETER
(developed by Torricelli
in 1643)
Hg rises in tube until force of Hg (down)
balances the force of atmosphere
(pushing up). (Just like a straw in a
soft drink)
P of Hg pushing down related to
• Hg density
• column height
Manometers
from Greek manos meaning sparse
CONVERSIONS
•
•
•
•
•
•
•
760 mm of Hg
760 torr
29.9 in. of Hg
1 Atmosphere
101.325 KPa (Kilopascals)
14.7 lbs. per sq. in.
about 34 feet of water!
All
Equal
Standard Temperature and
Pressure (STP)
P = 1 atmosphere
T = 0 °C
The molar volume of an ideal gas is 22.42 liters
at STP.
1 mol occupies 22.42 L at STP
sphygmomanometer
• sphygmometer
• Greek sphygmos meaning pulse (from sphyzein to
throb)
Measures to 300mm
Hg
This is the inner mechanisms of certain pressure gauges.
When a pressure cooker is used, what
causes the increased pressure?
PV=nRT
P=nRT
V
Temperature goes from 25oC to 100oC
Turn to Kelvin by adding 273 to Celsius
297K to 373K 75K/297K=25% increase in pressure
P1V1=n1RT1
P2V2=n2RT2
n1T1 n1T1
n2T2 n2T2
P1V1= R
P2V2= R
n1T1
P1V1= P2V2
n2T2
n1T1
n2T2
We can take advantage of the fact that
the R constant is the same even if the
conditions of the gas changes.
Change in Conditions Problem
1. What volume will 52.5 g of CH4 occupy at STP?
2. You heat 1.437 g NH3 in a stoppered 250 mL
flask until it explodes (425 °C). What was the
pressure inside the flask immediately prior to the
explosion?
The Gas Laws
a.k.a.
The Old Dead Guy’s with crazy hair
Laws
And so…. The gas laws
Boyle’s Law*
• Boyle’s LawAt constant
temperature,
volume is
inversely
proportional
to pressure.
Gay-Lussac’s Law
• At constant V :
Robert Boyle
*Holds precisely only at very low
pressures.
Joseph-Louis Gay-Lussac
There’s more….
Charles’ Law
• At constant pressure the
volume of a gas is directly
proportional to temperature,
and extrapolates to zero at
zero Kelvin.
Avogadro’s Law
• At constant V :
• For a gas at constant
temperature and pressure, the
volume is directly proportional
to the number of moles of gas
(at low pressures).
a = proportionality constant
V = volume of the gas
n = number of moles of gas
1 mol occupies 22.42 L at STP
Charles’ Law
Charles’ Law
Jacques Charles
• At constant pressure the
volume of a gas is directly
proportional to temperature,
and extrapolates to zero at
zero Kelvin.
Absolute Zero
There’s more….
Charles’ Law
• At constant pressure the
volume of a gas is directly
proportional to temperature,
and extrapolates to zero at
zero Kelvin.
Avogadro’s Law
• At constant V :
• For a gas at constant
temperature and pressure, the
volume is directly proportional
to the number of moles of gas
(at low pressures).
a = proportionality constant
V = volume of the gas
n = number of moles of gas
1 mol occupies 22.42 L at STP
Avogadro’s Law
Amedeo Avogadro
Avogadro developed this law after Joseph Louis GayLussac had published in 1808 his law on volumes (and
combining gases).
The greatest problem Avogadro had to resolve was
the confusion at that time regarding atoms and
molecules.
The scientific community did not give great attention to
his theory, so Avogadro's hypothesis was not
immediately accepted. André-Marie Ampère achieved
the same results three years later by another method
• At constant V :
• For a gas at constant
temperature and pressure, the
volume is directly proportional
to the number of moles of gas
(at low pressures).
a = proportionality constant
V = volume of the gas
n = number of moles of gas
1 mol occupies 22.42 L at STP
Molar Volume for Several Gases at STP
All of the laws can be summarized
nicely…
Combined Gas Law
What temperature is required to cause the pressure of a (steel)
cylinder of gas to increase from 350 to 500 mm Hg? The initial
temperature was 298 K.
A gas occupies a volume of 400. mL at 500. mm Hg pressure.
What will be its volume, at constant temperature, if the pressure
is changed to 250 torr?
Info. given:
Question: what is V2?
V1 = 400. mL
P1 = 500. mm
Temperature does not change
P2 = 250 torr (760 mm=760 torr) = 250 mm
We will use Boyle’s Law: P1V1 = P2V2
(500.mm)(400.mL)=(250.mm)(x)
x= (500)(400) = 800. mL
250
A gas occupies a volume of 410 mL at 27°C and 740 mm Hg
pressure. Calculate the volume the gas would occupy at STP.
Info. given:
Question: what is V2?
V1 = 410 mL
T1 = 27°C
T2 = 0 °C
P1 = 740 mm
P2 = 760 mm
PV P V

