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E13: The Gas E13: The Gas LawsLaws
E13: The Gas E13: The Gas LawsLaws
Properties of GasesProperties of GasesProperties of GasesProperties of GasesAll gases are influenced byAll gases are influenced by
VolumeVolume TemperatureTemperature PressurePressure AmountAmount
Gases respond to changes in Gases respond to changes in these variables in these variables in predictable ways.predictable ways.
All gases are influenced byAll gases are influenced by VolumeVolume TemperatureTemperature PressurePressure AmountAmount
Gases respond to changes in Gases respond to changes in these variables in these variables in predictable ways.predictable ways.
TEMPERATURETEMPERATUREIs measured in Kelvin (K) for Is measured in Kelvin (K) for gasesgases
The Kelvin scale is based on the Celsius The Kelvin scale is based on the Celsius scale, but scale, but zero is set at Absolute Zerozero is set at Absolute Zero (the theoretical coldest possible temp) (the theoretical coldest possible temp) instead of at the freezing point of Hinstead of at the freezing point of H22O like ˚C.O like ˚C.
Absolute Zero (0 K) is -273.15˚CAbsolute Zero (0 K) is -273.15˚C Therefore, we can easily convert measurements Therefore, we can easily convert measurements
between Celsius and Kelvin by adding or subtracting between Celsius and Kelvin by adding or subtracting that value:that value:
TempTempKelvin Kelvin ≃≃ Temp Temp˚Celsius˚Celsius + 273 + 273
Is measured in Kelvin (K) for Is measured in Kelvin (K) for gasesgases
The Kelvin scale is based on the Celsius The Kelvin scale is based on the Celsius scale, but scale, but zero is set at Absolute Zerozero is set at Absolute Zero (the theoretical coldest possible temp) (the theoretical coldest possible temp) instead of at the freezing point of Hinstead of at the freezing point of H22O like ˚C.O like ˚C.
Absolute Zero (0 K) is -273.15˚CAbsolute Zero (0 K) is -273.15˚C Therefore, we can easily convert measurements Therefore, we can easily convert measurements
between Celsius and Kelvin by adding or subtracting between Celsius and Kelvin by adding or subtracting that value:that value:
TempTempKelvin Kelvin ≃≃ Temp Temp˚Celsius˚Celsius + 273 + 273
TempTempK K Temp Temp˚C˚C + 273 + 273TempTempK K Temp Temp˚C˚C + 273 + 273
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QuickTime™ and a decompressor
are needed to see this picture.
We don’t use ˚F in science because it is
such a complicated conversion!
To go between Kelvin & ˚C, simply add or subtract 273!
TEMPERATURE PRACTICETEMPERATURE PRACTICE1.1. Water changes to steam at Water changes to steam at 100.00˚C100.00˚C..
What temperature is this in Kelvin?What temperature is this in Kelvin?
100˚C + 273 = 373 K100˚C + 273 = 373 K
2.2. A gas liquefies at A gas liquefies at -178˚C-178˚C. . What temperature is this in Kelvin?What temperature is this in Kelvin?
-178˚C + 273 = 95 K-178˚C + 273 = 95 K
1.1. Water changes to steam at Water changes to steam at 100.00˚C100.00˚C.. What temperature is this in Kelvin?What temperature is this in Kelvin?
100˚C + 273 = 373 K100˚C + 273 = 373 K
2.2. A gas liquefies at A gas liquefies at -178˚C-178˚C. . What temperature is this in Kelvin?What temperature is this in Kelvin?
-178˚C + 273 = 95 K-178˚C + 273 = 95 K
TempTempK K Temp Temp˚C˚C + 273 + 273
VOLUMEVOLUMEGases will spread to fill the Gases will spread to fill the size and shape of their size and shape of their container.container.
Measured in Measured in L, mL, or cubicL, mL, or cubic meter values meter values
If you need to convert, use these If you need to convert, use these known equalities:known equalities:
1L = 1dm1L = 1dm33 = 1000 cm = 1000 cm33 = 1000 = 1000 mLmL
Gases will spread to fill the Gases will spread to fill the size and shape of their size and shape of their container.container.
Measured in Measured in L, mL, or cubicL, mL, or cubic meter values meter values
If you need to convert, use these If you need to convert, use these known equalities:known equalities:
1L = 1dm1L = 1dm33 = 1000 cm = 1000 cm33 = 1000 = 1000 mLmL
CONVERSIONSCONVERSIONSYou MUST get used to canceling out by using You MUST get used to canceling out by using
CONVERSION FACTORSCONVERSION FACTORS (equalities) (equalities) Given unwanted unit x Given unwanted unit x equal value of desired unitequal value of desired unit
equal value of unwanted unitequal value of unwanted unit
Example: 200mL = ? dmExample: 200mL = ? dm33
200200mLmL x _ x _11dmdm3 3 = = .2 dm.2 dm33
10001000mL mL
Get in the habit of WRITING UNITS to keep track Get in the habit of WRITING UNITS to keep track of what you have cancelledof what you have cancelled..
