The Gas State
 Gases are everywhere – atmosphere, environmental
processes, industrial processes, bodily functions
 Gases have unique properties from liquids and solids
 Gases are compressible (very important for storage)
 Gas particles are widely separated and move at very
fast speeds
 Most gases have relatively low densities
 Gas have relatively low viscosity (resistance to
movement) allowing them move freely through pipes
and even small orifices
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The Gas State
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

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Chemical behavior of gases depends on composition
Physical behavior of all gases is similar
Gases are miscible mixing together in any proportion
Behavior of gases described by ideal gas law and
kinetic-molecular theory, the cornerstone of this
chapter
Gas volume changes greatly with pressure
Gas volume changes greatly with temperature
Gas volume is a function of the amount of gas
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Cylinders of Gas
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The Empirical Gas Laws
 Gas behavior can be described by pressure,
temperature, volume, and molar amount
 Holding any two constant allows relations between
the other two
 Boyle’s Law : The volume of a sample of gas at a
given temperature varies inversely with the applied
pressure
Vα 1/P
PV= constant
P1V41=P2V2
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The Empirical Gas Laws
Boyle’s Law
Gas Pressure-Volume Relationship
5
Practice Problem
Boyle’s Law
A sample of chlorine gas has a volume of 1.8 L at
1.0 atm. If the pressure increases to 4.0 atm (at
constant temperature), what would be the
new volume? using
P1V1=P2V2
V2 = P1V1/P2
atm) × (1.8 L) / (4.0 atm)
V2= 0.45 L
= (1.0
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The Empirical Gas Laws
Charles’s Law
The volume occupied by any sample of gas at
constant pressure is directly proportional to its
absolute temperature.
Vα T abs
(Tabs (K) = oC + 273.15)
Assumes constant moles and pressure
Temperature on absolute scale (oC + 273.15)
V/T = constant
V1/T1 = V2/T2
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The Empirical Gas Laws
Charles’s Law: Linear Relationship of Gas Volume and
Temperature at Constant Pressure
V = at + b
0 = a(-273.15) + b
b = 273.15b
V = at + 273.15a = a(t + 273.15)
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Practice Problem
Charles’s Law
A sample of methane gas that has a volume of 3.8
L at 5.0°C is heated to 86.0°C at constant
pressure. Calculate its new volume.
using V1/T1 = V2/T2
Convert temperature (°C ) to absolute (kelvin)
50 °C = 5.0 +273.15 = 278.15 K
86.0 °C = 86.0 + 273.15 = 359.15 K
V2 = (T2/T1) V1 = (359.15 K /278.15 K ) × 3.8 L
V2 = 4.9 L
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The Combined Gas Law
Combined Gas Law: In the event that all three parameters, P, V, and T,
are changing, their combined relationship is defined as follows
assuming the mass of the gas (number of moles) is constant.
Boyle’s Law
P1V1 = P2V2
V x P = const
(fixed T,n)
1662
Charles’ Law
V / T = const
V1 / V2 = T1 / T2
(fixed P,n)
1787
P1V1/T1 = P2V2/T2
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Avagadro’s Law
Avogadro’s Law
 The volume of a sample of gas is directly
proportional to the number of moles of gas, n
Vα n
V/n = constant
 Equal volumes of any two gases at the same
temperature and pressure contain the same
number of molecules (mol)
V1/n1 = V2/n2
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Avagadro’s Law
Avogadro’s Law
 The volume of one mole of gas is called the:
molar gas volume, Vm
 Volumes of gases are often compared at standard
temperature and pressure (STP), chosen to be
 0 oC (273.15 oK) and 1 atm pressure
 At STP, the molar volume, Vm, that is, the volume
occupied by one mole of any gas, is 22.4 L/mol
VSTP/NSTP = Vm = 22.4 L ( at STP )
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The Ideal Gas Law
From the empirical gas laws, we see that volume varies
in proportion to pressure, absolute temperature,
and the mass of gas (moles) present
Boyle law
Vα 1/P
V= constant x 1/P
Charles law Vα T abs
V = constant x T
Avogadros law Vα n
V = constant x n
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The Ideal Gas Law
This implies that there must exist a proportionality
constant governing these relationships
Combining the three proportionalities, we can obtain
the following relationship.
V m = “R” x n x (T abs/P)
where “R” is the proportionality constant referred to
as the Ideal Gas Constant, which relates Molar
Volume (V) to the ratio of Temperature to Pressure
T/P
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The Ideal Gas Law
The ideal gas equation is usually expressed in the
following form:
PV = nRT
P
=
Pressure (in atm)
V = Volume (in liters)
n = Number of atoms (in moles)
R = Universal gas constant 0.0821 L.atm/mol.K
T = Temperature (in 0 Kelvin = °C + 273.15)
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The Ideal Gas Law
PV = nRT
What is R, universal gas constant?
the R is independent of the particular gas studied
PV
(1atm)(22.414L)

