Transcript Gases

QUICK QUIZ
Clear your desk except for a pen/cil (& eraser?)
YOU GOT THIS
IMPORTANCE OF GASES ALL AROUND
N2 fills airbags in car accidents
Pressure cookers help us cook our meals
Atmospheric pressure influences the weather
Air fills our tires
CO2 puts fizz in our soda pop
HOW IS GAS UNIQUE?
Large amount of space between gas particles
Can expand indefinitely
 Fill any container completely
Diffuse and mix rapidly
Solid
Liquid
Gas
PUMP-MAKERS’ CONUNDRUM
Pump Makers for the Grand Dukes of Tuscany wanted to create a suction
pump that would work up to 12 meters
 Successful to about 10 meters (34 feet).
As the story goes, Torricelli (old dead Italian) replaced the top of the pipe
with glass
 Saw that water column had a maximum height
 Maximum height varied predictably with weather
TORRICELLI’S EXPERIMENT
Filled
Inverted
tube
with
Tube
Hg
2.5 ft (760 mm)
Torricelli decided to study the behavior of denser
liquids in similar columns
Hg Sank
in tube
due to
gravity
 Denser liquids = more compact & easier to study
 Mercury 13.6 times denser than water – should have
maximum height 13.6 times lower.
 It was!
 Water max ~34 feet; Hg max ~2.5 feet
 2.5 feet ≈ 760 mm
Filled
reservoir
with Hg
Hg
BAROMETER
On average, Hg settled at 760 mm height
 Balance between gravity (pulling down) &
air pressure (pushing up)
785 mm
760 mm
730 mm
Less air pressure = Hg not pushed as high in tube
More air pressure = Hg pushed higher in tube
Hg
STANDARD PRESSURE
Average atmospheric pressure at sea-level
 Pushes Hg to 760 mm height, unit is
Other equivalent units:
101.3 kPa
(kilo Pascals)
760 mmHg
(millimeters of mercury)
PRESSURE UNIT CONVERSIONS
Write the known quantity with unit
Multiply by conversion factor
 Unit you WANT on top; unit you HAVE on bottom
Multiply across and divide
 Units will cancel and WANTED unit remains
Known
Quantity
x Want
Have
= Unit you
Want
PRACTICE PROBLEMS?
What is 475 mmHg expressed in atm?
475 mmHg
x 1 atm
= 0.0625 atm
760 mmHg
Todd’s bicycle tires are filled to 125.0 psi. What is the equivalent pressure
in atm?
125.0 psi
x 1 atm
14.7 psi
= 8.50 atm
VARIABLES THAT AFFECT PRESSURE
P = Pressure (mmHg, atm, kPa, torr, psi, etc.)
V = Volume of a Gas (L or mL)
 Space gas takes up in a container – Size of container
T = Temperature (°C or K)
 You measure in Celsius, but will always end up converting to Kelvin
 How fast particles move
n = number of particles (moles)
PRESSURE VS TEMPERATURE GRAPH
(CONSTANT n & V)
Pressure (psi)
Pressure Relationship to Temperature (C)
18,0
16,0
14,0
12,0
10,0
8,0
6,0
4,0
2,0
0,0
0,0
10,0
0 °C =
Standard Temperature
20,0
30,0
40,0
Temperature (C)
50,0
60,0
PRESSURE VS TEMPERATURE GRAPH
(CONSTANT N & V)
Pressure Relationship to Temperature (C) Long
Trendline
15,0
10,0
-273 °C
Pressure (psi)
20,0
5,0
0,0
-300,0
-200,0
-100,0
Temperature (C)
Negative Temperatures on Celcius Scale
makes trend difficult to describe with math
0,0
100,0
Solution: Add 273 to the
whole temperature scale…
PRESSURE VS TEMPERATURE GRAPH
(CONSTANT N & V)
Pressure (psi)
Pressure Relationship to Temperature in Kelvin
On this scale
If a sample of gas at 100 K
Had a starting pressure of 5 psi…
18,0
16,0
14,0
12,0
10,0
8,0
6,0
4,0
2,0
0,0
…Then the sample increased
…Or
the sample increased
Temperature
(K) is to
to 200 K (DOUBLED)
300 K (TRIPLED)
Directly
Proportional
The pressure wouldThe
increase
pressure would increase
topsiPressure
to 10.0 psi (DOUBLED)
to 15.0
(TRIPLED)
0,0
100,0
200,0
Temperature (K)
-273 °C becomes Zero
degrees Kelvin (K) =
ABSOLUTE ZERO
300,0
400,0
GAY-LUSSAC’S LAW
(CONSTANT V, n)
33 °C – (300 K)
As Temperature increases, Pressure Increases
 As Temperature Doubles, Pressure Doubles
 As Temperature Triples, Pressure Triples
 As Temperature Halves, Pressure Halves
-123 °C – (150 K)
Temperature MUST BE Kelvin
Pressure and Temperature Directly Proportional
PT
 Increased Temperature – Particles hit container walls more
frequently because they are faster
𝑃1 𝑷𝟐
Mathematical Law:
=
𝑇1 𝑻𝟐
4 atm
2
GENERAL FORMAT FOR GAS PROBLEMS
G – Grid:
Organize variables into grid
U – Unknown:
Identify unknown variables
E – Equation:
Write the equation the problem requires
S – Solve:
Solve the equation (no numbers!)
S – Substitute: Substitute numbers into equation
GAY-LUSSAC LAW SAMPLE PROBLEM
Derek left his can of Axe Body Spray next to the fireplace. The gas inside the can
started at 22 °C and 55.5 psi. The temperature raised to 100 °C. What is the
new pressure?
1
2
P
V
T
n
55.5
psi
/
22 °C
295 K
/
/
100 °C
373 K
/
P2
P2T2
T2 P1 = ____
____
T1
T2
P2 =
Want P2 alone.
Multiply both sides of
equation by T2
373 K * 55.5 psi
295 K
Must convert all
temperatures to Kelvin
(+273)

