Gases

Elements that exist as gases at 250C and 1 atmosphere

5.1
5.1
 Physical Characteristics of Gases
• Gases assume the volume and shape of their containers.
• Gases are the most compressible state of matter.
• Gases will mix evenly and completely when confined to
the same container.
• Gases have much lower densities than liquids and solids.

5.1
 Force
Pressure =
Area
(force = mass x acceleration)

Units of Pressure

1 pascal (Pa) = 1 N/m2

1 atm = 760 mm Hg = 760 torr
= 101,325 Pa
= 14.7 psi
= 29.92 in. Hg
Barometer
5.2
10 miles 0.2 atm

4 miles 0.5 atm

Sea level 1 atm

5.2
 Boyle’s Law
P α 1/V
This means Pressure and
Volume are INVERSELY
PROPORTIONAL if moles
and temperature are
constant (do not change).
For example, P goes up as
V goes down. Robert Boyle
(1627-1691).
P1V1 = P2 V2 Son of Earl of
Cork, Ireland.
Charles’s Law
If n and P are constant,
then V α T
V and T are directly
proportional.
V1 V2
Jacques Charles (1746-
=
1823). Isolated boron
T1 T2 and studied gases.
Balloonist.
• If one temperature goes up, the
volume goes up!
 Gay-Lussac’s Law
If n and V are constant,
then P α T
P and T are directly
proportional.
P1 P2
=
T1 T2

If one temperature goes up,

the pressure goes up!
 Combined Gas Law
• The good news is that you don’t have to
remember all three gas laws!
• Since they are all related to each other, we can
combine them into a single equation.
• BE SURE YOU KNOW THIS EQUATION!
P1 V1 P 2 V2
=
T1 T2
 Avogadro’s Law
V  number of moles (n)
Constant temperature
V = constant x n Constant pressure

V1/n1 = V2/n2

5.3
 Ideal Gas Equation
1
Boyle’s law: V (at constant n and T)
P
Charles’ law: V  T(at constant n and P)
Avogadro’s law: V n(at constant P and T)

nT
V
P
nT nT
V = constant x =R R is the gas constant
P P

PV = nRT

5.4
 The conditions 0 0C and 1 atm are called standard
temperature and pressure (STP).
Experiments show that at STP, 1 mole of an ideal
gas occupies 22.414 L.

PV = nRT
PV (1 atm)(22.414L)
R= =
nT (1 mol)(273.15 K)

R = 0.082057 L • atm / (mol • K)

5.4
Density (d) Calculations

m PM m is the mass of the gas in g

d= =
V RT M is the molar mass of the gas

Or,
Molar Mass (M ) of a Gaseous Substance

dRT
M= d is the density of the gas in g/L
P
5.4
 A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm
and 27.00C. What is the molar mass of the gas?

PV = nRT

2.10 L X 1 atm
n=
L•atm
0.0821 mol•K x 300.15 K

n = 0.0852 mol

g M = 54.6 g/mol
n=
M
5.3
 Gas Stoichiometry

What is the volume of CO2 produced at 370 C and 1.00

atm when 5.60 g of glucose are used up in the reaction:
C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)

g C6H12O6 mol C6H12O6 mol CO2 V CO2

1 mol C6H12O6 6 mol CO2

5.60 g C6H12O6 x x = 0.187 mol CO2
180 g C6H12O6 1 mol C6H12O6

L•atm
0.187 mol x 0.0821 x 310.15 K
nRT mol•K
V= = = 4.76 L
P 1.00 atm
5.5
 Dalton’s Law of Partial Pressures

V and T
are
constant

P1 P2 Ptotal = P1 + P2
5.6
Consider a case in which two gases, A and B, are in a
container of volume V.

nART
PA = nA is the number of moles of A
V
nBRT nB is the number of moles of B
PB =
V
nA nB
PT = PA + PB XA = XB =
nA + nB nA + nB

PA = XA PT PB = XB PT

ni
Pi = Xi PT mole fraction (Xi) =
nT
5.6
 A sample of natural gas contains 8.24 moles of CH4,
0.421 moles of C2H6, and 0.116 moles of C3H8. If the
total pressure of the gases is 1.37 atm, what is the
partial pressure of propane (C3H8)?

