Gases
Properties of Gases
• Gases adopt the volume and shape of their containers
• Gases are easy compressed, whereas liquids and solids are
not
• Gases placed within the same container mix uniformly and
completely
• Gases have much lower densities than liquids and solids
Pressure
• Gas molecules exert a pressure when they hit the
surface of their container
• Unit of pressure:
force
N
 2  Pa (pascal)
surface m
• Atmospheric pressure (1 atm):
1 atm = 101 325 Pa = 1.01325 x 102 kPa
Atmospheric Pressure
• Atmospheric pressure is the
pressure exerted by the
column of air situated above
a surface
• We do not feel the
atmospheric pressure because
we are physiologically
adapted for it
Atmospheric Pressure
Gas Pressure
Boyle’s Law
• Boyle observed that the
volume of a gas decreases
(increases) when the pressure
exerted on the gas increases
(decreases)
• Boyle’s Law states that the
volume of a given mass of gas
at constant temperature is
inversely proportional to its
pressure
Boyle’s Law
• At constant temperature:
Charles’ and Gay-Lussac’s Law
• Charles and Gay-Lussac observed
that at constant pressure for a
given mass of gas, the volume
increases with increasing
temperature and decreases with
decreasing temperature
The Kelvin Scale
• Plots of V vs. T have different slopes
for different pressures, but the
extrapolation of each line to V = 0
crosses the T axis at T = -273.15 oC
• Kelvin proposed that -273.15 oC is
the lowest temperature that can be
achieved, i.e., the absolute zero
• The Kelvin Scale:
T(K) = t(oC) + 273.15 oC
• N.B. These formulas are only valid if we
express T in kelvins!!!
Avogadro’s Law
• Avogadro’s hypothesis (1811):
At the same temperature and same pressure, equal volumes
of different gases contain the same number of molecules
• Avogadro's law also says that, at constant pressure and
temperature, the volume of a gas is directly proportional to
the number of moles of gas present
Avogadro’s Law
• Avogadro 's law insists that
when two gases react together
and the product(s) is a gas :
-The ratio between the
volumes of reactants is a
simple number
- The ratio between the total
volume of reactants and the
total volume of products is
a simple number
Ideal Gas Equation
1
• Boyle’ Law: V 
P
(n, T constant)
• Charles’ Law: V  T (n, P constant)
• Avogadro’s Law: V  n (P, T constant)
Ideal Gas Equation
• We can put the three laws together:
nT
V
P
nT
VR
P
• Ideal Gas Equation: PV = nRT
• R is the ideal gas constant
Ideal Gases
• An ideal gas is a theoretical gas in which the pressure,
volume, and temperature obey the ideal gas equation
- There is no attraction or repulsion between the molecules
of an ideal gas
- The volume of the molecules in an ideal gas is negligible
compared to the volume of the container (i.e., the space
in the container is empty)
• The approximation of an ideal gas is better at high T and
low P
Ideal Gas Constant
• At 0 oC and 1 atm (STP: standard temperature and
pressure), most real gases behave like ideal gases
• It is observed that for one mole of any gas at STP, the
volume is around 22.414 L
PV (1 atm)(22.414 L)
R 

nT (1 mol)(273.15 K)
R  0.082 057 L  atm
• 8.3145
• R is the ideal gas constant
K  mol
Ideal Gas Equation
• Example: Calculate the volume (in litres) occupied by
2.12 moles of nitric oxide (NO) at 6.54 atm and 76oC.
• Solution:
nRT
V

P
T = 349 K
(2.12 mol)(0.0821 L  atm
6.54 atm
V = 9.29 L
)(349 K)
K  mol
Ideal Gas Equation
• Example: What volume is occupied by 49.8 g of HCl at
STP?
• Solution:
T = 273.15 K and P = 1 atm
n
(49.8 g)
 1.366 mol
(1.008 g/mol  35.45 g/mol)
L  atm
)(273.15 K)
nRT (1.366 mol)(0.0821
K

mol
V

P
1.00 atm
V = 30.6 L
Ideal Gas Equation
• A modified form of the ideal gas equation is sometimes
more useful to study variations of P , V, T for a fixed
amount of gas
P1V1
PV
PV P V
R 2 2  1 1  2 2
n1T1
n 2T2
n1T1 n 2T2
• If the moles of a gas does not change:
P1V1 P2 V2

