Sit down and get
ready for the
test, I will hand
them out when
the bell rings. You
DMA 3/21/11
have until 8:10
You can have a few
minutes to finish
your test, if you are
already done, work
on something quietly
DMA 3/21/11
When you are done with your test,
begin the following:
 In your book, read (it is important that you
This is
homework
if you don’t
get it done
today
actually read this) Ch 13-1 (pg. 417-423) and
take notes which include the following info:
 The 6 physical properties that are common to
all gases
 What it means when the terms “atom”
“diatomic molecule” or “polyatomic molecule”
are used when taking about gases
 The 6 postulates of the Kinetic-molecular
theory of gases
 Answer question #5 on pg. 423
DMA 3/22/11
Name 4 physical
properties of gases
The Nature of Gases-physical properties
 Gases expand to fill their containers
 Gases are fluid – they flow
 Gases have low density
 1/1000 the density of the equivalent liquid or
solid
 Gases are compressible
 Gases effuse and diffuse
Kinetic Molecular Theory
 Gases are made of particles, which have mass
 Particles of matter are ALWAYS in motion
 Volume of individual particles is  zero.
 Collisions of particles with container walls
cause pressure exerted by gas.
 Particles exert no forces on each other.
 Average kinetic energy proportional to
temperature in Kelvin of a gas.
Measuring Gases
 When you are measuring a gas, you
want to know four things about it:
 1. n --Amount of the gas-this means in
moles represented by the letter n
 2. V—volume
 3. T—temperature
 4. P—pressure
Temperature
 When using the gas laws, temperature
is measured in Kelvin
 To find Kelvin:
 Add 273 to Celcius.
 T(K)=T(oC) + 273
Converting Celsius to Kelvin
Gas law problems involving temperature require
that the temperature be in KELVINS!
Kelvins = C + 273
°C = Kelvins - 273
Celcius & Kelvin
Temperature Scales
Converting Temperature

Convert the following temperatures to
the Kelvin scale.




20 C
85 C
–15 C
–190 C
An Early Barometer
Atmospheric pressure– the pressure
exerted by the air in the atmosphere.
Barometer-measures air pressure
The normal pressure due to
the atmosphere at sea level
can support a column of
mercury that is 760 mm high.
Mercury Barometer
Pressure
force
pressure 
area
Which shoes create the most pressure?
Standard Temperature and Pressure
“STP”
P = 1 atmosphere (760 torr)
T = 0C (273 Kelvins)
The molar volume of an ideal gas is
22.42 liters at STP
Pressure
 Is caused by the collisions of molecules
with the walls of a container
 is equal to force/unit area
 SI units = Newton/meter2 = 1 Pascal (Pa)
 1 standard atmosphere = 101,325 Pa=
101.3 kPa
 1 standard atmosphere = 1 atm =
760 mm Hg = 760 torr
Converting Pressure Units
 Change 5 atmospheres (atm) into
millimeters mercury (mm Hg).
 1 atm = 760 mm Hg
5 atm x 760 mm Hg
1 atm
=
3800 mm Hg
Converting Pressure Units
 Change 1900 mm mercury (mm Hg) into
atmospheres (atm).
 1 atm = 760 mm Hg
1900 mm Hg x 1 atm
760 mm Hg
=
2.5 atm
Converting Pressure Units
 Change 560 kilopascals into millimeters
mercury (mm Hg).
 101.3 kPa = 760 mm Hg
560 kPa x 760 mm Hg =
101.3 kPa
4201 mm Hg
Converting Pressure Units
 Change 1013 kilopascals into millimeters
mercury (mm Hg).
 1 atm = 101.3 kPa
1013 atm x
1 atm
101.3 kPa
=
10 atm
Boyle’s Law*
Pressure is inversely proportional to volume
when temperature is held constant.
Pressure  Volume = Constant
P1V1 = P2V2
(T = constant)
A Graph of Boyle’s Law
 If pressure increases
then volume
decreases.
 If pressure decreases,
then volume
increases.
 Inversely proportional.
Charles’s Law
The volume of a gas is directly proportional to
temperature, and extrapolates to zero at zero
Kelvin. (P = constant)
V1
V2

T1
T2
( P  constant)
Practice Problem
 A gas at constant temperature occupies a
volume of 2.40 L and exerts a pressure of 710
mm Hg What volume will the gas occupy at a
pressure of 75 mm Hg?
 P1 = 710 mm Hg, V1 = 2.40 L, P2 = 75
mm Hg
 V2 = ?
 P1V1 = P2V2 (Boyle’s Law)
 710 mm Hg x 2.40 L = 75 mm Hg x V2
 V2 = 22.7 L
 The new volume will be 22.7 liters.
Practice Problem