Don’t forget sig figs!Don’t forget sig figs!
You MUST get used to canceling out by using You MUST get used to canceling out by using CONVERSION FACTORSCONVERSION FACTORS (equalities) (equalities)
Given unwanted unit x Given unwanted unit x equal value of desired unitequal value of desired unit equal value of unwanted unitequal value of unwanted unit
Example: 200mL = ? dmExample: 200mL = ? dm33
200200mLmL x _ x _11dmdm3 3 = = .2 dm.2 dm33
10001000mL mL
Get in the habit of WRITING UNITS to keep track Get in the habit of WRITING UNITS to keep track of what you have cancelledof what you have cancelled..
Don’t forget sig figs!Don’t forget sig figs!
Volume Conversion P.P.Volume Conversion P.P.Volume Conversion P.P.Volume Conversion P.P.3. 3. A gallon jug of milk holds A gallon jug of milk holds 3.78 L3.78 L.. How many How many mLmL is this? is this?3.78L3.78L x x 1000 mL1000 mL = 3780 mL = 3780 mL 1 1 L1 1 L
4. 4. A balloon contains A balloon contains 347 cm347 cm33 of air. of air. How many How many litersliters is this? is this?
347cm347cm3 3 x x 1L 1L = .347L= .347L 1 1000cm1 1000cm33
3. 3. A gallon jug of milk holds A gallon jug of milk holds 3.78 L3.78 L.. How many How many mLmL is this? is this?3.78L3.78L x x 1000 mL1000 mL = 3780 mL = 3780 mL 1 1 L1 1 L
4. 4. A balloon contains A balloon contains 347 cm347 cm33 of air. of air. How many How many litersliters is this? is this?
347cm347cm3 3 x x 1L 1L = .347L= .347L 1 1000cm1 1000cm33
Equalities: 1L = 1dmEqualities: 1L = 1dm33 = 1000 cm = 1000 cm33 = 1000 mL = 1000 mL
NOTE: look at the data for sig figs. Known equalities do not limit sig figs.
PRESSUREPRESSUREPRESSUREPRESSUREThe amount of The amount of FORCEFORCE exerted on a given area. exerted on a given area.Based on Pascals (Pa) but MANY different Based on Pascals (Pa) but MANY different
units can be used.units can be used.
Useful conversion equalitiesUseful conversion equalities::
101300 101300 Pa=Pa= 101.3kPa = 1atm = 760mm101.3kPa = 1atm = 760mmHgHg = 760.torr = 14.7psi = 1 bar = 760.torr = 14.7psi = 1 bar
All these known equalities mean we can move between All these known equalities mean we can move between all of these units by setting up more conversion factors.all of these units by setting up more conversion factors.
The amount of The amount of FORCEFORCE exerted on a given area. exerted on a given area.Based on Pascals (Pa) but MANY different Based on Pascals (Pa) but MANY different
units can be used.units can be used.
Useful conversion equalitiesUseful conversion equalities::
101300 101300 Pa=Pa= 101.3kPa = 1atm = 760mm101.3kPa = 1atm = 760mmHgHg = 760.torr = 14.7psi = 1 bar = 760.torr = 14.7psi = 1 bar
All these known equalities mean we can move between All these known equalities mean we can move between all of these units by setting up more conversion factors.all of these units by setting up more conversion factors.
Pressure Conversion Pressure Conversion P.P.P.P.Pressure Conversion Pressure Conversion P.P.P.P.
5. A radio announcer reports an atmospheric 5. A radio announcer reports an atmospheric pressure of pressure of 99.6 kiloPascals99.6 kiloPascals (kPa). (kPa).
What is the pressure in atmospheres (What is the pressure in atmospheres (atmatm)?)?
99.6 kPa99.6 kPa x x 1atm 1atm = .983 atm = .983 atm 1 101.3 kPa1 101.3 kPa
5. A radio announcer reports an atmospheric 5. A radio announcer reports an atmospheric pressure of pressure of 99.6 kiloPascals99.6 kiloPascals (kPa). (kPa).
What is the pressure in atmospheres (What is the pressure in atmospheres (atmatm)?)?
99.6 kPa99.6 kPa x x 1atm 1atm = .983 atm = .983 atm 1 101.3 kPa1 101.3 kPa
101300 101300 PaPa == 101.3101.3kPakPa == 11atmatm == 760760mmmmHHgg == 760.torr 760.torr == 14.714.7psipsi == 1 1 barbar
Pressure Conversion Pressure Conversion P.P.P.P.Pressure Conversion Pressure Conversion P.P.P.P.