G – Grid:
Grid:
Organize
Organize variables
variables into
into grid
grid

U – Unknown:
Identify unknown variables

E – Equation:
Equation:
Write
Write the
the equation
equation the
the problem requires
requires

S – Solve:
Solve the equation (no numbers!)

Substitute:
S – Substitute:
Substitute numbers
numbers into
into equation
equation
Substitute
CHARLES’ LAW (CONSTANT P, n)
As Temperature increases, Volume Increases
 As Temperature Doubles, Volume Doubles
 As Temperature Triples, Volume Triples
 As Temperature Halves, Volume Halves
Tube
Swimmer
deflates
witha
little
fully inflated
when it
hits cold
tubewater.
SAD HAPPY!
SWIMMER!
Temperature MUST BE Kelvin
Volume and Temperature Directly Proportional
 Increased Temperature – Volume increases so particles hit
walls same amount
VT
𝑉1 𝑽𝟐
Mathematical Law:
=
𝑇1 𝑻𝟐
CHARLES’ LAW SAMPLE PROBLEM
Lance’s bike tires have a volume of 2.0 L at 35 °C. He travels from the desert up to
the top of a mountain pass where the temperature is 14 °C. What is the volume
of the tires at the top of the pass?
1
2
P
V
T
n
/
2.0 L
35 °C
308 K
/
V2
14 °C
287 K
/
/
V2T2
T2 V1 = ____
____
T1
T2
P2 =
Want P2 alone.
Multiply both sides
of equation by T2
287 K * 2.0 L
308 K
Must convert all
temperatures to Kelvin
(+273)

G – Grid:
Grid:
Organize
Organize variables
variables into
into grid
grid

U – Unknown:
Identify unknown variables

E – Equation:
Equation:
Write
Write the
the equation
equation the
the problem requires
requires

S – Solve:
Solve the equation (no numbers!)

Substitute:
S – Substitute:
Substitute numbers
numbers into
into equation
equation
Substitute
BOYLE’S LAW (CONSTANT T, n)
As Volume increases, Pressure Decreases
 As Volume Doubles, Pressure Decreases by Half
 As Volume Triples, Pressure Decreases by Third
 As Volume cuts in half, Pressure Doubles
Pressure Proportional to Inverse of Volume
20 mL
 Pressure and volume Inversely Proportional
 Decreased Volume – Particles hit container walls more
frequently
10 mL
5 mL
P  1/V
Mathematical Law: P1V1 = P2V2
1
4 atm
BOYLE’S LAW SAMPLE PROBLEM
A scuba diver exhales bubbles of CO2 gas that have a volume of 2.0 mL at a pressure
of 3.0 atm. When the bubble reaches the surface, the pressure is 1.0 atm. What
is the volume of the bubble when it reaches the surface?
1
2
P
V
T
3.0
atm
2.0
mL
/
/
1.0 atm
V2
/
/
P____
P2 V2
1V1 = ____
P2
P2
n
V2 =
Want V2 alone.
Divide both sides
of equation by P2
3.0 atm* 2.0 mL
1.0 atm

G – Grid:
Grid:
Organize
Organize variables
variables into
into grid
grid

U – Unknown:
Identify unknown variables

E – Equation:
Equation:
Write
Write the
the equation
equation the
the problem requires
requires

S – Solve:
Solve the equation (no numbers!)

