section 3.7—gas behavior how does the behavior of gases affect airbags? what is pressure? force of...
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Section 3.7—Gas Behavior How does the behavior of gases affect airbags?
What is PRESSURE?
Force of gas particles running into a surface.
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Pressure is measured by a Barometer
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If pressure is molecular collisions with the container…
As # of moles increase, pressure increases Think about blowing up a balloon!
Pressure and Moles (# of Molecules)
As number of molecules increases, there will be more molecules to collide with the wall
• Collisions between molecules and the wall increase
Pressure increases
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# of Gas Particles vs. Pressure
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Pressure & Volume
If pressure is molecular collisions with the container…
As volume increases, pressure decreases. Think about how your lungs work! http://www.youtube.com/watch?v=q6-oyxnkZC0
As volume increases, molecules can travel farther before hitting the wall• Collisions between molecules & the wall decrease
Pressure decreases
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What is “Temperature”?
Temperature – measure of the average kinetic energy of the molecules
Energy due to motion(Related to how fast the molecules are moving)
As temperature increases, Average Kinetic Energy Increases and Molecular motion increases
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Pressure and Temperature
If temperature is related to molecular motion…and pressure is molecular collisions with the container…
As temperature increases, pressure increases
As temperature increases, molecular motion increases
• Collisions between molecules & the wall increase
Pressure increases
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Volume and Temperature
If temperature is related to molecular motion…and volume is the amount of space the gas occupies…
As temperature increases, volume increases Think of liquid nitrogen and the balloon. http://www.youtube.com/watch?v=QEpxrGWep4E
As temperature increases, molecular motion increases• molecules will move farther away from each other Volume increases
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Pressure In Versus Out
Example: A bag of chips is bagged at sea level. What happens if the bag is then brought up to the top of a mountain.
A container will expand or contract until the pressure inside equals atmospheric pressure outside
The internal pressure is higher than the external pressure.
The bag will expand in order to reduce the internal pressure.
The internal pressure of the bag at low altitude is high
At high altitude there is lower pressureHigher pressure
Lowerpressure
Lower pressure
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CanExplodes!
When Expansion Isn’t Possible
Example: An aerosol can is left in a car trunk in the summer. What happens?
Rigid containers cannot expand
The internal pressure is higher than the external pressure.
The can is rigid—it cannot expand, it explodes!
The temperature inside the can begins to rise.
As temperature increases, pressure increases.
Higher pressure
Lowerpressure
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Air Pressure Crushing Cans
http://www.csun.edu/scied/4-discrpeant-event/the_can_crush/index.htm
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http://www.youtube.com/watch?v=Zz95_VvTxZM
Another cool videohttp://www.youtube.com/watch?v=JsoE4F2Pb20
Air Pressure Crushing “Cans”
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Kinetic Molecular Theory(KMT): explains gas behavior based upon
the motion of molecules based on an ideal gas
IDEAL gases are IMAGINARY gases that follow the assumptions of the KMT
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1
Assumptions of the KMTAll gases are made of atoms or molecules that are in constant, rapid, random motion
Gas particles are not attracted nor repelled from one another ***All gas particle collisions are perfectly elastic (no kinetic energy is lost to other forms)
The volume of gas particles is so small compared to the space between the particles, that the volume of the particle itself is insignificant***
2
3
4
5
The temperature of a gas is proportional to the average kinetic energy of the particles
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So what is a “REAL” gas?Real gases, (like nitrogen), will eventually condense into a liquid when the temperature gets too low or the pressure gets too high BECAUSE:
Assumption #3
Assumption #5
Gas particles do have attractions and repulsions towards one another
Gas particles do take up space
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Real Gases Deviate from Ideal Gas Behavior when at high pressure
The gas molecules are compressed making the volume they take up more significant than if they were spread out
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Real Gases Deviate from Ideal Gas Behavior when at low temperature.
The lower kinetic energy causes the molecules to move slower and ATTRACTIVE FORCES that really exist start to take effect---------------------------
Polar gases (HCl) deviate more than nonpolar gases (He or H2
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At Lower Temperature
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Gas Movement: Effusion vs Diffusion
Effusion –gas escapes from a tiny hole in the container
Effusion is why balloons deflate over time!
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Diffusion –gas moves across a space from high to low concentration
Diffusion is the reason we can smell perfume across the room
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Effusion, Diffusion & Particle Mass
How are particle size (mass) and these concepts related?
As mass of the particles increases, rate of effusion and diffusion is lowered.
As particle size (mass) increases, the particles move slower it takes them more time to find the hole or to go across the room
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Rate of Diffusion & Particle Mass
Watch as larger particles take longer to get to your nose
H2
CO2
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Section 3.8-3.9 —Gas Laws
How can we calculate Pressure, Volume and Temperature of our airbag?
