1 BEHAVIOR OF GASES BEHAVIOR OF GASES Chapter 14 Chapter 14
Jan 23, 2016
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BEHAVIOR OF GASESBEHAVIOR OF GASESChapter 14Chapter 14
BEHAVIOR OF GASESBEHAVIOR OF GASESChapter 14Chapter 14
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Importance of Importance of GasesGases
Importance of Importance of GasesGases
Airbags fill with NAirbags fill with N22 gas in an accident. gas in an accident.
Gas is generated by the decomposition Gas is generated by the decomposition of sodium azide, NaNof sodium azide, NaN33..
2 NaN2 NaN3(s)3(s) ---> 2 Na ---> 2 Na(s)(s) + 3 N + 3 N2(g)2(g)
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THREETHREESTATES OF STATES OF
MATTERMATTER
THREETHREESTATES OF STATES OF
MATTERMATTER
Click picture above to view movie on states of matter
Click here for water production
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General Properties of GasesGeneral Properties of GasesThere is a lot of "free" There is a lot of "free"
space in a gas.space in a gas.
Gases can be expanded Gases can be expanded infinitely.infinitely.
Gases occupy containers Gases occupy containers uniformly and completely.uniformly and completely.
Gases diffuse and mix Gases diffuse and mix rapidly.rapidly.
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Kinetic Theory Revisited1. Gases consist of hard, spherical
particles (usually molecules or atoms)
2. Small- so the individual volume is considered to be insignificant
3. Large empty space between them
4. Easily compressed and expanded
5. No attractive or repulsive forces
6. Move rapidly in constant motion
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Kinetic Theory Revisited Recall: that the average kinetic
energy of a collection of gas particles is directly proportional to the Kelvin temperature of the gas.
Click for movie No Kinetic energy lost during
collisions. All particles have same energy at
same temperature.
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Properties of GasesProperties of GasesProperties of GasesProperties of GasesGas properties can be modeled Gas properties can be modeled
using math. Model depends using math. Model depends on-on-
VV = volume of the gas ( = volume of the gas (LL)) TT = temperature ( = temperature (KK)) nn = amount ( = amount (molesmoles)) PP = pressure ( = pressure (atm or atm or
kilopascalkilopascal))
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PressurePressurePressurePressurePressure of air is Pressure of air is
measured with a measured with a BAROMETERBAROMETER (developed by (developed by TorricelliTorricelli in 1643) in 1643)
Barometer calibrated for Barometer calibrated for column width and column width and pool width/depthpool width/depth
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PressurePressurePressurePressureHg rises in tube until Hg rises in tube until
gravitational force gravitational force of Hg (down) of Hg (down) balances the force balances the force of atmosphere of atmosphere (pushing up). (pushing up).
P of Hg pushing down P of Hg pushing down related to related to
Hg densityHg density column heightcolumn height
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PressurePressurePressurePressureColumn height measures Column height measures
PressurePressure of atmosphere of atmosphere
1 standard atm1 standard atm
= 760 mm Hg= 760 mm Hg
= 76 cm Hg= 76 cm Hg
= 760 torr= 760 torr
= 29.9 inches= 29.9 inches
= about 33 feet of water= about 33 feet of water
SI unit is SI unit is PASCALPASCAL, Pa, , Pa, where 1 atm = 101.325 kPawhere 1 atm = 101.325 kPa
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Gas -Volume, Temp, & Pressure
Click to view movie
1. Amount of Gas When we inflate a ball, we are
adding gas molecules. Increasing the number of gas
particles increases the number of collisions
• thus, the pressure increases If temp. is constant- doubling the
number of particles doubles pressure
Pressure and the Number of Molecules are Directly Related
Fewer molecules means fewer collisions.
Gases naturally move from areas of high pressure to low pressure because there is empty space to move in - spray can is example.
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Expanding Gas Uses?
The bombardier The bombardier beetle uses beetle uses decomposition of decomposition of hydrogen peroxide hydrogen peroxide to defend itself. The to defend itself. The gas acts as a gas acts as a propellant.propellant.
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The Shuttle Uses a Solid Booster and Uncontrolled Expanding
Gases Can Be Disastrous
1 atm
If you double the number of molecules
If you double the number of molecules
You double the pressure.
2 atm
As you remove molecules from a container
4 atm
As you remove molecules from a container the pressure decreases
2 atm
As you remove molecules from a container the pressure decreases
Until the pressure inside equals the pressure outside
Molecules naturally move from high to low pressure
1 atm
Changing the Size of the Container
In a smaller container molecules have less room to move.
Hit the sides of the container more often.
