TOPIC 7: GASES
Contents
• Properties of Gases
• The Simple Gas Laws
• The Ideal Gas Equation
• Gases in Chemical Reactions
• Mixture of Gases
• Kinetic-Molecular Theory of Gases
• Gas Properties Relating to the KineticMolecular Theory
• Nonideal(Real) Gases
1
PROPERTIES OF GASES
• Pressure : A force per unit area
Force(N)
P(Pa) =
Area(m2)
Liquid Pressure
P= g x h x d
g: acceleration of gravity
h: height of the liquid column
d: density
2
BAROMETRIC PRESSURE
Evangelista Torricelli, 1643
Standard Atmospheric
(Barometric) Pressure
Atmospheric
(Barometric)
pressure)
1,00 atm= 760 mmHg,
760 torr
101,325 kPa
1,01325 bar
1013,25 mbar
dHg = 13,5951 g/cm3 (0°C)
g = 9,80665 m/s2
3
Manometers
Measurement of Gas Pressure using an open sided
manometer
Gas pressure
equal to
barometric
pressure
Gas pressure
greater than
barometric
pressure
Gas pressure less
than barometric
pressure
4
The Simple Gas Laws
For a fixed amount of gas at a constant temperature, gas volume is
inversely proportional to a gas pressure.
• Boyle’s Law
1
Pα
V
PV = Constant
P1 V1 = P2 V2
5
Charles’s Law
Charles 1787 Gay-Lussac 1802
The volume of a fixed amount of gas at a constant pressure is directly
proportional to the Kelvin(absolute) temperature.
VαT
V=bT
Absolute temperature scale:
273,15oC or 0 K,
T(K)= t(oC)+ 273,15
Volume
(mL)
Volume (mL)
Temperature (oC)
V1
V2
=
T1
T2
Temperature (K)
6
Avogadro’s Law
• Gay-Lussac 1808
Gases react by volumes in the ratio of small whole
numbers.
Avogadro 1811
Equal volumes of different gases compared at the
same temperature and pressure contain equal number
of molecules.
Equal numbers of molecules of different gases
compared at the same temperature and the same
pressure occupy equal volumes.
At a constant pressure and temperature:
Vαn
or V = c n
Standard Conditions (0 C= 273,15 K and 1atm= 760 mm Hg)
1 mol gas = 22,414 L at STP
7
Combination of all Gas Laws: The ideal
Gas Equation
• Boyle’s Law
• Charles’s Law
• Avogadro’s Law
V α 1/P
VαT
Vαn
Vα
nT
P
PV = nRT
8
Gas Constant
PV = nRT
PV
R=
nT
= 0,082057 L atm mol-1 K-1
= 8.3145
8,3145 m3 Pa mol-1 K-1
= 8,3145 J mol-1 K-1
9
Example: What is the volume occupied by 13,7 g Cl2(g) sample at 45
C and 745 mm Hg?
1 atm = 760 mmHg; R = 0,08206 L atm /(mol K); Cl:35,5
Solution
T  45  273,15  318,15
1 atm
P  745m m Hg
 0,980
760m m Hg
n  13,7 g
1 m ol Cl2
 0,193
71 g Cl2
nRT
V
P
0,193m ol 0,08206L atm /(m olK )  318,15 K
V
0,980atm
V  5,14 L
Practice: What is the pressure exerted by 1,00 x 1020 molecules
of N2 in a 350 ml volume of container at 175 C ?
10
General Gas Equation
P1V1
P2V2
R=
=
n1T1
n2T2
PiVi
niTi
=
PsVs
nsTs
We often apply it in cases in which one or two of
the gas properties are held constant and we can
simplify the equation by eliminating these
constants
In the cases of the comparison of two gases,
General Gas Equation must be used. In other cases
the ideal gas equation is rather relevant.
11
Applications of the Ideal Gas Equations
Is the amount of gas
given or asked?
Yes
If the mass of gas is constant use the
Ideal Gas Equation PV=nRT
If the mass of gas is variable use the
No
General Gas Equation.
PiVi = PsVs
Use the General Gas Equation
by comparing the initial and
final conditions
PiVi = PsVs
Ti
Ts
niTi
Ti=Ts
nsTs
Boyle’s Law
PiVi = PsVs
Vi=Vs
Pi = Ps
Ti
Ts
Pi = Ps
Vi = Vs
Ti
Ts
12
Molar Mass Determination
Propylene is an important commercial chemical. It is used in the
synthesis of other organic chemicals and in plastic production. A glass
vessel weighs 40,1305 g, when clean,dry and evacuated ; 138,2410 g
when filled with water at 25°C (density of water δ= 0,9970 g/cm3)
and 40.2959 g when filled with propylene gas at 740,3 mm Hg and
24,0°C . What is the molar mass of propylene?