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Pressure Units
Several units are used when describing pressure
Unit Symbol
atmospheres atm
Pascals, kiloPascals
millimeters of mercury
pounds per square inch
Pa, kPa
mm Hg
psi
1 atm = 101300 Pa = 101.3 kPa = 760 mm Hg = 14.7 psi
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Conversions Between Different Pressure Units
1 atm = 760 mmHg = 101.3 kPa
Examples
1.Convert 654 mm Hg to atm
2.Convert 879 mm Hg to kPa
3.Convert 15.6 atm to kPa
654 mmHg x 1atm = 760 mmHg
.861 atm
879 mmHg x 101.3 Kpa = 760 mmHg
1.16 Kpa
15.6 atm x 101.3 Kpa = 1atm
1580 Kpa
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Temperature Unit used in Gas Laws
Kelvin (K)– temperature scale with an absolute zero
Temperatures cannot fall below an absolute zero
KC 273
Examples1.Convert 15.6 °C into K
2. Convert 234 K into °C
15.6 + 273 = K 288.6 289 K
°C + 273 = 234 -39 °C
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Standard Temperature & Pressure (STP)
the conditions of:1 atm (or the equivalent in another unit) 0°C (273 K)
Problems often use “STP” to indicate quantities…don’t forget this “hidden” information when making your list!
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GAS LAWS: “Before” and “After” This section has 5 gas laws which have “before” and “after” conditions.
For example:2
2
1
1
n
P
n
P
1= initial amount2= final amount
P= PressureV= VolumeT=Temperaturen= moles(molecules)
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Boyle’s Law
Pressure Increases as Volume Decreases
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Boyles’ LawVolume & Presssure are INVERSELY proportional when temperature and moles are held constant
P = pressureV = volume2211 VPVP
The two pressure units must match and the two volume units must match!
Example: A gas sample is 1.05 atm when at 2.5 L. What volume is it if the pressure is changed to 0.980 atm?
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Boyles’ Law 2211 VPVP ***The two pressure units must match & the two volume units must match!
Example: A gas sample is 1.05 atm when 2.5 L. What volume is it if the pressure is changed to 0.980 atm?
P1 = 1.05 atm
V1 = 2.5 L
P2 = 0.980 atm
V2 = ? L
V2 = 2.7 L
2980.05.205.1 VatmLatm
2980.0
5.205.1V
atm
Latm
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Boyles Law: Graph
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Charles’ Law
Volume Increases as Temperature Increases
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Charles’ LawVolume & Temperature are DIRECTLY proportional when pressure and moles are held constant.
V = VolumeT = Temperature
2
2
1
1
T
V
T
V
The two volume units must match & temperature must be in Kelvin!
Example: What is the final volume if a 10.5 L sample of gas is changed from 25C to 50C?
V1 = 10.5 L
T1 = 25C
V2 = ? L
T2 = 50C
Temperature needs to be in Kelvin!
25C + 273 = 298 K
50C + 273 = 323 K
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Charles’ Law2
2
1
1
T
V
T
V
***The two volume units must match & temperature must be in Kelvin!
Example: What is the final volume if a 10.5 L sample of gas is changed from 25C to 50C?
V1 = 10.5 L
T1 = 25C
V2 = ? L
T2 = 50C
V2 = 11.4 L
= 298 K
= 323 K
K
V
K
L
323298
5.10 2
2298
5.10323V
K
LK
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Charles Law: Graph
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Gay-Lussac’s Law
Temperature decreases as Pressure decreases
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Gay-Lussac’s LawPressure & temperature are DIRECTLY proportional when moles and volume are held constant
P = PressureT = Temperature
The two pressure units must match and temperature must be in Kelvin!
Example: A sample of hydrogen gas at 47C exerts a pressure of .329 atm. The gas is heated to 77C at constant volume and moles. What will the new pressure be?
P1 = .329 atm
T1 = 47C
P2 = ? atm
T2 = 77C
Temperature needs to be in Kelvin!
47C + 273 = 320 K
77C + 273 = 350 K
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Gay-Lussac’ Law
Example: A sample of hydrogen gas at 47C exerts a pressure of .329 atm. The gas is heated to 77C at constant volume and moles. What will the new pressure be?
P1 = .329 atm
T1 = 47C
P2 = ? atm
T2 = 77C
P2 = .360 atm
= 320 K
= 350 K
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Gay Lussac Law: Graph
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Avogadro’s LawMoles and Volume are directly proportional when temp. & pressure are held constant
V = Volumen = # of moles of gas
2
2
1
1
n
V
n
V
Example: A sample with 0.15 moles of gas has a volume of 2.5 L. What is the volume if the sample is increased to 0.55 moles?