As volume decreases pressure increases. Think air pump
1 atm
4 Liters
As the pressure on a gas increases
2 atm
2 Liters
As the pressure on a gas increases the volume decreases
Pressure and volume are inversely related
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What happens to the air in a diver’s lungs the deeper
they go?
Temperature Raising the temperature
of a gas increases the pressure if the volume is held constant.
The molecules hit the walls harder.
The only way to increase the temperature at constant pressure is to increase the volume.
If you start with 1 liter of gas at 1 atm pressure and 300 K
and heat it to 600 K one of 2 things happens
300 K
Either the volume will increase to 2 liters at 1 atm
300 K600 K
300 K 600 K
•Or the pressure will increase to 2 atm.•Or someplace in between
The Gas Laws Describe HOW gases behave. Can be predicted by theory. Amount of change can be calculated
with mathematical equations.
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A. Boyle’s Law
P
VPV = k
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A. Boyle’s Law The pressure and volume of
a gas are inversely related
• at constant mass & temp
P
V
PV = k
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A. Boyle’s Law
Click to view movie
Boyle’s Law At a constant temperature pressure
and volume are inversely related. As one goes up the other goes down P x V = K (K is some
constant) Easier to use P1 x V1=P2 x V2
Click for movie
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GIVEN:
V1 = 100. mL
P1 = 150. kPa
V2 = ?
P2 = 200. kPa
WORK:
P1V1 = P2V2
Boyle’s Gas Law Problems A gas occupies 100. mL at 150. kPa.
Find its volume at 200. kPa.
BOYLE’S LAW
P V
(150.kPa)(100.mL)=(200.kPa)V2
V2 = 75.0 mL
A balloon is filled with 25 L of air at 1.0 atm pressure. If the pressure is change to 1.5 atm what is the new volume?
A balloon is filled with 73 L of air at 1.3 atm pressure. What pressure is needed to change to volume to 43 L?
Examples
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kT
VV
T
B. Charles’ Law
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kT
VV
T
B. Charles’ Law The volume and absolute
temperature (K) of a gas are directly related
• at constant mass & pressure
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Charles’ Law
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B. Charles’ Law
Click for movie
B. Charles’ Law The volume of a gas is directly
proportional to the Kelvin temperature if the pressure is held constant.
V = K x T(K is some constant) V/T= K V1/T1= V2/T2
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GIVEN:
V1 = 473 cm3
T1 = 36°C = 309K
V2 = ?
T2 = 94°C = 367K
WORK:
V1T2 = V2T1
Gas Law Problems A gas occupies 473 cm3 at 36°C. Find
its volume at 94°C.
CHARLES’ LAW
T V
(473 cm3)(367 K)=V2(309 K)
V2 = 562 cm3
Examples What is the temperature (ºC) of a gas
that is expanded from 2.5 L at 25ºC to 4.1L at constant pressure.
What is the final volume of a gas that starts at 8.3 L and 17ºC and is heated to 96ºC?
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kT
PP
T
C. Gay-Lussac’s Law
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kT
PP
T
C. Gay-Lussac’s Law The pressure and absolute
temperature (K) of a gas are directly related
• at constant mass & volume
C. Gay-Lussac’s Law The temperature and the pressure
of a gas are directly related at constant volume.
P = K x T(K is some constant) P/T= K P1/T1= P2/T2
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GIVEN:
P1 = 765 torr
T1 = 23°C = 296K
P2 = 560. torr
T2 = ?
WORK:
P1T2 = P2T1
E. Gas Law Problems A gas’ pressure is 765 torr at 23°C.
At what temperature will the pressure be 560. torr? GAY-LUSSAC’S LAW
P T
(765 torr)T2 = (560. torr)(296K)
T2 = 217 K = -56°C
Examples What is the pressure inside a 0.250 L
can of deodorant that starts at 25ºC and 1.2 atm if the temperature is raised to 100ºC?
At what temperature will the can above have a pressure of 2.2 atm?
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Too Much Pressure and Not Enough
Volume!!!!
Putting the pieces together The Combined Gas Law Deals with
the situation where only the number of molecules stays constant.
(P1 x V1)/T1= (P2 x V2)/T2
Allows us to figure out one thing when two of the others change.
The combined gas law contains all the other gas laws!
If the temperature remains constant.
P1 V1
T1
x=
P2 V2
T2
x
Boyle’s Law
The combined gas law contains all the other gas laws!
If the pressure remains constant.
P1 V1
T1
x=
P2 V2
T2
x
Charles’ Law
The combined gas law contains all the other gas laws!
If the volume remains constant.
