Contributed by:
Properties of gases, Gas laws, Ideal gas equation, Kinetic theory of gases
2.
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
3.
Pressure
• Gas molecules exert a pressure when they hit the surface
of their container
• Unit of pressure: force N
2 Pa (pascal)
• Atmospheric pressure (1 surface
atm): m1 atm = 101 325
Pa = 1.01325 x 102 kPa
4.
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
5.
Atmospheric Pressure
• We measure atmospheric
pressure with a barometer
• The mercury column exerts a
pressure equal to the
atmospheric pressure:
1 atm
= 760 mm Hg
= 760 torr
6.
Gas Pressure
• Gas pressure is measured using a
manometer
• We use Hg since it is very dense
(13.6 g/mL) and permits the
manufacturing of compact
barometers and manometers, if
not 1 atm= 760 mm x 13.6 =
10 300 mm H2O = 10.3 m H2O!!!
7.
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
8.
Boyle’s Law
• At constant temperature:
9.
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
10.
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
11.
Charles’ and Gay-Lussac’s Law
• N.B. These formulas are only valid if we
express T in kelvins!!!
12.
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
13.
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
14.
Ideal Gas Equation
1
• Boyle’ Law: (n,VT
constant)
P
• Charles’ Law: (n,V
Pconstant)
T
• Avogadro’s Law: (P,VTconstant)
n
15.
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
16.
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
17.
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
• R is the ideal gas constant
PV (1 atm)(22.414 L)
R
nT (1 mol)(273.15 K)
R 0.082 057 L atm
K mol
18.
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: T = 349 K
nRT (2.12 mol)(0.0821 L atm )(349 K)
V K mol
P 6.54 atm
V = 9.29 L
19.
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 V = 30.6 L
(49.8 g)
n 1.366 mol
(1.008 g/mol 35.45 g/mol)
L atm
nRT (1.366 mol)(0.0821 K mol
)( 273.15 K)
V
P 1.00 atm
20.
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 P2 V2 P1V1 P2 V2
R
n1T1 n 2T2 n1T1 n 2T2
• If the moles of a gas does not change:
P1V1 P2 V2
T1 T2
21.
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 T2 = 315 K
P1V1 T2
P2
T1 V2
(1.2 atm)(4.0 L) 315 K
P2 2.6 atm
339 K 1.7 L
22.
The Density and Molar Mass of a Gas
• The ideal gas law allows us to determine the density 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
23.
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: ρRT
M
P
(3.38 g )(0.0821 L atm )(313 K)
M L mol K
1.97atm
M 44.1 g/mol
24.
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
25.
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 ) n T
V V
26.
Dalton’s Law of Partial Pressures
• The mole fraction, XA, is given by
nA nA
XA
n T n A n B n C
• According to the Law of Partial Pressures:
n A RT
PA V nA
X A
PT n T RT nT
V
PA X A PT
27.
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?
• Solution: n T 8.24 0.421 0.116 8.78 mol
• In the same way, we can calculate the partial pressures of C2H6 and C3H8
8.24 mol
X CHatm,
to be 0.0657 atm and 0.0181 0.938
respectively
4
8.78 mol
PCH 4 X CH 4 PT 0.938 1.37 atm 1.29 atm
28.
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
29.
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
30.
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:
1atm
PH 2 988 mm Hg 31.82 mm Hg 956 mm Hg 956 mm Hg 1.258 atm
760 mm Hg
PH 2 V (1.258 atm)(0.641L)
PH 2 V n H 2 RT n H2 0.0324 mol
RT (0.0821 L atm )(303 K)
K mol
2.016 g
mH2 0.0324 mol 0.0653 g
1 mol
31.
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
32.
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
33.
The Kinetic Theory of Gases
• The average kinetic energy of gas molecules is given by
where is the average velocity squared
1 ____2
Ec m v
____
2
2
v
• The last assumption says that:
____ 2 2 2
v v v
v2 1 2 N
N
where k is the Boltzmann constant
1 ____2 1 ____2
m v T m v kT
2 2
34.
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
35.
The Kinetic Theory of Gases
• With the Kinetic Theory of Gases and the Maxwell distribution, we can derive the following
equation: ____
1
PV nM v 2
• But we know empirically that PV = nRT, thus 3
____ ____
1 3RT
nM v 2 nRT 2
v
3 M
• The average quadratic velocity, v quadr, increases when T increases
____ or the molar mass, M,
3RT
2
decreases v v quad
M
• 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)
36.
The Kinetic Theory of Gases
• Example: Calculate the average velocity of chlorine molecules (Cl 2) in
metres/seconds at 20oC.
• Solution: T = 293 K and M = 70.90 g/mol = 0.07090 kg/mol
(3)(8.314 J )(293 K)
v quadr K mol 321 m
kg s
0.07090
mol
• N.B. The Kinetic Theory of Gases allows us to calculate v quadr 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.
37.
• 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).
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