The Kinetic Theory,
Pressure & Gas Laws
GAS
The four quantities needed to describe a gas:
1) number of particles
2) temperature
3) pressure
4) volume
The Kinetic Theory
The Kinetic Theory explains the effects of
temperature & pressure on matter.
• Idea that all particles of matter are constantly in motion
4 Assumptions: of gases for IDEAL GASES
1) All gases are composed of small particles and
have no volume
2) These small particles move in continual, random,
and rapid motion with no attraction or repulsion
forces
3) All collisions are perfectly elastic
4) The average kinetic energy of the particles is
directly proportional to its Kelvin temperature
Real vs. Ideal Gases
An ideal gas is one that follows the gas laws at all
conditions of pressure and temperature and
follows all the assumptions of the KMT
An ideal gas does not really exist!!!!!
Real gases can be liquefied and sometimes solidified
The behavior of real gases under many conditions is
similar to ideal gases. Each gas is different.
Real gases behave like ideal gases at most conditions
except for extremely high pressures and low
temperatures
Kinetic Theory con’t
The physical behavior of a gas depends on its volume,
temperature, and pressure.
Temperature: Kelvin
the average kinetic energy of the particles in an object
ie. 25oC, O2 molecules 443 m/s; 1700 km/h;
1057 mi/h
Absolute Zero:
temperature at which all molecular motion should
cease; lowest possible temperature
-273.15oC or 0K
Kinetic Theory con’t
Temperature is not the total amount of thermal energy a
substance has absorbed.
Question 1:
How is the average kinetic energy of water molecules
affected when hot water from a kettle is poured into
cups at the same temperature as the water?
IT IS UNAFFECTED
Kinetic Theory con’t
Question 2:
By what factor does the average kinetic energy of the
molecules of a gas in an aerosol container increase
when the temperature is raised from 300K (27oC) to
900K?
The average kinetic energy triples
Pressure
Pressure:
the force per unit area 1
object exerts on another
SI Unit – Pascal (Pa)
Gas Pressure:
the pressure exerted by a
gas
it is created by collisions
of gas particles with an
object
What happens to the pressure when you add
more gas particles without changing its
volume or temperature? Why?
What happens to the pressure when you
decrease the volume of the container for a
fixed mass of a gas at a constant
temperature? Why?
What must happen to a gas’s temperature in
order for its pressure to decrease? Why?
Pressure
Pascal: (Pa)
• the SI unit for pressure
• equivalent to 1N/m2
Millimeter of Mercury: (mmHg)
pressure needed to support a column of
mercury 1 mm high
Pressure con’t
Atmosphere: (atm)
the pressure required to support 760 mm of
mercury in a mercury barometer
Torr: (torr)
named after Evangelista Torricelli who
invented the barometer
Barometer:
a closed-arm manometer used to measure
pressure
Pressure con’t
1 atm = 760 mmHg
1 atm = 101.325 kPa
760 mmHg = 101.325 kPa
1 torr = 1 mmHg
1 atm = 760 torr
1 atm = 760 mmHg = 760 torr = 101.325 kPa
Question 3: Convert the following:
a. 4.328 atm to kPa
b. 328 kPa to mmHg
c. 3290 Pa to atm
4.328 atm x 101.325 kPa
1.0 atm
= 438.5 kPa
328 kPa x 760 mmHg
101.325 kPa
= 2460 mmHg
3290 Pa x 1.0 atm
101325 Pa
= 0.0325 atm
manometer:
a devise used to measure pressure of a gas
Two Types of Manometers
1) closed-arm:
used to measure the actual or “absolute” gas
pressure
*known as a barometer
2) open-arm manometer:
one arm of the manometer is open to the
atmospheric air to measure the gas
pressure of a confined gas
Avogadro’s Hypothesis
Equal volumes of gases at the same
temperature and pressure contain equal
numbers of particles
States of Matter
1) GASES:
are independent of one another moving in straight lines
until they collide with something that affects its
direction and possibly its speed
have no definite shape or volume
gases assume the shape of the container
2) Liquids:
have a definite volume but will take the shape of their
container
particles are attracted to one another but have enough