Substitute:
S – Substitute:
Substitute numbers
numbers into
into equation
equation
Substitute
COMBINED GAS LAW (CONSTANT n)
So far we’ve looked at P, V, and T relationships, all at constant n
 Majority of gas law problems - in closed containers, n rarely changes
Sometimes more than one variable changes
 We can combine Gay-Lussac, Boyle, and Charles’ Laws to account for this
𝑃1𝑉1 𝑷𝟐𝑽𝟐
Mathematical Law:
=
𝑇1
𝑻𝟐
Magic of the Combined Gas Law
 Just by knowing this one, you know Gay-Lussac, Boyle, and Charles’ Laws
 If one of the variables is constant, it will cancel out of the equation:
Gay-Lussac
Combined
Charles
Boyle
P1 V1 = ____
P2 V2
____
T1
T2
COMBINED GAS LAW SAMPLE PROBLEM
In a diesel engine, fuel is ignited when it is injected into hot compressed air. In a
typical high-speed diesel engine, the cylinder chamber has a volume of 1.20L, a
temperature of 49 °C, and a pressure of 1.00 atm (equal to atmosphere). After
compression, the cylinder has a volume of 0.0629 L and a pressure of 38.7 atm.
What is the temperature of the compressed gas?
Want T
1
2
P
V
T
n
1.00
atm
1.20
L
49 °C
322 K
/
38.7
atm
0.0629
L
T2
/
2 in
numerator.
Want Tentire
2 alone.
2 Invert
Must
multiply
both
equation
sides of equation
by P2V2
P2____
VP21V1 = P____
2V2P2V
T1
T2
38.7 atm * 0.0629 L * 322 K
T2 =
1.00 atm * 1.20 L
Must convert all
temperatures to Kelvin
(+273)

G – Grid:
Grid:
Organize
Organize variables
variables into
into grid
grid

U – Unknown:
Identify unknown variables

E – Equation:
Equation:
Write
Write the
the equation
equation the
the problem requires
requires

S – Solve:
Solve the equation (no numbers!)

Substitute:
S – Substitute:
Substitute numbers
numbers into
into equation
equation
Substitute
AVOGADRO’S LAW (CONSTANT P, T)
n actually changes!
As number of particles increases, Volume Increases
 As number of particles Doubles, Volume Doubles
 As number of particles Triples, Volume Triples
 As number of particles Halves, Volume Halves
Volume and number of particles Directly Proportional
 Increased number of particles – Volume increases so particles hit walls same
amount
Vn
𝑉1 𝑽𝟐
Mathematical Law:
=
𝑛1 𝒏𝟐
Solved similarly to Gay-Lussac and Charles’ Laws
g  mol
MODIFIED COMBINED GAS LAW
Can modify Combined Gas Law to include n
Modified Combined
Gas Law
P1 V1 = ____
P2 V2
____
T1n1
T2n2
IDEAL GAS LAW
Can find the value of any variable at a single point in time
 Variables don’t change
 Set ratios of variables in combined gas law equal to constant: R
R is always the same: 0.08206
𝑳 ∗𝒂𝒕𝒎
𝑲∗𝒎𝒐𝒍
 Variables’ units must match units of R
 V = L;
P = atm;
T = K;
PV = R
____
Tn
amount = mol (not g)
Since we don’t like fractions, the equation is typically rearranged:
PV = nRT
IDEAL GAS LAW SAMPLE PROBLEM
Air is mostly nitrogen. How much N2 is required to fill a small room (27,000 L) to 745
mmHg at 25 °C?
P
V
745
27,000
mmHg
L
0.98 atm
1
Must
convert all
Pressures
to atm
745
mmHg
x1
atm
760
mmHg
Must
convert all
Volumes
to L
T
n
25 °C
298 K
n
PV
n RT
__ =____
RT
RT
n =
0.98 atm * 27,000L
0.08206 (L*atm/(K*mol))* 298K
Must convert
temperature to
Kelvin (+273)
= 0.98 atm
Want n alone.
Must divide both
sides of equation
by RT

G – Grid:
Grid:
Organize
Organize variables
variables into
into grid
grid

U – Unknown:
Identify unknown variables

E – Equation:
Equation:
Write
Write the
the equation
equation the
the problem requires
requires

S – Solve:
Solve the equation (no numbers!)