T1
T2
Ideal Gas Equation
• Example: A sample of radioactive radon gas that had an
initial volume of 4.0 L, an initial pressure of 1.2 atm and a
temperature of 66oC, undergoes a change that changes its
volume and temperature to 1.7 L and 42oC . What is the
final pressure? The number of moles remains constant.
• Solution:
T1 = 339 K
and
P2 
P2 
T2 = 315 K
P1V1 T2

T1 V2
(1.2 atm)(4.0 L) 315 K

 2.6 atm
339 K
1.7 L
The Density and Molar Mass of a Gas
• The ideal gas law allows us to determine the density (r) or
molar mass (M) of a gas:
n
P
PV  nRT 

V RT
m
m
P
n


M
MV RT
m PM
ρ 
V RT
mRT ρRT
M

PV
P
The Density and Molar Mass of a Gas
• Example: The density of a gaseous organic compound is
3.38 g/L, at 40oC et at 1.97 atm. What is its molar mass?
• Solution:
M
ρRT
P
(3.38 g )(0.0821 L  atm
)(313 K)
L
mol

K
M
1.97atm
M  44.1 g/mol
Dalton’s Law of Partial Pressures
• The equations we have seen so far are for pure gases
• Dalton’s Law of Partial Pressures states that the total
pressure of a mixture of gases is the sum of the partial
pressures of the gases that make up the mixture
• The partial pressure of a gas in a mixture is the pressure
that the gas would exert if it was alone
• The Law of Partial Pressures is consistent with the lack of
attractions/repulsion in an ideal gas
Dalton’s Law of Partial Pressures
• The partial pressure of gas A, PA, in a mixtures of gases is
n A RT
PA 
V
• The Law of Partial Pressures states that the total pressure,
PT, is given by
PT  PA  PB  PC  
n A RT n B RT n C RT
PT 



V
V
V
RT
RT
PT  (n A  n B  n C  )
 nT
V
V
Dalton’s Law of Partial Pressures
nA
nA
XA 

nT nA  nB  nC 
n A RT
PA
nA
V


 XA
n
RT
T
PT
nT
V
PA  X A PT
Dalton’s Law of Partial Pressures
• Example: A sample of natural gas contains 8.24 mol of CH4, 0.421
mol of C2H6 and 0.116 mol of C3H8. If the total pressure is 1.37
atm, what is the partial pressure of each gas?
n T  8.24  0.421  0.116  8.78 mol
• Solution:
PCH 4
8.24 mol
X CH 4 
 0.938
8.78 mol
 X CH 4  PT  0.938  1.37 atm  1.29 atm
• In the same way, we can calculate the partial pressures of C2H6 and
C3H8 to be 0.0657 atm and 0.0181 atm, respectively
Dalton’s Law of Partial Pressures
• We often collect a gas
produced within a reaction by
the displacement of water
• e.g.; KClO3(s) decomposes
to give KCl(s) and O2(g), and
because O2(g) is not very
soluble in water, the O2(g)
displaces the water in the
inverted bottle
Dalton’s Law of Partial Pressures
• In the inverted bottle, above
water, one should not forget the
pressure due to water vapour
PT  PO 2  PH 2O

PO 2  PT  PH 2O
• The partial pressure due to
water vapour is well known as a
function of temperature
• The same principle applies to
each gas which is insoluble in
water
Dalton’s Law of Partial Pressures
• Example: Hydrogen is prepared by reacting calcium with water.
Hydrogen is collected using an assembly like the one just seen. The
volume of gas collected at 30 oC and at 988 mm Hg is 641 mL.
What is the mass of hydrogen that was produced? The pressure of
water vapour at 30oC is 31.82 mm Hg.
• Solution:
PH 2  988 mm Hg  31.82 mm Hg  956 mm Hg  956 mm Hg 
PH 2 V  n H 2 RT