R
nT (1.00 mol)(273.15 K)
R  0.082057 L atm mol K
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Practice Problem
A steel tank has a volume of 438 L and is filled
with 0.885 kg of O2.
Calculate the pressure of oxygen in the tank at
21oC use PV = nRT
V = 438 L, R = 0.0821 L.atm/mol.K, T = 21 +
273.15 = 294.15 K , n = 885/32 = 27.7 mol.
So the pressure = nRT/V =
27.7 mol x 0.0821 L.atm/mol.K x 294.15 K/483 L
= 1.53 atm
Mixtures of Gases
Dalton’s Law of Partial Pressures
The total pressure of a mixture of gases equals
the sum of the partial pressures of the
individual gases.
n 3 RT
n1RT
n 2 RT
P1 
, P2 
, P3 
V
V
V
RT
Ptotal  P1  P2  P3  n1  n 2  n 3 
V
 RT 
Ptotal  n Total 

 V 
The Kinetic Molecular Theory of Gases
The Ideal Gas Law is an empirical
relationship based on
experimental observations.
Boyle, Charles and Avogadro.
Kinetic Molecular Theory is a
simple model that attempts to
explain the behavior of gases.
The Kinetic Molecular Theory of Gases
1. A pure gas consists of a large number of identical molecules separated by
distances that are large compared with their size. The volumes of the
individual particles can be assumed to be negligible (zero).
2. The molecules of a gas are constantly moving in random directions with a
distribution of speeds. The collisions of the particles with the walls of the
container are the cause of the pressure exerted by the gas.
3. The molecules of a gas exert no forces on one another except during
collisions, so that between collisions they move in straight lines with constant
velocities. The gases are assumed to neither attract or repel each other. The
collisions of the molecules with each other and with the walls of the
container are elastic; no energy is lost during a collision.
4. The average kinetic energy of a collection of gas particles is assumed to
be directly proportional to the Kelvin temperature of the gas.
Diffusion:
is the transfer of a gas through space or another
gas over time.
Effusion:
Is the process in which individual molecules flow
through a hole without collisions between
molecules.
According to Graham's law, the rate at which
gases effuse (i.e., how many molecules pass
through the hole per second) is dependent on
their molecular weight; gases with a lower
molecular weight effuse more quickly than
gases with a higher molecular weight.
The equation for effusion is given as
Rate of effusion of gas1
M2

Rate of effusion of gas 2
M1
Where M1 and M2 are molecular masses of
gases 1 and 2.
Real Gases:
Deviations from Ideality
Real gases behave ideally at ordinary temperatures and
pressures.
At low temperatures and high pressures real gases do
not behave ideally.
 The reasons for the deviations from ideality are:
1. The molecules are very close to one another, thus
their volume is important.
2. The molecular interactions also become important.
Real Gases:
Deviations from Ideality
 van der Waals’ equation accounts for the behavior of
real gases at low temperatures and high pressures.

n 2a 
V  nb  nRT
P +
2 
V 

 The van der Waals constants a and b take into account two things:
a accounts for intermolecular attraction
For nonpolar gases the attractive forces are London Forces
For polar gases the attractive forces are dipole-dipole attractions or
hydrogen bonds.
b accounts for volume of gas molecules

At large volumes a and b are relatively small and van der Waal’s
equation reduces to ideal gas law at high temperatures and low
pressures.
Real Gases:
Deviations from Ideality
 Calculate the pressure exerted by 84.0 g of ammonia, NH3, in a
5.00 L container at 200. oC using the ideal gas law.
PV = nRT
P = nRT/V n = 84.0g * 1mol/17 g T = 200 + 273
P = (4.94mol)(0.08206 L atm mol K )(473K)
(5 L)
P = 38.3 atm
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Real Gases: Deviations from Ideality
 Calculate the pressure exerted by 84.0 g of ammonia, NH3, in a 5.00 L
container at 200. oC using the van der Waal’s equation. The van der
Waal's constants for ammonia are: a = 4.17 atm L2 mol-2 b =3.71x10-2 L
mol-1

n 2a 
V  nb  nRT
P +
2 
V 

nRT n 2a
P
- 2
V - nb V
n = 84.0g * 1mol/17 g T = 200 + 273
P = (4.94mol)(0.08206 L atm mol-1 K-1)(473K) (4.94 mol)2*4.17 atm L2 mol-2
(5 L)2
5 L – (4.94 mol*3.71E-2 L mol-1)
P = 39.81 atm – 4.07 atm = 35.74
P = 38.3 atm
7% error