Substitute:
S – Substitute:
Substitute numbers
numbers into
into equation
equation
Substitute
REAL GASES
Ideal Gases are not real.
 They are mathematical models or “ideals” for how gases behave
The Gas laws assume that gas molecules have no volume
 Gas molecules have volume
The Gas laws assume that gases don’t interact with each other
 Gases interact
The discrepancies between the Gas Laws and Real life are very small for most gases
 Gas Laws work very well for most situations, so we use them
It’s reasonable to state that 1 mole of
any gas at STP is 22.4 L
This is called the MOLAR VOLUME
DALTON’S LAW OF PARTIAL PRESSURES
Air is a mixture of O2, N2, Ar, negligible other gases
 Atmospheric pressure is due to pressure from all of the gases
 78.1% N2 = 594 mmHg
 21.0% O2 = 159 mmHg
 0.9% Ar = 7 mmHg
Pair
=
PN2
+
PO2
+
PAr
760 mmHg = 594 mmHg + 159 mmHg + 7 mmHg
These pressures of the
individual gases in the mixture
are called Partial Pressures
The total pressure of a mixture of gases is the sum of
the partial pressures of the gases in the mixture:
PTOTAL = P1 + P2 + P3 + …
PARTIAL PRESSURE SAMPLE PROBLEM
A SCUBA dive tank is filled with compressed air. What is the total pressure of the
mixture of gases in the tank?
 Partial pressure of the gases in the tank
 78.1% N2 = 1953 psi
 21.0% O2 = 525 psi
 0.9% Ar = 22 psi
PTANK
PTANK
PTANK
=
PN2
= 1953 psi
= 2500 psi
+
+
PO2
+
PAr
525 psi + 22 psi
COLLECTING A GAS BY DISPLACEMENT
When a tab of Alka-Seltzer reacts with water, CO2 forms and
bubbles out of the water. It is possible to collect this gas by
displacement (through water).
 Sample of CO2 also contains a small amount of H2O
Ptotal=PCO2 + PH2O
So…PCO2 = Ptotal – PH2O
 Total Pressure = Air pressure in room
 Can look up PH2O on a chart
Some H2O
gas gets
into the
sample
PARTIAL PRESSURE OF WATER SAMPLE PROBLEM
A student collects some H2 gas over water at a temperature of 20 °C and an
atmospheric pressure of 768.0 torr. When he looks at the chart, he finds that the
vapor pressure of water at 20 °C is 17.5 torr. What is the pressure of the H2 gas
alone?
PTOTAL
=
PH2
+
PH2O
PTOTAL
-
PH2O
PH2
=
PH2
=
768.0 torr
PH2
=
750.5 torr
- 17.5 torr
DALTON’S LAW WITH VOLUME & MOLES
This law applies to VOLUME of a gas as well.
VTOTAL
VH2
=
=
VH2
=
VH2
VTOTAL
+
-
So, if the total volume is 10 mL and the volume of the H2O is 1 mL
10 mL - 1 mL = 9 mL
It also applies to MOLES
nTOTAL
nH2
=
=
nH2
+
nTOTAL -
So, if the total sample is 20 mol and H2O makes up 3 mol:
nH2
VH2O
VH2O
nH2O
nH2O
= 20 mL - 3 mL = 17 mol
Can use together with other gas laws.
LAW VS. THEORY
Law is the actual observation.
 As temperature increases, volume increases (Charles’ Law)
 As volume increases, pressure decreases (Boyle’s Law)
Theory is explanation of law.
 Does NOT become law
 Becomes more reliable as it is modified to match continued
observation
KINETIC MOLECULAR THEORY
The Kinetic Molecular Theory (KMT) is an explanation of how the gas laws work.
 Gas is composed of particles in constant motion
 Temperature is the average kinetic energy of the particles
 Increased kinetic energy = increased motion = increased temperature
 Compared to the space they take up, the particles are so small, the
volume of the particles is negligible
 Particles aren’t attracted to or repelled by one another
 When particles collide with each other or the container, their kinetic
energy doesn’t change