energy to slide past each other
no bond is formed between the particles
3) Solids:
particles possess relatively fixed positions and
vibrate around that fixed point
attractive forces hold the particles extremely
close but the particles of the solid are still
traveling in straight paths between colliding with
its exceedingly close neighboring particles
the physical state of a substance at STP depends
on the attractive forces verses the energy of the
particles
Ionic compounds:
tend to be solids with strong electric charges
Molecular compounds
are attracted by van der Waals forces
high molecular mass compounds tend to be solids
nonpolar molecules of low molecular mass tend to be
gases
*the greater the mass and polarity, the more likely the
compound will be a solid or liquid
4) Plasma:
matter at temperatures greater than 5000oC causing a
state where the matter is composed of electrons and
positive ions
magnetohydrodynamics
Vaporization
Vaporization:
the conversion of a liquid to a gas or vapor below its
boiling point (bp)
Evaporation:
the vaporization of an uncontained liquid
Vapor Pressure:
pressure created by a vapor in equilibrium with its
liquid
Vaporization con’t
Boiling Point: (bp)
temperature at which the vapor pressure of the
liquid is equal to the external pressure
Normal Boiling Point:
boiling point for a substance at 1 atm
* Boiling is a cooling process
Vaporization con’t
evaporation
Liquid
Vapor (gas)
condensation
Heat of Fusion:
the additional amount of energy required to cause
a phase change from a solid to liquid once a
substance reaches its melting/freezing point
Heat of Vaporization:
the additional amount of energy required to cause
a phase change from a gas to liquid once a
substance reaches its boiling/condensation point
Phase Changes
Phase Diagram
Triple Point:
temperature and pressure where all three phases
are at equilibrium
Critical Point
at the critical temperature and critical pressure of
the substance, beyond this point the liquid and gas
phases become indistinguishable
JOHN DALTON
DALTON’S LAW OF PARTIAL
PRESSURES
At a constant volume and temperature, the total pressure
exerted by a mixture of gases is equal to the sum of the
partial pressures of the gases
PARTIAL PRESSURE:
the pressure(contribution) each gas in a mixture makes to
the total pressure of the mixture
Ptotal = P1 + P2 + P3 …
Example:
Air contains oxygen, nitrogen, carbon dioxide, and trace
amounts of other elements. What is the partial pressure
of O2 at 1.00 atm if PN2 = 593.4 mmHg, PCO2 = 0.3
mmHg and Ptrace = 7.1 mmHg?
Ptotal = PO2 + PCO2 + PN2 + Ptrace
PO2 = Ptotal - (PCO2 + PN2 + Ptr)
= 760mmHg - (593.4 mmHg + 0.3 mmHg + 7.1 mmHg)
= 159 mmHg
Your Turn:
Determine the total pressure of a mixture of gases if the
partial pressures of the gases are PO2 = 242.5 mmHg,
PHe = 27.3 kPa, and PNe = 0.021 atm.
PT = PO2 + PHe + PNe
Look for unit agreement!!!!
We will use mmHg as the common unit for our answer
= 27.3 kPa X 760 mmHg = 205 mmHg
101.325 kPa
= 0.021 atm X 760 mmHg = 16 mmHg
1 atm
PT = 242.5 mmHg + 205 mmHg + 16 mmHg
464 mmHg
Robert Boyle
When the number of particles and temperature are
constant, pressure and volume are inversely
proportional. (opposite)
P1 V1 = P2 V2
P1 =
initial pressure
V1 =
initial volume
P2 =
final pressure
V2 =
final volume
Example:
A balloon is filled with 30.0 L of helium gas at 1.00 atm.
What is the volume when the balloon rises to an altitude
where the pressure is only 185.3 mmHg? Assume
temperature remains constant.
30.0 L = V1
1.00 atm = P1
185.3 mmHg = P2
? L = V2
Look for unit agreement
1.00 atm = 760 mmHg
P1V1 = P2V2
Divide both sides by P2
V2 =
P1V1
P2
V2 = (760 mmHg)(30.0 L) =
185.3 mmHg
= 123 L
Jacques Charles
Charles’ Law
states that the volume of a fixed mass of gas is directly
proportional to its Kelvin temperature if the pressure is
kept constant
V1 = V2
T1 T2
*temperature must always be in Kelvin
A 225 cm3 volume of gas is collected at 58.0oC. What
volume, in liters, would this sample of gas occupy at
standard temperature?
V1 = 225 cm3
V2 = ?