P1 V1
T1
x=
P2 V2
T2
x
Gay-Lussac’s Law
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= kPVPTVT
PVT
D. Combined Gas Law
P1V1
T1
=P2V2
T2
P1V1T2 = P2V2T1
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GIVEN:
V1 = 7.84 cm3
P1 = 71.8 kPa
T1 = 25°C = 298 K
V2 = ?
P2 = 101.325 kPa
T2 = 273 K
WORK:
P1V1T2 = P2V2T1
(71.8 kPa)(7.84 cm3)(273 K)
=(101.325 kPa) V2 (298 K)
V2 = 5.09 cm3
E. Gas Law Problems A gas occupies 7.84 cm3 at 71.8 kPa &
25°C. Find its volume at STP.
P T VCOMBINED GAS LAW
Examples A 15 L cylinder of gas at 4.8 atm
pressure at 25ºC is heated to 75ºC and compressed to 17 atm. What is the new volume?
If 6.2 L of gas at 723 mm Hg at 21ºC is compressed to 2.2 L at 4117 mm Hg, what is the temperature of the gas?
The Fourth Part Avagadro’s Hypothesis click for movie V is proportional to number of
molecules at constant T and P V is proportional to moles One mole = 6.02 x 1023 particles One mole = 22.4 L at STP movie V = K n ( n ) is the number of moles Gets put into the Ideal Gas Law
Ideal Gases Remember in this chapter we assume
the gases behave ideally. Ideal gases don’t really exist, but
assuming it makes the math easier. So we get a close approximation.
Particles have no volume. No attractive forces. However, real gases do behave like
ideal gases at high temperature and low pressure.
The Ideal Gas Law P x V = n x R x T Pressure times Volume equals the
number of moles times the Ideal Gas Constant (R) times the temperature in Kelvin.
This time R does not depend on anything, it is really constant
R = 0.0821 (L atm)/(mol K) or R = 62.4 (L mm Hg)/(K mol) We now have a new way to count
moles. By measuring T, P, and V. We aren’t restricted to STP.
n = PV/RT
The Ideal Gas Law
Examples How many moles of air are there in a
2.0 L bottle at 19ºC and 747 mm Hg? What is the pressure exerted by 1.8 g
of H2 gas exert in a 4.3 L balloon at 27ºC?
Density The Molar mass of a gas can be
determined by the density of the gas. D= mass = m
Volume V Molar mass = mass = m
Moles n n = PV
RT
Molar Mass Formulas Molar Mass = m
(PV/RT) Molar mass = m RT
V P Molar mass = DRT
P
Dalton’s Law of Partial Pressures The total pressure inside a container
is equal to the partial pressure due to each gas.
The partial pressure is the contribution by that gas.
PTotal = P1 + P2 + P3
For example
We can find out the pressure in the fourth container.
By adding up the pressure in the first 3.2 atm
1 atm
3 atm
6 atm
Examples What is the total pressure in a balloon
filled with air if the pressure of the oxygen is 170 mm Hg and the pressure of nitrogen is 620 mm Hg?
In a second balloon the total pressure is 1.3 atm. What is the pressure of oxygen if the pressure of nitrogen is 720 mm Hg?
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B. Dalton’s Law The total pressure of a mixture
of gases equals the sum of the partial pressures of the individual gases.
Ptotal = P1 + P2 + ...When a H2 gas is collected by water displacement, the gas in the collection bottle is actually a mixture of H2 and water vapor.
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GIVEN:
PH2 = ?
Ptotal = 94.4 kPa
PH2O = 2.72 kPa
WORK:
Ptotal = PH2 + PH2O
94.4 kPa = PH2 + 2.72 kPa
PH2 = 91.7 kPa
B. Dalton’s Law Hydrogen gas is collected over water at
22.5°C. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa.
Look up water-vapor pressure for 22.5°C.
Sig Figs: Round to least number of decimal places.
The total pressure in the collection bottle is equal to atmospheric pressure and is a mixture of H2 and water vapor.
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GIVEN:
Pgas = ?
Ptotal = 742.0 torr
PH2O = 42.2 torr
WORK:
Ptotal = Pgas + PH2O
742.0 torr = PH2 + 42.2 torr
Pgas = 699.8 torr
A gas is collected over water at a temp of 35.0°C when the barometric pressure is 742.0 torr. What is the partial pressure of the dry gas?
Look up water-vapor pressure for 35.0°C.
Sig Figs: Round to least number of decimal places.
B. Dalton’s Law
The total pressure in the collection bottle is equal to barometric pressure and is a mixture of the “gas” and water vapor.
Diffusion
Effusion Gas escaping through a tiny hole in a container.
Depends on the speed of the molecule.
Molecules moving from areas of high concentration to low concentration.