Sulfur dioxide gas at a pressure of 3.4 atm and
a volume of 14 liters increases pressure to 5
atm. What is the resulting volume?
P1 = 3.4 atm, V1 = 14 L, P2 = 5 atm
V2 = ?
P1V1 = P2V2 (Boyle’s Law)
3.4 atm x 14 L = 5 atm x V2
V2 = 9.5 L
The new volume will be 9.5 liters.
Practice Problem

Carbon monoxide at a pressure of 11.8
kPa in a vessel of 420 ml expands to a
new volume of 630 ml. What is the new
pressure in kPa?

P1 = 11.8 kPa, V1 = 420 ml, V2 = 630
ml





P2 = ?
P1V1 = P2V2 (Boyle’s Law)
11.8 kPa x 420 ml = P2 x 630 ml
P2 = 7.9 kPa
The new pressure is 7.9 kilopascals.
Practice Problem

Given 90 ml of H2 gas collected when the
temperature is 27 C, how many ml will H2 occupy
at 42 C?




V1 = 90 ml, T1 = 27 C, T2 = 42 C, V2 = ?
T1 = 27 C + 273 = 300 K
T2 = 42 C + 273 = 315 K
V1 = V2 (Charles’ Law)
T1

T2
90 ml = V2
300 K

315 K
V2 = 94.5 ml
V2 = 90 ml x 315 K
300 K
Why Don’t I Get a Constant Value for
PV = k?
1. Air is not made
of ideal gases
2. Real gases
deviate from
ideal behavior
at high pressure
What you need to do
DMA 3.24.11
Using the conversions you have in
your notes, convert the following
pressure units:
1. 20 mm Hg = ? atm
2. 3 atm = ? kPa
3. 3 atm = ? mm Hg
4. 500 mm Hg = ? kPa
5. 200 kPa = ? atm
DMA 3.25.11
What is the kinetic
molecular theory?
Gay Lussac’s Law
The pressure and temperature of a gas are
directly related, provided that the volume
remains constant.
P1 P2

T1 T2
The Combined Gas Law
The combined gas law expresses the relationship
between pressure, volume and temperature of a
fixed amount of gas.
P1V1 P2V2