T1 = 58.0oC : 273.15 + 58.0oC = 331.2K
T2 = 0.0oC : 273.15 + 0.0oC = 273.2K
Charles’ Law
V2 = V1T2
T1
= (225 cm3)(273.2 K)
331.2 K
= 186 cm3 · 1 L =
1000 cm3
= 0.186 L
Your Turn:
A balloon, inflated in an air-conditioned room at 27.0oC,
has a volume of 4.0 L. It is heated to a temperature
of 57.3oC. What is the new volume of the balloon if
the pressure remains constant?
T1 = 27.0oC
V1 = 4.0 L
T2 = 57.3oC
V2 = ????
Convert all temperatures to Kelvin
V 1 = V2
T1 T2
V2 = V1 T2
T1
V2 = (4.0 L)(330.5K)
300.2 K
= 4.4 L
Joseph Louis Gay-Lussac
Gay-Lussac’s Law
the pressure of a given mass of gas is directly
proportional to the Kelvin temperature if the
volume is held constant
P1 = P2
T1 T2
An acetylene gas cylinder has a pressure of 24350
mmHg at a temperature of 19.49oC. What would
the internal pressure be if the temperature was
increased to 100.0oC? Assume no change the
volume of the cylinder.
P1 = 24350 mmHg
T1 = 19.49oC + 273.15 = 292.64 K
P2 = X
T2 = 100.00oC + 273.15 = 373.15K
P1
T1
P2
P2
T2
= P2T1
T2
X = (24350 mmHg)(373.15 K)
294.64 K
24350 mmHg =
X
292.64 K
373.15K
=
X = 31050 mmHg
Your Turn:
A gas has a pressure of 50.0 atm at 540 K. What will
the temperature, in Celsius, be if the pressure is
increased to 8330 kPa?
P1 = 50.0 atm
T1 = 540 K
P2 = 8330 kPa
T2 = X
Unit agreement must be kept. Choose atm or kPa
50.0 atm x 101.325 kPa =
1.00 atm
= 5070 kPa
5070 kPa = 8330 kPa
540 K
X
890 K
oC
= 890 – 273 = 617oC
Combined Gas Law
Combines Boyle’s, Charles’ and Gay-Lussac’s laws
together
P1V1 = P2V2
T1
T2
all three laws can be derived from the combined gas law
by removing the variable with a constant value
The volume of a gas measured at 75.6 kPa pressure and
60.0oC is to be corrected to correspond to the volume
it would occupy at STP. The measured volume of the
gas is 10.0 cm3.
P1 = 75.6 kPa
V1 = 10.0 cm3
T1 = 60.0oC = 333.2 K
P2 = 101.325 kPa
T2 = 0.0oC = 273.2 K
V2 = ????? cm3
P1V1 = P2V2
T1
T2
V2 = P1V1T2
P2 T1
V2 = (75.6 kPa)(273.2 K)(10.0 cm3)
(101.325 kPa)(333.2 K)
= 6.12 cm3
Your Turn:
A cylinder of compressed oxygen gas has a volume of
30.0 L and 100.0 atm pressure at 300.0 K. The
cylinder is cooled until the pressure is 5.00 atm.
What is the new temperature, in Celsius, of the gas in
the cylinder?
V1 = 30.0 L
P1 = 100.0 atm
T1 = 27.0oC = 300.2 K
V2 = 30.0 L
P2 = 5.00 atm
T2 = ???? K
T2 = P2T1
P1
T2 = (5.00 atm)(300.2 K)
100.0 atm
= 15.0 K
15.0 K = 273.15 + oC
oC
= - 258.2
IDEAL GAS LAW
AVOGADRO’S HYPOTHESIS:
gases at the same temperature,
pressure and volume contain the same
number of particles
IDEAL GAS LAW
PV = nRT
PerVNeRT
P = Pressure
V = Volume in liters or dm3
n = moles
R = ideal gas constant
determined by the unit of pressure
T = temperature in Kelvin
R = 0.0821 atm·L
mol·K
= 8.31 kPa·L
mol·K
= 62.4 mmHg·L
mol·K
Ideal Gas Law
A rigid steel cylinder with a volume of 20.0 L is
filled with nitrogen gas to a final pressure of 200.0
atm at 27.0oC. How many moles of nitrogen gas
does the cylinder contain?
V = 20.0 L
P = 200.0 atm
T = 27.0oC + 273.2 = 300.2K
n = ? mol
CAN ONLY BE PV=nRT
R=?