6. Express a pressure of 6. Express a pressure of 729 mm729 mmHgHg in in atmospheresatmospheres..
6. Express a pressure of 6. Express a pressure of 729 mm729 mmHgHg in in atmospheresatmospheres..
101300 101300 PaPa == 101.3101.3kPakPa == 11atmatm == 760760mmmmHHgg == 760.torr 760.torr == 14.714.7psipsi == 1 1 barbar
729mm729mmHgHg x x 1atm 1atm = .959 atm = .959 atm 1 760.mm1 760.mmHgHg
KINETIC THEORY of MATTERKINETIC THEORY of MATTERSOLIDSSOLIDS:: FIXED volume & shape. Tightly packed FIXED volume & shape. Tightly packed
atoms (atoms (most dense statemost dense state). Molecules vibrate in place.). Molecules vibrate in place.
LIQUIDSLIQUIDS:: FIXED volume, FLUID shape. FIXED volume, FLUID shape. Molecules vibrate AND have limited rotation.Molecules vibrate AND have limited rotation.
GASESGASES:: NO FIXED VOLUME or SHAPE. NO FIXED VOLUME or SHAPE. Least Least densely packed atoms, take the shape of container. densely packed atoms, take the shape of container. Molecules vibrate, rotate, & travel freely.Molecules vibrate, rotate, & travel freely.
SOLIDSSOLIDS:: FIXED volume & shape. Tightly packed FIXED volume & shape. Tightly packed atoms (atoms (most dense statemost dense state). Molecules vibrate in place.). Molecules vibrate in place.
LIQUIDSLIQUIDS:: FIXED volume, FLUID shape. FIXED volume, FLUID shape. Molecules vibrate AND have limited rotation.Molecules vibrate AND have limited rotation.
GASESGASES:: NO FIXED VOLUME or SHAPE. NO FIXED VOLUME or SHAPE. Least Least densely packed atoms, take the shape of container. densely packed atoms, take the shape of container. Molecules vibrate, rotate, & travel freely.Molecules vibrate, rotate, & travel freely.
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As TEMPERATURE goes UP
Kinetic ENERGY goes Up
(warm molecules move more!)
““IDEAL” GASES:IDEAL” GASES:““IDEAL” GASES:IDEAL” GASES:Gases consist of small particles with massGases consist of small particles with massCompared to the size of their container, the Compared to the size of their container, the
volume of the gas particles is tiny.volume of the gas particles is tiny.The particles can collide without losing energy The particles can collide without losing energy
(= “elastic”)(= “elastic”)Collisions w/ container walls cause PRESSURECollisions w/ container walls cause PRESSUREThe speed of their movement is directly related The speed of their movement is directly related
to temperature (to temperature (↑↑T T ↑↑kinetic energy)kinetic energy)
Gases consist of small particles with massGases consist of small particles with massCompared to the size of their container, the Compared to the size of their container, the
volume of the gas particles is tiny.volume of the gas particles is tiny.The particles can collide without losing energy The particles can collide without losing energy
(= “elastic”)(= “elastic”)Collisions w/ container walls cause PRESSURECollisions w/ container walls cause PRESSUREThe speed of their movement is directly related The speed of their movement is directly related
to temperature (to temperature (↑↑T T ↑↑kinetic energy)kinetic energy)
STP:STP: Standard Temp. & PressureStandard Temp. & PressureSTP:STP: Standard Temp. & PressureStandard Temp. & Pressure
Unless otherwise stated, gases will be Unless otherwise stated, gases will be assumed to be at Standard Temperature assumed to be at Standard Temperature and Pressure (STP).and Pressure (STP).
STP is based on air pressure at sea level STP is based on air pressure at sea level (760.mm(760.mmHgHg) and zero ˚C (= 273K)) and zero ˚C (= 273K)
Unless otherwise stated, gases will be Unless otherwise stated, gases will be assumed to be at Standard Temperature assumed to be at Standard Temperature and Pressure (STP).and Pressure (STP).
STP is based on air pressure at sea level STP is based on air pressure at sea level (760.mm(760.mmHgHg) and zero ˚C (= 273K)) and zero ˚C (= 273K)
S.T.P. = 273 K & 760. mmmmHgHg
BOYLE’S LAW:The inverse relationship between Volume & Pressure
BOYLE’S LAW:The inverse relationship between Volume & Pressure
We explored this in E11:We explored this in E11:As you pressed the syringe, you As you pressed the syringe, you
decreased the volumedecreased the volume inside. inside. It got harder to push, because It got harder to push, because as theas the
volume decreased, the pressure of the volume decreased, the pressure of the gas inside increasedgas inside increased..
Inverse relationship: Inverse relationship: VV↓↓PP↑↑ ( (or or VV↑↑PP↓↓))
Boyle’s Law assumes that temperature isn’t changed (is constant).Boyle’s Law assumes that temperature isn’t changed (is constant).
We explored this in E11:We explored this in E11:As you pressed the syringe, you As you pressed the syringe, you
decreased the volumedecreased the volume inside. inside. It got harder to push, because It got harder to push, because as theas the
volume decreased, the pressure of the volume decreased, the pressure of the gas inside increasedgas inside increased..