T1
T2
Why would you need this law when you
already have all the others? What is
different here?
Combined Gas Law
If you should only need one of the other gas
laws, you can cover up the item that is
constant and you will get that gas law!
P1 V1 =
T1
P2 V2
T2
Boyle’s Law
Charles’ Law
Gay-Lussac’s
Law
Combined Gas Law Problem
A sample of helium gas has a volume of 0.180 L, a pressure of
0.800 atm and a temperature of 29°C.
What is the
new temperature(°C) of the gas at a
volume of
90.0 mL and a pressure of 3.20atm?
 What do you need to do first?
Set up Data Table
P1 = 0.800 atm
V1 = 180 mL
T1 = 302 K
P2 = 3.20 atm
V2= 90 mL
T2 = ??
Calculation
P1 = 0.800 atm
P2 = 3.20 atm
V1 = 180 mL
V2= 90 mL
P1 V1
T1 = 302 K
T2 = ??
P2 V2
=
T1
P1 V1 T2 = P2 V2 T1
T2
T2 = P2 V2 T1
P1 V1
T2 = 3.20 atm x 90.0 mL x 302 K
0.800 atm x 180.0 mL
T2 = 604 K - 273 = 331 °C
= 604 K
Learning Check
A gas has a volume of 675 mL at 35°C and 0.850 atm
pressure. What is the temperature in °C when the gas
has a volume of 0.315 L and a pressure of 802 mm Hg?
One More Practice Problem
A balloon has a volume of 785 mL on
a fall day when the temperature is
21°C. In the winter, the gas cools to
0°C. What is the new volume of the
balloon?
To Do Today
 Finish Boyle’s Law and Charles’ Law
homework-due today!
 Combined gas law practice problems-due
Monday
 In book-review Ch 13 to help your
understanding 
DMA 3/28/11
 A balloon has a volume of 785 mL
on a fall day when the temperature
is 21°C. In the winter, the gas cools
to 0°C. What is the new volume of
the balloon?
Dalton’s Law
John Dalton
1766-1844
Dalton’s Law of Partial Pressures
For a mixture of gases in a container,
PTotal = P1 + P2 + P3 + . . .
This is particularly useful in calculating the pressure
of gases collected over water.
Dalton’s Law of Partial
Pressures
2 H2O2 (l) ---> 2 H2O (g) + O2 (g)
0.32 atm
0.16 atm
What is the total pressure in the flask?
Ptotal in gas mixture = PA + PB + ...
Therefore,
Ptotal = PH2O + PO2 = 0.48 atm
Dalton’s Law: total P is sum of PARTIAL
pressures.
Avogadro’s Hypothesis
Equal volumes of gases at the same T and P
have the same number of molecules.
V = n (RT/P) = kn
V and n are directly related.
twice as many
molecules
Avogadro’s Hypothesis and
Kinetic Molecular Theory
The gases in this
experiment are
all measured at
the same T and
V.
P proportional to n
IDEAL GAS LAW
PV=nRT
Brings together gas
properties.
Can be derived from
experiment and theory.
BE SURE YOU KNOW
THIS EQUATION!
Using PV = nRT
P = Pressure
V = Volume
T = Temperature
L • atm
R = 0.0821
Mol • K
n = number of moles
R is a constant, called the Ideal Gas Constant
Instead of learning a different value for R for all the
possible unit combinations, we can just memorize one
value and convert the units to match R.
Using PV = nRT
How much N2 is required to fill a small room with a
volume of 960 cubic feet (27,000 L) to 745 mm Hg at
25 oC?
Solution
1. Get all data into proper units
V = 27,000 L
T = 25 oC + 273 = 298 K
P = 745 mm Hg (1 atm/760 mm Hg)
= 0.98 atm
And we always know R, 0.0821 L atm / mol K
Using PV = nRT
How much N2 is req’d to fill a small room with a volume of 960
cubic feet (27,000 L) to P = 745 mm Hg at 25 oC?
Solution
2. Now plug in those values and solve for the
unknown.
PV = nRT
RT
RT
4
(0.98 atm)(2.7 x 10 L)
n =
(0.0821 L• atm/K • mol)(298 K)
n = 1.1 x 103 mol (or about 30 kg of gas)
Learning Check
Dinitrogen monoxide (N2O), laughing gas,
is used by dentists as an anesthetic. If
2.86 mol of gas occupies a 20.0 L tank at
23°C, what is the pressure (mm Hg) in
the tank in the dentist office?
Try This One
A sample of neon gas used in a neon sign has a
volume of 15 L at STP. What is the volume (L) of
the neon gas at 2.0 atm and –25°C?
 A sample of neon gas used in a
neon sign has a volume of 15 L at
STP. What is the volume (L) of the
neon gas at 2.0 atm and –25°C?
The first thing you need to do is
decide which law to use!
At STP
 At STP determining the amount of gas required
or produced is easy.
22.4 L = 1 mole
For example
What is the volume at STP of 4.00 g of CH4?
How many grams of He are present in 8.0 L
of gas at STP?
Gas Stoichiometry: Practice!
How many liters of O2 at STP are required
to produce 20.3 g of H2O?
Gases and Stoichiometry
2 H2O2 (l) ---> 2 H2O (g) + O2 (g)
Decompose 1.1 g of H2O2 in a flask with
a volume of 2.50 L. What is the volume
of O2 at STP?
Bombardier beetle
uses decomposition of
hydrogen peroxide to
defend itself.
Gases and Stoichiometry
2 H2O2 (l) ---> 2 H2O (g) + O2 (g)
Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L.
What is the volume of O2 at STP?
Solution
1.1 g H2O2
1 mol H2O2
34 g H2O2
1 mol O2
22.4 L O2
2 mol H2O2
1 mol O2
= 0.36 L O2 at STP
Not At STP
 Chemical reactions happen in MOLES.
 If you know how much gas - change it
to moles
 Use the Ideal Gas Law n = PV/RT
 If you want to find how much gas - use
moles to figure out volume
V = nRT/P
Example #1
 HCl(g) can be formed by the following reaction
 2NaCl(aq) + H2SO4 (aq)2HCl(g) + Na2SO4(aq)
 What mass of NaCl is needed to produce 340 mL of
HCl at 1.51 atm at 20ºC?
Example #2

2NaCl(aq) + H2SO4 (aq)
2HCl(g) + Na2SO4
(aq)

What volume of HCl gas at 25ºC and 715
mm Hg will be generated if 10.2 g of NaCl
react?
GAS DIFFUSION AND
EFFUSION
 diffusion is the gradual  effusion is the
mixing of molecules of
different gases.
movement of
molecules through a
small hole into an
empty container.
Diffusion
Diffusion: describes the
mixing of gases. The
rate of diffusion is the
rate of gas mixing.
Effusion
Effusion: describes the passage of gas into an
evacuated chamber.