Chapter 5
The Gaseous State
An Overview of the Physical States of Matter
The Distinction of Gases from Liquids and Solids
1. Gas volume changes greatly with pressure.
2. Gas volume changes greatly with temperature.
3. Gases have relatively low viscosity.
4. Most gases have relatively low densities under normal conditions.
5. Gases are miscible.
The Pressure of a Gas
Gas pressure is a result of
the constant movement of
the gas molecules and their
collisions with the surfaces
around them.
The pressure of a gas depends on
1) number of gas particles
2) volume of the container
3) speed of the gas particles
A Mercury Barometer
We measure air pressure
with a barometer.
Column of mercury
supported by air
pressure
Force of the air on the
surface of the mercury
counter balances the
force of gravity on the
column of mercury.
gravity
Pressure explains how you
can drink from a straw.
a) When a straw is put
into a glass of orange
soda, the pressure inside
and outside of the straw
are the same, so the liquid
levels inside and outside
of the straw are the same.
b) When a person sucks
on the straw, the pressure
inside the straw is
lowered. The greater
pressure on the surface of
the liquid outside of the
straw pushes the liquid up
the straw.
A Drinking Straw’s
Limitations.
Even if you formed a
perfect vacuum,
atmospheric pressure
could only push orange
soda to a total height of
about 10 m. This is
because a column of
water 10 m high exerts
the same pressure (14.7
lb/in2) as the gas
molecules in our
atmosphere.
Common Units of Pressure
1 atm = 14.7 psi, 1 atm = 760 mmHg
psi
atm
mmHg
1 atm = 14.7 psi, 1 atm = 101.325 kPa
psi
atm
kPa
The Open-End Manometer
The pressure of a gas trapped in a container can be measured with an
instrument called a manometer.
A competition is established between the pressures of the atmosphere
and the gas.
The difference in the liquid levels is a measure of the difference in
pressure between the gas and the atmosphere.
Boyle’s Law
Pressure of a gas is inversely proportional to its
volume (constant T and amount of gas).As P
increases, V decreases by the same factor.
Boyle’s Law
Robert Boyle (1627–1691)
Pressure of a gas is inversely proportional to its
volume (constant T and amount of gas)
As P increases, V decreases by the same factor.
V ∝ 1/P
P x V = constant
V = k/P
P1 x V1 = P 2 x V2
Boyle’s Experiment
Boyle’s Experiment, P x V
Boyle’s Law
V ∝ 1/P
P x V = constant
V = k/P
P1 x V1 = P 2 x V2
A cylinder with a movable piston has a volume of
7.25 L at 4.52 atm. What is the volume at 1.21 atm?
V1 =7.25 L, P1 = 4.52 atm, P2 = 1.21 atm,
V1, P1, P2
P1 · V1 = P2 · V2
V2
V2= ?
A balloon is put in a bell jar and the pressure is reduced from
782 torr to 0.500 atm. If the volume of the balloon is now
2.78 x 103 mL, what was it originally?
V2 =2780 mL, P1 = 762 torr, P2 = 0.500 atm
V2, P1, P2
V1 = ?
V1
P1 · V1 = P2 · V2 , 1 atm = 760 torr (exactly)
A balloon is put in a bell jar and the pressure is reduced from
782 torr to 0.500 atm. If the volume of the balloon is now
2.78 x 103 mL, what was it originally?
D = m/V
Boyle’s Law: A Molecular View
Pressure is caused by the molecules striking the sides of the container.
When you decrease the volume of the container with the same
number of molecules in the container, more molecules will hit the
wall at the same instant.
This results in increasing the pressure
Charles’s Law
At constant n and P, the volume of a gas increases
proportionately as its absolute temperature increases. If the
absolute temperature is doubled, the volume is doubled.
Charles’s Law
Jacques Charles (1746–1823)
Volume is directly proportional to temperature
(constant P and amount of gas)
As T increases, V also increases
Kelvin T = Celsius T + 273
V ∝ T
V = kT
Charles’s Law
V ∝ T
V = kT
If you plot volume vs.
temperature for any gas
at constant pressure, the
points will all fall on a
straight line.
If the lines are
extrapolated back to a
volume of “0,” they all
show the same
temperature, −273.15 °C,
called absolute zero.
Charles’s Law – A Molecular View
The pressure of gas
inside and outside the
balloon are the same.
At low temperatures, the
gas molecules are not
moving as fast, so they
don’t hit the sides of the
balloon as hard.
Therefore the volume is
small.
Charles’s Law – A Molecular View
The pressure of gas
inside and outside the
balloon are the same.
At high temperatures,
the gas molecules are
moving faster, so they
hit the sides of the
balloon harder –
causing the volume to
become larger.
A gas has a volume of 2.57 L at 0.00 °C. What was
the temperature when the volume was 2.80 L?
T1 = ?