Inverse relationship: Inverse relationship: VV↓↓PP↑↑ ( (or or VV↑↑PP↓↓))
Boyle’s Law assumes that temperature isn’t changed (is constant).Boyle’s Law assumes that temperature isn’t changed (is constant).
SOLVING GAS LAW PROBLEMSSOLVING GAS LAW PROBLEMS If you’re given a problem that If you’re given a problem that involves more involves more
than one gas variablethan one gas variable (vol., pressure, temp., (vol., pressure, temp., or amount), you’ll need to figure out which are or amount), you’ll need to figure out which are involved and how they will influence each other.involved and how they will influence each other.
STEP 1:STEP 1: While reading the problem fill this in: While reading the problem fill this in:(BEFORE)(BEFORE) (AFTER)(AFTER)
VV11 = = VV22 = =
PP11 = = PP22 = =
TT11 = = TT22 = =You may need to You may need to convert unitsconvert units before moving onbefore moving on
(like units should match, always use K for temp)(like units should match, always use K for temp)
If you’re given a problem that If you’re given a problem that involves more involves more than one gas variablethan one gas variable (vol., pressure, temp., (vol., pressure, temp., or amount), you’ll need to figure out which are or amount), you’ll need to figure out which are involved and how they will influence each other.involved and how they will influence each other.
STEP 1:STEP 1: While reading the problem fill this in: While reading the problem fill this in:(BEFORE)(BEFORE) (AFTER)(AFTER)
VV11 = = VV22 = =
PP11 = = PP22 = =
TT11 = = TT22 = =You may need to You may need to convert unitsconvert units before moving onbefore moving on
(like units should match, always use K for temp)(like units should match, always use K for temp)
BOYLE’S LAW P.P. #1BOYLE’S LAW P.P. #1A sample gas occupies aA sample gas occupies a volume of 2.00L volume of 2.00L atat
27˚C 27˚C andand 1.00 atm pressure. 1.00 atm pressure. What will theWhat will the volumevolume be at be at same temperature same temperature
but abut a pressure of 2.75 atm? pressure of 2.75 atm?
STEP 1:STEP 1: Identify knowns & unknowns. Identify knowns & unknowns.
(BEFORE)(BEFORE) (AFTER)(AFTER)VV11 = = 2.00 L2.00 L VV22 = = ??
PP11 = = 1.00 atm1.00 atm PP22 = = 2.75 atm2.75 atm
TT11 = = 27˚C27˚C TT22 = = 27˚C27˚C
A sample gas occupies aA sample gas occupies a volume of 2.00L volume of 2.00L atat 27˚C 27˚C andand 1.00 atm pressure. 1.00 atm pressure.
What will theWhat will the volumevolume be at be at same temperature same temperature but abut a pressure of 2.75 atm? pressure of 2.75 atm?
STEP 1:STEP 1: Identify knowns & unknowns. Identify knowns & unknowns.
(BEFORE)(BEFORE) (AFTER)(AFTER)VV11 = = 2.00 L2.00 L VV22 = = ??
PP11 = = 1.00 atm1.00 atm PP22 = = 2.75 atm2.75 atm
TT11 = = 27˚C27˚C TT22 = = 27˚C27˚C
SOLVING GAS LAW PROBLEMSSOLVING GAS LAW PROBLEMSSTEP 1:STEP 1: Identify knowns & unknowns, convert. Identify knowns & unknowns, convert.STEP 2:STEP 2: Identify relationship involvedIdentify relationship involved
Is it V-P (Boyle’s Law)? T-V (Charles Law)?Is it V-P (Boyle’s Law)? T-V (Charles Law)?or is it T-P (Guy Lussac’s Law)or is it T-P (Guy Lussac’s Law)
STEP 3:STEP 3: Set up problem Set up problemSTEP 4:STEP 4: Solve & check against known Solve & check against known
relationship.relationship.
STEP 1:STEP 1: Identify knowns & unknowns, convert. Identify knowns & unknowns, convert.STEP 2:STEP 2: Identify relationship involvedIdentify relationship involved
Is it V-P (Boyle’s Law)? T-V (Charles Law)?Is it V-P (Boyle’s Law)? T-V (Charles Law)?or is it T-P (Guy Lussac’s Law)or is it T-P (Guy Lussac’s Law)
STEP 3:STEP 3: Set up problem Set up problemSTEP 4:STEP 4: Solve & check against known Solve & check against known
relationship.relationship.