The two volume units must match!
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Avogadro’s Law2
2
1
1
n
V
n
V
Example: A sample with 0.15 moles of gas has a volume of 2.5 L. What is the volume if the sample is increased to 0.55 moles?
The two volume units must match!
n1 = 0.15 moles
V1 = 2.5 L
n2 = 0.55 moles
V2 = ? L
mole
V
mole
L
55.015.0
5.2 2
215.0
5.255.0V
mole
Lmole
V2 = 9.2 L
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Combined Gas Law
P = PressureV = Volumen = # of molesT = Temperature22
22
11
11
Tn
VP
Tn
VP
Each “pair” of units must match and temperature must be in Kelvin!
Example: What is the final volume if a 15.5 L sample of gas at 755 mmHg and 298 K is changed to STP?
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Combined Gas Law
P = PressureV = VolumeT = Temperature
Moles is not mentioned so remove it from equation!
Example: What is the final volume if a 15.5 L sample of gas at 755 mmHg and 298K is changed to STP?
P1 = 755 mmHg
V1 = 15.5 L
T1 = 298 K
P2 = 760mmHg
V2 = ? L
T2 = 273 K V2 = 14.1 L
STP is standard temperature (273 K) and pressure (1 atm)
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22
22
11
11
Tn
VP
Tn
VP
22
12
11
11
Tn
VP
Tn
VP
The combined gas law can be used for all “before” and “after” gas law problems!
For example, if volume is held constant, then
and the combined gas law becomes:
21 VV
22
2
11
1
Tn
P
Tn
P
Why you really only need 1 of these
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Watch as variables are held constant and the combined gas law “becomes” the other 3 laws
22
22
11
11
Tn
VP
Tn
VPHold pressure and
temperature constantAvogadro’s Law
22
22
11
11
Tn
VP
Tn
VPHold moles and
temperature constantBoyles’ Law
22
22
11
11
Tn
VP
Tn
VPHold pressure and moles
constantCharles’ Law
Transforming the Combined Law
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Dalton’s Law
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Dalton’s Law
• Each gas in a mixture exerts its own pressure called a partial pressure = P1, P2….
• Total Pressure = PT
Example: A gas mixture is made up of oxygen(2.3 atm) and nitrogen(1.7 atm) gases. What is the total pressure?
....321 PPPPT
PT = 2.3 atm + 1.7 atm
4.0 atm
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2KClO3 (s) 2KCl (s) + 3O2 (g)
PT = PO + PH O2 2
Modified Dalton’s Law
When a gas is Collected over water, the total pressure of the mixture collectedis a combination of water vapor and the gas you are collecting!
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Modified Dalton’s Law
Example: What is the pressure of the water vapor if the total pressure of the flask is 17.5 atm and the pressure of the oxygen gas is 16.1 atm?
watergas PPPT
17.5 = 16.1 atm + PH2O
1.4 atm
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The Ideal Gas Law (an“AT NOW”equation)The volume of a gas varies directly with the number of moles and its Kelvin temperature
P = PressureV = Volume n = molesR = Gas Law ConstantT = Temperature
nRTPV
There are three possibilities for “R”!
Kmole
atmL
*
*0821.0
Choose the one with units that match your pressure units!
Volume must be in Liters when using “R” to allow the unit to cancel!
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The Ideal Gas Law nRTPV Example: A sample with 0.55 moles of gas is at 105.7 kPa
and 27°C. What volume does it occupy?
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The Ideal Gas Law
Example: A sample with 0.55 moles of gas is at 105.7 kPa and 27°C. What volume does it occupy?
n = 0.55 moles
P = 105.7 kPa
T = 27°C + 273 = 300 K
V = ?
R = 8.31 L kPa / mole K
)300(**31.8)55.0()7.105( KkmolekPaLmoleVkPa
V2 = 13 L
The Ideal Gas Law does not compare situations—it describes a gas in one situation.
)7.105(
)300(**31.8)55.0(
kPa
KkmolekPaLmole
V
nRTPV
Chosen to match the kPa in the “P” above
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The Ideal Gas LawExample 2:
What mass of hydrogen gas in grams is contained in a 10.0 L tank at 27°C and 3.50 atm of pressure?
n = ?
P = 3.50 atm
T = 27°C + 273 = 300 K
V = 10.0 L
R = .0821 L atm /mole K
)300(**0821.)0.10()50.3( KkmoleatmLnLatm
)*
*0821(.)300(
0.10)50.3(
kmoleatmLxK
Latmn
nRTPV
Chosen to match the atm in the “P” above
n = 1.42 mol 1.42 mol x 2.02 g = 2.87 g 1 mol