(1.258 atm)(0.641 L)
 0.0324 mol
L

atm
RT
(0.0821
)(303 K)
K  mol
2.016 g
 0.0324 mol 
 0.0653 g
1 mol
n H2 
mH2
PH 2 V
1atm
 1.258 atm
760 mm Hg

The Kinetic Theory of Gases
• We arrived at the ideal gas equation empirically, i.e., we don’t
know why PV = nRT
• Maxwell and Boltzmann tried to explain the physical properties
of a gas from the movements of its individual molecules
• In an ideal gas, there are no attractions or repulsion between the
gas molecules, so the energy of the gas comes entirely from the
kinetic energy of individual molecules
• The kinetic energy of a molecule is dependent only on the mass
and velocity of that molecule
Kinetic Theory of Gases: Assumptions
• A gas is formed of molecules separated from one another
distances much greater than their own dimensions, i.e., the
volume of a molecule is negligible
• Gas molecules are in constant motion in all directions, and they
frequently collide and these collisions are perfectly elastic, i.e.,
the total (kinetic) energy of all the molecules of a system is
constant
• Gas molecules exert no attractive or repulsive force between
each other
• The average kinetic energy of gas molecules is proportional to
the temperature of the gas in Kelvins, and two gases at the same
temperature have the same average kinetic energy
The Kinetic Theory of Gases
• The average kinetic energy of gas molecules is given by
____
2
1 ____2
Ec  m v
2
where v is the average velocity squared
____
2
2
2
v

v



v
2
N
v2  1
N
• The last assumption says that:
1 ____2
m v T
2

where k is the Boltzmann constant
1 ____2
m v  kT
2
The Kinetic Theory of Gases
• The Maxwell distribution
describes the probability of
finding a molecule with a
given speed at a given
temperature
• The most probable speed
increases as the temperature
is increased
• There is a greater dispersion
in speeds at high temperature
The Kinetic Theory of Gases
• With the Kinetic Theory of Gases and the Maxwell distribution, we
____
1
can derive the following equation:
PV  nM v 2
3
• But we know empirically that PV = nRT, thus
____
1
nM v 2  nRT 
3
____
2
v  v quad 
____
2
v 
3RT
M
3RT
M
• The average quadratic velocity, vquadr, increases when T increases or
the molar mass, M, decreases
• So that vquadr is given in m/s (i.e., SI units), M must be given in
kg/mol (SI) and R must be expressed as 8.314 J/K (SI units)
The Kinetic Theory of Gases
• Example: Calculate the average velocity of chlorine molecules
(Cl2) in metres/seconds at 20oC.
• Solution: T = 293 K and M = 70.90 g/mol = 0.07090 kg/mol
v quadr 
(3)(8.314 J
)(293 K)
K  mol
 321 m
s
0.07090 kg
mol
• N.B. The Kinetic Theory of Gases allows us to calculate vquadr
for He and H2 and we observe that it approaches Earth’s escape
velocity (1.1 x 104 m/s). Both gases can thus escape our
atmosphere.
• In a 5.00 L container, we have 8.22 g of O2(g) and a pressure of
1.00 atm. In another 5.00 L container, we have 8.22 g of N2(g) and
a pressure of 1.00 atm. What is the average velocity (or quadratic
velocity) of the molecules in each container? We put all of the
O2(g) and all of the N2(g) in a third 5.00 L container that is
maintained at 25.0 oC. What is the total pressure of this container?
On average, do the molecules of O2(g) have more kinetic energy,
the same kinetic energy, or less kinetic energy than the molecules
of N2(g)? You do not need to explain your reasoning (that is to
say, just provide the answer).