Boyle’s Law P.P. #1Boyle’s Law P.P. #1STEP 1:STEP 1: Identify knowns & unknowns Identify knowns & unknowns(BEFORE)(BEFORE) (AFTER)(AFTER)VV11 = 2.00L = 2.00L VV22 = ? = ?PP11 = 1.00atm = 1.00atm PP22 = 2.75atm = 2.75atmTT11 = 27˚C = 27˚C TT22 = same = same
STEP 2:STEP 2: Identify relationship involved Identify relationship involvedIs it V-P (Boyle’s Law)? Is it V-P (Boyle’s Law)? YES!YES! V V ↓ ↓ P P ↑↑T-V (Charles Law)? T-V (Charles Law)? T V T Vor is it T-P (Guy Lussac’s Law) or is it T-P (Guy Lussac’s Law) T P T P
STEP 1:STEP 1: Identify knowns & unknowns Identify knowns & unknowns(BEFORE)(BEFORE) (AFTER)(AFTER)VV11 = 2.00L = 2.00L VV22 = ? = ?PP11 = 1.00atm = 1.00atm PP22 = 2.75atm = 2.75atmTT11 = 27˚C = 27˚C TT22 = same = same
STEP 2:STEP 2: Identify relationship involved Identify relationship involvedIs it V-P (Boyle’s Law)? Is it V-P (Boyle’s Law)? YES!YES! V V ↓ ↓ P P ↑↑T-V (Charles Law)? T-V (Charles Law)? T V T Vor is it T-P (Guy Lussac’s Law) or is it T-P (Guy Lussac’s Law) T P T P
SOLVING GAS LAW PROBLEMSSOLVING GAS LAW PROBLEMSSTEP 1:STEP 1: Identify knowns & unknowns, convert. Identify knowns & unknowns, convert.STEP 2:STEP 2: Identify relationship involved Identify relationship involvedSTEP 3:STEP 3: Set up problemSet up problemYou can use Algebraic method You can use Algebraic method oror conversions conversions I will use conversions, because in upcoming I will use conversions, because in upcoming
concepts you will HAVE to be familiar with itconcepts you will HAVE to be familiar with it..The conversion fraction is set up to cause the The conversion fraction is set up to cause the
expected result. expected result. STEP 4:STEP 4: Solve & check against the known Solve & check against the known
relationship. relationship. Also check your sig figs!Also check your sig figs!
STEP 1:STEP 1: Identify knowns & unknowns, convert. Identify knowns & unknowns, convert.STEP 2:STEP 2: Identify relationship involved Identify relationship involvedSTEP 3:STEP 3: Set up problemSet up problemYou can use Algebraic method You can use Algebraic method oror conversions conversions I will use conversions, because in upcoming I will use conversions, because in upcoming
concepts you will HAVE to be familiar with itconcepts you will HAVE to be familiar with it..The conversion fraction is set up to cause the The conversion fraction is set up to cause the
expected result. expected result. STEP 4:STEP 4: Solve & check against the known Solve & check against the known
relationship. relationship. Also check your sig figs!Also check your sig figs!
Boyle’s Law P.P. #1Boyle’s Law P.P. #1 STEP 1:STEP 1: Identify knowns & unknowns Identify knowns & unknowns
VV11 = 2.00L = 2.00L VV22 = ? = ?PP11 = 1.00atm = 1.00atm PP22 = 2.75atm = 2.75atm
STEP 2:STEP 2: Identify relationship:Identify relationship: VV↓↓ P P↑↑ STEP 3:STEP 3: Set up problemSet up probleml Put your Put your “lonely” known in front“lonely” known in front ( (VV11 here here))l Arrange the paired variable in a fractionArrange the paired variable in a fraction to to
match the known relationship (match the known relationship (VV↓↓ P P↑↑)) by by putting theputting the larger value on the top or bottom larger value on the top or bottom of the fraction, whichever the of the fraction, whichever the ARROW of ARROW of the unknownthe unknown points to! points to! (in this case, V should go (in this case, V should go ↓, so we put the ↓, so we put the larger P value on bottom: 1.00 atm / 2.75 atm)larger P value on bottom: 1.00 atm / 2.75 atm)
STEP 1:STEP 1: Identify knowns & unknowns Identify knowns & unknownsVV11 = 2.00L = 2.00L VV22 = ? = ?PP11 = 1.00atm = 1.00atm PP22 = 2.75atm = 2.75atm
STEP 2:STEP 2: Identify relationship:Identify relationship: VV↓↓ P P↑↑ STEP 3:STEP 3: Set up problemSet up probleml Put your Put your “lonely” known in front“lonely” known in front ( (VV11 here here))l Arrange the paired variable in a fractionArrange the paired variable in a fraction to to
match the known relationship (match the known relationship (VV↓↓ P P↑↑)) by by putting theputting the larger value on the top or bottom larger value on the top or bottom of the fraction, whichever the of the fraction, whichever the ARROW of ARROW of the unknownthe unknown points to! points to! (in this case, V should go (in this case, V should go ↓, so we put the ↓, so we put the larger P value on bottom: 1.00 atm / 2.75 atm)larger P value on bottom: 1.00 atm / 2.75 atm)
Boyle’s Law P.P. #1Boyle’s Law P.P. #1 STEP 1:STEP 1: Identify knowns & unknowns Identify knowns & unknowns
VV11 = 2.00L = 2.00L VV22 = ? = ?PP11 = 1.00atm = 1.00atm PP22 = 2.75atm = 2.75atm
STEP 2:STEP 2: Identify relationship:Identify relationship: V V↓↓ P P↑↑ STEP 3:STEP 3: Set up problemSet up problem
2.00L x 2.00L x 1.00 atm1.00 atm = = .727 atm.727 atm 2.75 atm2.75 atm
STEP 4:STEP 4: Solve & check against the known Solve & check against the known relationship. relationship. Also check your sig figs!Also check your sig figs!