V2 =2.57 L, T2 = 0.00 °C, V1 = 2.80 L
V1, V2, T2
T1
(2.80 L)
(2.57 L)
The temperature inside a balloon is raised from 25.0 °C
to 250.0 °C. If the volume of cold air was 10.0 L, what is the
volume of hot air?
V1 =10.0 L, t1 = 25.0 °C L, t2 = 250.0 °C
V1, T1, T2
V2
V2 = ?
Avogadro’s Law
At constant T and P, the volume of a gas increases
proportionately as its molar amount increases. If the
molar amount is doubled, the volume is doubled.
Avogadro’s Law
Amedeo Avogadro (1776–1856)
Volume is directly proportional to the number of
gas molecules (constant P and T)
More gas molecules = larger volume
Equal volumes of gases contain equal numbers of
molecules.
The gas doesn’t matter.
Avogadro’s Law
V ∝ n
V = kn
If 1.00 mole of a gas occupies 22.4 L and the temperature
and pressure are not changed, what volume would
0.750 moles occupy?
V1 =22.4 L, n1 = 1.00 mol, n2 = 0.750 mol
V1, n1, n2
V2
V2 = ?
Amonton’s Law
Guillaume Amontons (1663–1705)
Pressure is directly proportional to the
absolute temperature of gas molecules
(constant n and V)
Higher temperature = Higher pressure
Amonton’s Law
P ∝ T
P = kT
P1
=
T1
P2
T2
Boyle’s Law
V ∝ 1/P
P x V = constant
V = k/P
P1 x V1 = P 2 x V2
Charles’s Law
V ∝ T
V = kT
Avogadro’s Law
V ∝ n
V = kn
Amonton’s Law
P ∝ T
P = kT
P1
=
T1
P2
T2
1
V ∝P
V∝T
nT
V∝ P
PV
k= nT = R =
V∝n
nT
V =(k)
P
(1.00 atm)(22.414 L)
(1.00 mole)(273.15 K)
R = 0.0821(atm)(L)/(mol)(K)
Ideal Gas Law
By combining the gas laws we can write a general equation:
R is called the gas constant.
The value of R depends on the units of P and V.
We will use 0.0821 and convert P to atm and V to L.
The other gas laws are found in the ideal gas law if two
variables are kept constant.
Boyle’s Law
Charles’s Law
PV = k
V/T = k
n,R,T constant
n,R,P constant
Ideal
Gas Law
PV = nRT
n,R,V constant
T,R,P constant
Amonton’s Law
Avogadro’s Law
P/T = k
V/n = k
Ideal
Gas Law
PV = nRT
n and R constant
Combined Gas Law
PV/T = k
P1V1 P2V2
=
T1
T2
Standard Conditions
Because the volume of a gas varies with
pressure and temperature, chemists
have agreed on a set of conditions to
report our measurements so that
comparison is easy – we call these
standard conditions - STP
Standard pressure = 1 atm
Standard temperature = 273 K, 0°C
A gas occupies 10.0 L at 44.1 psi and 27 °C. What
volume will it occupy at standard conditions?
V1 = 10.0L, P1 = 44.1 psi, t1 = 27 °C, P2 = 1.00 atm, t2 = 0 °C,
P1, V1, T1, R
n
P2, n, T2, R
V2 = ?
V2
A gas occupies 10.0 L at 44.1 psi and 27 °C. What
volume will it occupy at standard conditions?
P1, V1, T1, R
n =1.219 mol
n
P2, n, T2, R
V2
A gas occupies 10.0 L at 44.1 psi and 27 °C. What
volume will it occupy at standard conditions?
P1 = 3.00 atm
V1 = 10.0 L
T1 = 300 K
P2 = 1.00 atm
T2 = 273 K
V2 = ?
P1V1 P2V2
=
T1
T2
P1V1T2
V2 =
T1P2
(3.00 atm)(10.1L)(273 K)
V2 =
(300 K)(1.00 atm)
V2 =
27.3 L
Calculate the volume occupied by 637 g of SO2 (MM 64.07) at
6.08 x 104 mmHg and –23 °C.
mSO2 = 637 g, P = 6.08 x 104 mmHg, t = −23 °C,
g
n
K = 273+ºC=273+(-23)=250
P, n, T, R
V=?
V
Molar Volume
Solving the ideal gas equation for the volume
of 1 mol of gas at STP gives 22.4 L
for 6.022 x 1023 molecules of any gas
We call the volume of 1 mole of gas at STP the
molar volume.
It is important to recognize that one mole
measures of different gases have different
masses, even though they have the same volume.
Molar Volume
How many liters of O2 @ STP can be made from the
decomposition of 100.0 g of PbO2?
2 PbO2(s) → 2 PbO(s) + O2(g)
(PbO2 = 239.2, O2 = 32.00)
100.0 g PbO2,
g PbO2
2 PbO2 → 2 PbO + O2,
mol PbO2
mol O2
L=?
L O2