Perfume molecules spreading across the room.
Graham’s Law The rate of effusion and diffusion is
inversely proportional to the square root of the molar mass of the molecules.
Kinetic energy = 1/2 mv2
m is the mass v is the velocity.
Chem Express
Bigger molecules move slower at the same temp. (by Square root)
Bigger molecules effuse and diffuse slower
Helium effuses and diffuses faster than air - escapes from balloon.
Graham’s Law
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C. Graham’s Law Graham’s LawGraham’s Law
• Rate of diffusion of a gas is inversely related to the square root of its molar mass.
• The equation shows the ratio of Gas A’s speed to Gas B’s speed.
A
B
B
A
m
m
v
v
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Determine the relative rate of diffusion for krypton and bromine.
1.381
Kr diffuses 1.381 times faster than Br2.
Kr
Br
Br
Kr
m
m
v
v2
2
A
B
B
A
m
m
v
v
g/mol83.80
g/mol159.80
C. Graham’s Law
The first gas is “Gas A” and the second gas is “Gas B”. Relative rate mean find the ratio “vA/vB”.
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An unknown gas diffuses 4.0 times faster than O2. Find its molar mass.
Am
g/mol32.00 16
A
B
B
A
m
m
v
v
A
O
O
A
m
m
v
v2
2
Am
g/mol32.00 4.0
16
g/mol32.00 mA
2
Am
g/mol32.00 4.0
g/mol2.0
C. Graham’s Law
The first gas is “Gas A” and the second gas is “Gas B”. The ratio “vA/vB” is 4.0.
Square both sides to get rid of the square
root sign.
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A molecule of oxygen gas has an average speed of 12.3 m/s at a given temp and pressure. What is the average speed of hydrogen molecules at the same conditions?
A
B
B
A
m
m
v
v
2
2
2
2
H
O
O
H
m
m
v
v
g/mol 2.02
g/mol32.00
m/s 12.3
vH 2
C. Graham’s Law
3.980m/s 12.3
vH 2
m/s49.0 vH 2
Put the gas with the unknown
speed as “Gas A”.
You may need to review Chapter 12 click for movie
At STP At STP determining the amount of
gas required or produced is easy. 22.4 L = 1 mole For example
How many liters of O2 at STP are required to produce 20.3 g of H2O?
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
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A. Gas Stoichiometry Moles Moles Liters of a Gas: Liters of a Gas:
• STP - use 22.4 L/mol • Non-STP - use ideal gas law
Non-Non-STPSTP• Given liters of gas?
–start with ideal gas law• Looking for liters of gas?
–start with stoichiometry conversions.
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1 molCaCO3
100.09g CaCO3
B. Gas Stoichiometry Problem What volume of CO2 forms from 5.25 g of
CaCO3 at 103 kPa & 25ºC?
5.25 gCaCO3
= 0.0525 mol CO2
CaCO3 CaO + CO2
1 molCO2
1 molCaCO3
5.25 g ? Lnon-STP
Looking for liters: Start with stoich and calculate moles of CO2.
Plug this into the Ideal Gas Law to find liters.
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WORK:
PV = nRT
(103 kPa)V=(.0525mol)(8.314 LkPa/molK) (298K)
V = 1.26 L CO2
B. Gas Stoichiometry Problem What volume of CO2 forms from 5.25 g
of CaCO3 at 103 kPa & 25ºC?
GIVEN:
P = 103 kPaV = ?
n = 0.0525 molT = 25°C = 298 KR = 8.314 LkPa/molK
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WORK:
PV = nRT
(97.3 kPa) (15.0 L)= n (8.314 LkPa/molK) (294K)
n = 0.597 mol O2
B. Gas Stoichiometry Problem How many grams of Al2O3 are formed from 15.0
L of O2 at 97.3 kPa & 21°C?
GIVEN:
P = 97.3 kPaV = 15.0 L
n = ?T = 21°C = 294 KR = 8.314 LkPa/molK
4 Al + 3 O2 2 Al2O3 15.0 L
non-STP ? g
Given liters: Start with Ideal Gas Law and
calculate moles of O2.
NEXT
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2 mol Al2O3
3 mol O2
B. Gas Stoichiometry Problem How many grams of Al2O3 are formed
from 15.0 L of O2 at 97.3 kPa & 21°C?
0.597mol O2
= 40.6 g Al2O3
4 Al + 3 O2 2 Al2O3
101.96 g Al2O3
1 molAl2O3
15.0Lnon-STP
? gUse stoich to convert moles of O2 to grams Al2O3.
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?
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Too Little Pressure and Too Little Volume!!!!
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What is a Turbocharger?