STEP 1:STEP 1: Identify knowns & unknowns Identify knowns & unknownsVV11 = 2.00L = 2.00L VV22 = ? = ?PP11 = 1.00atm = 1.00atm PP22 = 2.75atm = 2.75atm
STEP 2:STEP 2: Identify relationship:Identify relationship: V V↓↓ P P↑↑ STEP 3:STEP 3: Set up problemSet up problem
2.00L x 2.00L x 1.00 atm1.00 atm = = .727 atm.727 atm 2.75 atm2.75 atm
STEP 4:STEP 4: Solve & check against the known Solve & check against the known relationship. relationship. Also check your sig figs!Also check your sig figs!
BOYLE’S LAW P.P. #2BOYLE’S LAW P.P. #2A balloon is inflated to a volume of 12.6L on a A balloon is inflated to a volume of 12.6L on a
day when atmospheric pressure is 674mmday when atmospheric pressure is 674mmHgHg. . What’s th volume on a day when the pressure What’s th volume on a day when the pressure drops to 651mmdrops to 651mmHgHg? (Temp is constant)? (Temp is constant)
(BEFORE)(BEFORE) (AFTER)(AFTER)VV11 = 12.6 L = 12.6 L VV22 = = ??
PP11 = 674mm = 674mmHgHg PP22 = 651mm = 651mmHgHg
Relationship = inverse: Relationship = inverse: V V ↑↑ P P ↓ (so we will put ↓ (so we will put larger P on top to make volume increase)larger P on top to make volume increase)
A balloon is inflated to a volume of 12.6L on a A balloon is inflated to a volume of 12.6L on a day when atmospheric pressure is 674mmday when atmospheric pressure is 674mmHgHg. . What’s th volume on a day when the pressure What’s th volume on a day when the pressure drops to 651mmdrops to 651mmHgHg? (Temp is constant)? (Temp is constant)
(BEFORE)(BEFORE) (AFTER)(AFTER)VV11 = 12.6 L = 12.6 L VV22 = = ??
PP11 = 674mm = 674mmHgHg PP22 = 651mm = 651mmHgHg
Relationship = inverse: Relationship = inverse: V V ↑↑ P P ↓ (so we will put ↓ (so we will put larger P on top to make volume increase)larger P on top to make volume increase)
BOYLE’S LAW P.P. #2BOYLE’S LAW P.P. #2VV11 = 12.6 L = 12.6 L VV22 = ? = ? Relationship:
PP11 = = 674mm674mmHgHg PP22 = = 651mm651mmHgHg V V ↑↑ P P ↓↓
Solve for VSolve for V22::
12.6L12.6L x x 674mm674mmHg Hg = = 8,492.4L = 13.0 L13.0 L 11 651mm651mmHgHg 651
VV11 = 12.6 L = 12.6 L VV22 = ? = ? Relationship:
PP11 = = 674mm674mmHgHg PP22 = = 651mm651mmHgHg V V ↑↑ P P ↓↓
Solve for VSolve for V22::
12.6L12.6L x x 674mm674mmHg Hg = = 8,492.4L = 13.0 L13.0 L 11 651mm651mmHgHg 651
1) Put lonely known over 1
2) Follow the arrow! V V ↑↑
VV22 should go should go
UP, so put UP, so put larger P on top larger P on top of fraction!of fraction!
3) Do the math: cancel out units and multiply
4) Check your answer! Did it go up or down as expected? Units? SIG FIGS should match given DATA
HOMEWORK:HOMEWORK:
BACK OF NOTESHEET: Do #1-3!
WORKBOOK: Read pgs. 65 & 66: Charles’ Law & Guy Lussac’s Law
Do 1st Practice Problem pg.65 & pg.66
BACK OF NOTESHEET: Do #1-3!
WORKBOOK: Read pgs. 65 & 66: Charles’ Law & Guy Lussac’s Law
Do 1st Practice Problem pg.65 & pg.66
BOYLE’S LAW problemsBOYLE’S LAW problems1.1. A sample of COA sample of CO22 gas occupies a volume of gas occupies a volume of
8.75L8.75L at at 0.940 atm0.940 atm. If the temperature . If the temperature remains constant, calculate the volume remains constant, calculate the volume when pressure is when pressure is 1.00atm1.00atm..
VV11== VV22= V P= V P
PP11== PP22==
________ x __________ =________ x __________ = 11
1.1. A sample of COA sample of CO22 gas occupies a volume of gas occupies a volume of 8.75L8.75L at at 0.940 atm0.940 atm. If the temperature . If the temperature remains constant, calculate the volume remains constant, calculate the volume when pressure is when pressure is 1.00atm1.00atm..