R = 0.0821 atm·L/K·mol
PV = nRT
n = PV
RT
= (200.0 atm)(20.0 L)
(0.0821 atm· L)(300.2 K)
mol· K
all units cancel except mol
= 162 mol
Your turn:
Determine the pressure that 453.38 g of oxygen gas
would exert if it is put into a container with a
volume of 25.0 mL and a temperature of 17.4oC.
m = 453.38 g O2
V = 25.0 mL
T = 17.4oC
P=?
Must be PV=nRT
LOOK FOR UNIT AGREEMENT
n = 453.38 g O2 X 1 mol O2 = 14.2 mol
31.9988 g O2
V = 25.0 mL = 0.0250 L
T = 17.4oC + 273.2 = 290.6 K
P = ? atm
R = 0.0821 L atm/K mol
P = nRT
V
P = (14.2 mol)(0.0821 L atm/K mol)(290.6K)
0.0250 L
P = 13600 atm
Graham’s Law of Effusion
Diffusion:
the tendency of atoms, ions or molecules to move
toward areas of lower concentration until there is a
uniform composition
Effusion:
occurs as a gas escapes through a tiny hole in a
container
Graham’s Law
the rate of effusion of a gas in inversely
proportional to the square root of its molar mass
RateA =
RateB
molar mass B
molar mass A
Example
Which gas effuses faster, carbon dioxide or neon?
By how much faster?
Remember, lighter moves faster
CO2 = 44.0098 g/mol
Ne = 20.179 g/mol
Ne is faster because its molar mass is smaller
RateNe =
RateCO2
=
44.0098 g/mol
20.179 g/mol
1.477
neon gas is 1.477 times faster than carbon dioxide
Example 2
Determine the molar mass for a gas that is 0.6372
times as fast as oxygen gas.
Rateunknown =
Rateoxygen
Molar mass oxygen
molar mass unknown
Rateunknown = 0.6372
Rateoxygen
2
0.6372 =
31.9988 g/mol
X
0.4060 = 31.9988 g/mol
X
X = 78.81 g/mol
2
14.83 g zinc reacts with an excess of
hydrochloric acid, HCl, in the production of
hydrogen gas. Calculate the pressure, in
kPa, of the gas collected if it has a volume
of 2.150 L at a temperature of 25.0oC.
Assume 100% yield.
Zn + 2 HCl → ZnCl2 + H2
mass Zn = 14.83 g
PH2 = ? kPa
VH2 = 2.150 L
TH2 = 25.0oC + 273.2 = 298.2K
All means Gas Law Stoichiometry
Zn + 2HCl → ZnCl2 + H2
14.83 g Zn X 1mol Zn X 1 mol H2 =
65.39 g Zn 1 mol Zn
= 0.2268 mol H2
P = nRT
V
P = (0.2268 mol)(8.31 kPa·L/K·mol)(298.2 K)
2.150 L
P = 261.4 kPa
Acetylene, C2H2, is a gas used in cutting
metal. If 1.70 x 103 g of acetylene gas
undergoes a complete combustion reaction,
calculate the mass of oxygen gas that would
be required to react with all the acetylene
gas. Second, calculate the pressure of the
oxygen gas if the container the acetylene is
in has a volume of 55.00 L and a
temperature of 84.56oC.
2 C2H2 + 5 O2 → 4 CO2 + 2 H2O
1.70 x 103 g C2H2
X
1 mol C2H2
X
5 mol O2
26.0379 g C2H2 2 mol C2H2
163 mol O2
X
31.998 g O2
1 mol O2
5220 g O2
=
=
n = 163 mol
V = 55.00 L
T = 84.56 + 273.15 = 357.71 K
P = (163 mol)(0.0821 L· atm/ K·mol)(357.71
K)
55.00 L
87.0 atm
Calculate the mass of nitrogen gas required to
react hydrogen gas with a pressure of
356.33 kPa, a temperature of 288.3 K and a
volume of 377.4 mL in the production of
ammonia.
N2 + 3 H2 → 2 NH3
nH2 = PV
RT
(356.33 kPa)(0.3774 L)
(8.31 kPa·L)(288.3 K)
K·mol
= 0.05613 mol H2
= 0.05613 mol H2 X 1 mol N2
3 mol H2
= 0.5241 g N2
X
28.0134 g N2
1 mol N2
=