We will use the combined gas law:
T
T
 P1V1  T2   (740mm)( 410mL)  0C 
V2  
   

  0?
(27C)
 760mm 
 T1  P2  
1
1
1
Oops…use Kelvin
2
2
2
27°C=300K; 0°C=273K
 P1V1  T2   (740mm)( 410mL)  273K 
V2  
   

  363mL
(300K )
 760mm 
 T1  P2  
Suppose you have 856 mL of a gas. A weather front comes
through, and the barometric pressure changes from 780 mm Hg
to 720 mmHg. Along with this, the temperature changes
from 86 °F (30. °C) to 72 °F (22 °C). What is
the new volume of your gas?
Gas Stoichiometry
• When gases are involved in a reaction, das properties
must be combined with stoichiometric relationships.
E.g. Determine the volume of gas evolved at 273.15 K and
1.00 atm if 1.00 kg of each reactant were used. Assume
complete reaction (i.e. 100% yield)
CaO(s) + 3C(s)  CaC2(s) + CO(g).
• Strategy:
– Determine the number of moles of each reactant to which this
mass corresponds.
– Use stoichiometry to tell us the corresponding number of moles
of CO produced.
– Determine the volume of the gas from the ideal gas law.
Kinetic Molecular Theory
• The first kinetic interpretation of gases was Robert Hooke
in 1676.
• At the time, Isaac Newton’s picture of a gas was the
accepted view. Newton suggested that gas particles exert
pressure on the walls of a container because of repulsive
forces between the molecules.
Isaac Newton
Enter Maxwell and Boltzmann
• James Clerk Maxwell in 1859 and Ludwig
Boltzmann in the 1870s finally got people
to listen to a kinetic theory of gasses.
Postulates of the Theory
• Gases are composed of molecules which are small
compared to the average distance between them.
d(N2,g) = 0.00125 g/L (273°C)
d(N2,liq) = 0.808 g/mL (-195.8°C)
• Molecules move randomly, but in straight lines until they
collide with other molecules, or the walls of the container.
• Forces of attraction and repulsion between gas molecules
are negligible (except during collisions!)
• All collisions between gas molecules are elastic.
• The average kinetic energy of a molecule in a gas sample is
proportional to the absolute temperature.
Kinetic Molecular Theory
and the Ideal Gas Law
• Consider that pressure is due to the large
number of collisions of gas molecules with the
walls of the container.
p  (frequency of collisions)·(average force)
 1

p   u   N mu 
 V 
u:
m:
N:
average speed
mass
number of molecs.
Kinetic Molecular Theory
and the Ideal Gas Law
 1

p   u   N mu 
 V 
pV  Nmu
2
pV  NT
pV  nT
pV  nRT
½mu2 is the kinetic
energy and  T
N is proportional to
the number of moles.
Insert a constant
of proportionality.
Root Mean Square Speed
• The root mean square speed of gas molecules depends on the
temperature and the molar mass.
3RT
u
M
What is the rms speed of O2 molecules at 21 oC and
15.7 atm?
3 R T
u
M


3  8.314  294  2 2
u
m s
3
32.0 x 10



3  8.314 kg m 2 s  2 mol 1 K 1  294 K 
u
32.0 x 10 3 kg mol 1

u  489 m s
1
Graham’s Law of Effusion
• The rate of effusion of gas from a system is
proportional to the rms speed of the
molecules.
3RT
u
M
At constant temperature:
rate  1
M
An Example
• What is the ratio of rates of effusion of CO2
and SO2 from the same container at the
same temperature and pressure?
M SO2
rate of effusion for CO2

rate of effusion for SO 2
M CO2
rate of effusion for CO2
64.1 g / mol

rate of effusion for SO2
44.0 g / mol
This is because CO2 molecules move
1.21 times faster than SO2 molecules!
rate of effusion for CO 2
 1.21
rate of effusion for SO 2
Another Example
• If it takes 4.69 times as long for a particular
gas to effuse as it takes hydrogen under the
same conditions, what is the molecular
weight of the gas?
The time of effusion is inversely proportional
to the rate of effusion.
4.69 
M
2.0 g/mol
M
22.0 
2.0 g/mol
44.0 g/mol  M
Dalton’s Law of Partial Pressures
• John Dalton, in 1801,
suggested that each gas
in a mixture exerts a
pressure and that the
total pressure is the
sum of these partial
pressures.
ptot  p A  pB  pC  
More on partial pressures
• Each component gas has a partial pressure
which can be found using the ideal gas law.
p AV  nA RT
Where nA is the number of
moles of component A.
And the mole fraction is given by:
nA
pA
XA 

ntot ptot
An example on partial pressures
• A 1.00 L sample of dry air at 786 Torr and 25 oC
contains 0.925 g N2 plus other gasses (such as O2,
Ar and CO2.) a) What is the partial pressure of
N2? b) What is the mole fraction of N2?
mol
0.925 g 
 0.0330 mol
28.0 g
0.0330 mol 0.0821atm L mol 1 K 1 298 K   0.807 atm
1.00 L
613Torr
760 Torr
 0.780
0.807 atm 
 613 Torr X N 2 
786 Torr
atm