VV11== VV22= V P= V P
PP11== PP22==
________ x __________ =________ x __________ = 11
BOYLE’S LAW problemsBOYLE’S LAW problems2.2. At STP a balloon has a volume of 1.75L. At STP a balloon has a volume of 1.75L.
When the balloon is moved to higher elev., When the balloon is moved to higher elev., the gas expands to 1900mL. What is the air the gas expands to 1900mL. What is the air pressure of the higher elevation?pressure of the higher elevation?
VV11== VV22== PP11== PP22== V P V P
________ x __________ =________ x __________ = 11
2.2. At STP a balloon has a volume of 1.75L. At STP a balloon has a volume of 1.75L. When the balloon is moved to higher elev., When the balloon is moved to higher elev., the gas expands to 1900mL. What is the air the gas expands to 1900mL. What is the air pressure of the higher elevation?pressure of the higher elevation?
VV11== VV22== PP11== PP22== V P V P
________ x __________ =________ x __________ = 11
BOYLE’S LAW problemsBOYLE’S LAW problems3. What will the new volume be if a 4.25mL 3. What will the new volume be if a 4.25mL
sample of gas collected at 0.75 atm is sample of gas collected at 0.75 atm is subjected to a new pressure of 1.25 atm?subjected to a new pressure of 1.25 atm?
VV11== VV22== PP11== PP22== V V
PP
________ x __________ =________ x __________ = 11
3. What will the new volume be if a 4.25mL 3. What will the new volume be if a 4.25mL sample of gas collected at 0.75 atm is sample of gas collected at 0.75 atm is subjected to a new pressure of 1.25 atm?subjected to a new pressure of 1.25 atm?
VV11== VV22== PP11== PP22== V V
PP
________ x __________ =________ x __________ = 11
CHARLES’ LAW NOTESCHARLES’ LAW NOTESIn the E12 computer simulation lab, you In the E12 computer simulation lab, you
raised the temperature of a confined gas, raised the temperature of a confined gas, and recorded its effects on the volume.and recorded its effects on the volume.
There is a There is a DIRECTDIRECT relationship between relationship between temperature and volume = Charles’ Lawtemperature and volume = Charles’ Law
When the temperature increases, the gas When the temperature increases, the gas expanded (volume also expanded (volume also increasesincreases). ). TT↑↑ V V↑↑
In the E12 computer simulation lab, you In the E12 computer simulation lab, you raised the temperature of a confined gas, raised the temperature of a confined gas, and recorded its effects on the volume.and recorded its effects on the volume.
There is a There is a DIRECTDIRECT relationship between relationship between temperature and volume = Charles’ Lawtemperature and volume = Charles’ Law
When the temperature increases, the gas When the temperature increases, the gas expanded (volume also expanded (volume also increasesincreases). ). TT↑↑ V V↑↑
CHARLES’ LAWCHARLES’ LAW
The same steps are followed when solving The same steps are followed when solving a Charles’ Law problem (dealing with a Charles’ Law problem (dealing with volume & temp at a constant pressure) volume & temp at a constant pressure) as you do for Boyle’s Law.as you do for Boyle’s Law.
Just remember to Just remember to convert temps to Kelvin!convert temps to Kelvin!Also Also Boyle’s Law is the only gas law that is Boyle’s Law is the only gas law that is
INVERSEINVERSE ( (VV↑↑ P P↓)↓)Charles Law & the upcoming Guy-Lussac’s Charles Law & the upcoming Guy-Lussac’s
Law are DIRECT relationships. (Law are DIRECT relationships. (TT↑↑ V V↑)↑)
The same steps are followed when solving The same steps are followed when solving a Charles’ Law problem (dealing with a Charles’ Law problem (dealing with volume & temp at a constant pressure) volume & temp at a constant pressure) as you do for Boyle’s Law.as you do for Boyle’s Law.
Just remember to Just remember to convert temps to Kelvin!convert temps to Kelvin!Also Also Boyle’s Law is the only gas law that is Boyle’s Law is the only gas law that is
INVERSEINVERSE ( (VV↑↑ P P↓)↓)Charles Law & the upcoming Guy-Lussac’s Charles Law & the upcoming Guy-Lussac’s
Law are DIRECT relationships. (Law are DIRECT relationships. (TT↑↑ V V↑)↑)
CHARLES’ LAW exampleCHARLES’ LAW example
10.L of H10.L of H22 gas at 20.˚C was heated to 100˚C. gas at 20.˚C was heated to 100˚C.
What was the new volume?What was the new volume?
VV11== 10.L10.L V V22= = ?? TT↑↑ V V↑↑
TT11== 20.˚C= 20.˚C= 293K293K T T22= 100˚C = = 100˚C = 373K373K
__10.L_____10.L___ x x ___293K_______293K____ = = 7.9L7.9L (2s.f. in data)(2s.f. in data)
11 373K 373K
10.L of H10.L of H22 gas at 20.˚C was heated to 100˚C. gas at 20.˚C was heated to 100˚C.