Pi = Xi PT PT = 1.37 atm

0.116
Xpropane = = 0.0132
8.24 + 0.421 + 0.116

Ppropane = 0.0132 x 1.37 atm = 0.0181 atm

5.6
 Bottle full of oxygen
gas and water vapor

2KClO3 (s) 2KCl (s) + 3O2 (g)

PT = PO2 + PH 2O
5.6
5.6
 Chemistry in Action:
Scuba Diving and the Gas Laws
Depth (ft) Pressure
(atm)
0 1

33 2

66 3

P V

5.6
 Kinetic Molecular Theory of Gases
1. A gas is composed of molecules that are separated from
each other by distances far greater than their own
dimensions. The molecules can be considered to be points;
that is, they possess mass but have negligible volume.
2. Gas molecules are in constant motion in random directions.
Collisions among molecules are perfectly elastic.
3. Gas molecules exert neither attractive nor repulsive forces
on one another.
4. The average kinetic energy of the molecules is proportional
to the temperature of the gas in kelvins. Any two gases at
the same temperature will have the same average kinetic
energy

5.7
 Kinetic theory of gases and …
• Compressibility of Gases
• Boyle’s Law
P  collision rate with wall
Collision rate  number density
Number density  1/V
P  1/V
• Charles’ Law
P  collision rate with wall
Collision rate  average kinetic energy of gas molecules
Average kinetic energy  T
PT
5.7
 Kinetic theory of gases and …
• Avogadro’s Law
P  collision rate with wall
Collision rate  number density
Number density  n
Pn
• Dalton’s Law of Partial Pressures
Molecules do not attract or repel one another
P exerted by one type of molecule is unaffected by the
presence of another gas
Ptotal = Pi

5.7
 Deviations from Ideal Behavior

1 mole of ideal gas

Repulsive Forces
PV = nRT
PV = 1.0
n=
RT
Attractive Forces

5.8
Effect of intermolecular forces on the pressure exerted by a gas.

5.8
Van der Waals equation
nonideal gas

an 2
( P + V2 )(V – nb) = nRT
}

corrected corrected
pressure volume

5.8
 Velocity of a Gas

The distribution of speeds

of three different gases
at the same temperature

The distribution of speeds

for nitrogen gas molecules
at three different temperatures

urms =  3RT
M
5.7
Gas diffusion is the gradual mixing of molecules of one gas
with molecules of another by virtue of their kinetic properties.

NH4Cl

NH3 HCl
17 g/mol 36 g/mol
5.7
 GAS DIFFUSION AND
EFFUSION
• diffusion is the gradual • effusion is the
mixing of molecules movement of molecules
through a small hole
of different gases. into an empty container.
 GAS DIFFUSION AND
EFFUSION
Graham’s law governs
effusion and diffusion of
gas molecules. KE=1/2 mv2

Rate for A M of B
Rate for B M of A

Rate
Rate ofof effusion
effusion is
is
inversely
inversely proportional
proportional Thomas Graham, 1805-1869.
Professor in Glasgow and London.
to
to its
its molar
molar mass.
mass.
 GAS DIFFUSION AND
EFFUSION
Molecules effuse thru holes in a rubber
balloon, for example, at a rate (=
moles/time) that is
• proportional to T He
• inversely proportional to M.
Therefore, He effuses more rapidly
than O2 at same T.
 Gas
Gas Diffusion
Diffusion
relation
relation of
of mass
mass to
to rate
rate of
of diffusion
diffusion
•• HCl
HCland
andNH
NH33diffuse
diffuse
from
fromopposite
oppositeends
endsof
of
tube.
tube.
•• Gases
Gasesmeet
meetto
toform
form
NH
NH44Cl
Cl
•• HCl
HClheavier
heavierthan
thanNH
NH33
•• Therefore,
Therefore,NH
NH44Cl
Cl
forms
formscloser
closerto
toHCl
HCl
end
endof
oftube.
tube.
 Graham’s Law Problem 1
1 mole of oxygen gas and 2 moles of ammonia are
placed in a container and allowed to react at 850
degrees celsius according to the equation:

4 NH3(g) + 5 O2(g) --> 4 NO(g) + 6 H2O(g)

Using Graham's Law, what is the ratio of the

effusion rates of NH3(g) to O2(g)?
 Graham’s Law Problem 2
What is the rate of effusion for H2 if 15.00 ml
of CO2 takes 4.55 sec to effuse out of a
container?
 Graham’s Law Problem 3
What is the molar mass of gas X if it effuses
0.876 times as rapidly as N2(g)?