Strategy:
Find out Vves , mgas ; Use the Ideal Gas Equation
Vves = mH2O / dH2O = (138,2410 g – 40,1305 g) / (0,9970 g cm-3)
= 98,41 cm3 = 0,09841 L
mgas:
mgas = m - mempty= (40,2959 g – 40,1305 g)
= 0,1654 g
13
Ideal Gas Equation:
PV = nRT
M=
m
RT
PV =
M
m RT
M=
PV
(0,6145 g)(0,08206 L atm mol-1 K-1)(297,2 K)
(0,9741 atm)(0,09841 L)
M = 42,08 g/mol
14
Gas Densities
PV = nRT
m
RT
PV =
M
ve
m
m
, n=
d=
M
V
MP
m
=d=
V
RT
Gas densities differ from solid and liquid densities in two
important ways:
1- Gas densities depend strongly on temperature and pressure;
increasing as the gas pressure increases, decreasing as the
temperature increases. Densities of liquids and solids also depend
somewhat on temperature, but they depend far less on pressure
2- The density of a gas is directly proportional to its molar mass.
No simple realtionship exists between density and molar mass for
liquids and solids.
15
Gases in Chemical Reactions
• Use the stoichiometric factors to relate the
amount of a gas to amounts of other
reactants or products.
• Use the ideal gas equation to relate the
amount of gas to volume,temperature and
pressure.
• Law of combining volumes can be modified
with the other laws
16
Law of Combining Volumes
• At times, if the reactants and/or products involved in a stoichiometric
calculation are gases, we can use a particularly simple approach:
2NO(g) + O2 (g)
2NO2 (g)
2 mol NO + 1 mol O2 (g)
2 mol NO2(g)
Suppose the gases are compared at the same T and P , in this case one mol
of gas occupies a particular volume 1V , 2 mol of gas 2V and 3 mol
of gas 3V of liters.
2NO(g) + O2 (g)
2 L NO(g) + 1 L O2 (g)
2NO2 (g)
2 L NO2(g)
17
Example: The decomposition of sodium azide, NaN3, produces
N2(g). Together with the necessary devices to initiate the reaction
and trap the sodium metal formed, this reaction is used in air bag
safety systems. What volume of N2(g) measured at 735 mm Hg
and 26°C is produced when 70,0 g NaN3 is decomposed?
2 NaN3(s) → 2 Na(l) + 3 N2(g)
Calculate the mole of N2 :
nN2 = 70 g N3 x
1 mol NaN3
3 mol N2
= 1,62 mol N2
x
65,01 g N3/mol N3 2 mol NaN3
Calculate the volume of N2
nRT
(1,62 mol)(0,08206 L atm mol-1 K-1)(299 K)
V=
=
P
1.00 atm
(735 mm Hg)
760 mm Hg
= 41,1 L
18
Mixtures of Gases
• Simple gas laws and ideal gas equation are
applicable to a mixture of gases such as air.
• A simple approach to working with gas
mixtures is to use ntot, total amount in moles
• Partial pressure
– Dalton’s law of partial pressures:
The total pressure of a mixture of gases is the
sum of partial pressures of the components of the
mixture. Ptop= Pa + Pb + Pc + …
19
Dalton’s Law of Partial Pressures
Ptot = Pa + Pb +…
Va = naRT/Ptot
ve
na
naRT/Ptot
Va
=
=
ntopRT/Ptot
ntot
Vtot
Vtot = Va + Vb+…
na
Remember
ntot
= a (Mole
Fraction)
na
naRT/Vtot
Pa
=
=
ntotRT/Vtot
ntot
Ptot
20
Kinetic-Molecular Theory of Gases
• A gas is composed of a very large number of extremely
small particles in constant, random, straight line
motion. Molecules of gas are seperated by great
distances(The molecules are treated as if they have a mass
but no volume,so called point masses.
• Molecules collide with one another and with the walls of
their container very rapidly. There are assumed to be no
forces between molecules. That is each molecule acts
independently of all others.In a collection of molecules
at constant temperature the total energy remains
constant.
21
Gas Properties relating to the Kinetic
Molecular Theory
Diffusion
- The migration of molecules
of different substances as a
result of random molecular
motion.
Effusion
The escape of gas molecules
from their container through a
tiny orifice or pin hole.
22
Graham’s Law
Rate of effusion, urms=
3RT
M
(urms ) A
3RT/M A
MB


(urms )B
3RT/MB
MA
Graham’s Law:
The rates of effusion of two different gases are inversely proportional
to the square roots of their molecular mass.
Graham’s Law applies only if certain conditions are met. For effusion the
gas pressure must be very low, not as a jet of gas.
3 RT=NAmv2,Note that the product NAm represents the mass of 1 mol of molecules,
The molar mass M,so:
EK = 3/2RT
1
3
M v 2  RT
2
2
3RT
v2 
M
v 2  urms 
3RT
M
23
Effusion
•
24
Example
•
25
Real Gases
• Compressibility factor: PV/nRT = 1
• PV= nRT – Ideal gas behaviour
– PV/nRT > 1 – At very high
pressures
– PV/nRT < 1 – Where
intermolecular forces of attraction
exist
26
Real Gases
•
27