What was the new volume?What was the new volume?
VV11== 10.L10.L V V22= = ?? TT↑↑ V V↑↑
TT11== 20.˚C= 20.˚C= 293K293K T T22= 100˚C = = 100˚C = 373K373K
__10.L_____10.L___ x x ___293K_______293K____ = = 7.9L7.9L (2s.f. in data)(2s.f. in data)
11 373K 373K
GUY-LUSSAC’S LAWGUY-LUSSAC’S LAW
The The DIRECTDIRECT relationship between relationship between gas gas temperaturetemperature and and pressure.pressure.
As temperature increases, pressure As temperature increases, pressure also also increasesincreases (T (T↑↑ P P↑).↑).
Again, don’t forget to convert any ˚C Again, don’t forget to convert any ˚C to Kelvin by adding 273 before to Kelvin by adding 273 before using!using!
The The DIRECTDIRECT relationship between relationship between gas gas temperaturetemperature and and pressure.pressure.
As temperature increases, pressure As temperature increases, pressure also also increasesincreases (T (T↑↑ P P↑).↑).
Again, don’t forget to convert any ˚C Again, don’t forget to convert any ˚C to Kelvin by adding 273 before to Kelvin by adding 273 before using!using!
GUY-LUSSAC’S exampleGUY-LUSSAC’S example
A sample of Ne gas with a pressure of 600.mmA sample of Ne gas with a pressure of 600.mmHgHg and a temperature of 100.˚C was allowed to drop and a temperature of 100.˚C was allowed to drop to half the pressure. What was the new to half the pressure. What was the new temperature in K? ˚C?temperature in K? ˚C?
PP11== 600.mm600.mmHgHg P P22= = 300.mm300.mmHgHg
TT11== 100.˚C= 100.˚C= 373K373K T T22= = ?? TT↓↓ P P ↓↓
__373K_____373K___ x x _300.mm_300.mmHgHg = = 187 K187 K = = 460.˚C460.˚C
11 600.mm 600.mmHgHg (3s.f. in all given data) (3s.f. in all given data)
A sample of Ne gas with a pressure of 600.mmA sample of Ne gas with a pressure of 600.mmHgHg and a temperature of 100.˚C was allowed to drop and a temperature of 100.˚C was allowed to drop to half the pressure. What was the new to half the pressure. What was the new temperature in K? ˚C?temperature in K? ˚C?
PP11== 600.mm600.mmHgHg P P22= = 300.mm300.mmHgHg
TT11== 100.˚C= 100.˚C= 373K373K T T22= = ?? TT↓↓ P P ↓↓
__373K_____373K___ x x _300.mm_300.mmHgHg = = 187 K187 K = = 460.˚C460.˚C
11 600.mm 600.mmHgHg (3s.f. in all given data) (3s.f. in all given data)
Unknown ↓, follow the arrow!
COMBINED GAS LAWCOMBINED GAS LAWThe previous 3 laws (Boyle’s, Charles’, & G-L) The previous 3 laws (Boyle’s, Charles’, & G-L)
require one variable to remain constant.require one variable to remain constant.Obviously, all 3 main variables can influence Obviously, all 3 main variables can influence
eachother. If temperature is changed it will eachother. If temperature is changed it will usually not only influence volume, but also usually not only influence volume, but also pressure.pressure.
The relationship among all 3 variables is the The relationship among all 3 variables is the Combined Gas LawCombined Gas Law: : PP11VV11 = = PP22VV22
TT11 T T22 This can also be solved by conversion….This can also be solved by conversion….We’ll multiply by We’ll multiply by 2 2 fractions, fractions, 1 for each known pair1 for each known pair
The previous 3 laws (Boyle’s, Charles’, & G-L) The previous 3 laws (Boyle’s, Charles’, & G-L) require one variable to remain constant.require one variable to remain constant.
Obviously, all 3 main variables can influence Obviously, all 3 main variables can influence eachother. If temperature is changed it will eachother. If temperature is changed it will usually not only influence volume, but also usually not only influence volume, but also pressure.pressure.
The relationship among all 3 variables is the The relationship among all 3 variables is the Combined Gas LawCombined Gas Law: : PP11VV11 = = PP22VV22
TT11 T T22 This can also be solved by conversion….This can also be solved by conversion….We’ll multiply by We’ll multiply by 2 2 fractions, fractions, 1 for each known pair1 for each known pair
COMBINED GAS LAW practiceCOMBINED GAS LAW practice
10. L of Ar gas with a pressure of 200.torr & a 10. L of Ar gas with a pressure of 200.torr & a temperature of 400.K was altered to 450. torr temperature of 400.K was altered to 450. torr and 600.K. What’s the new volume?and 600.K. What’s the new volume?
VV11 = 10.L = 10.L VV22 = = ??
PP11 = 200.torr = 200.torr PP22 = 450.torr = 450.torr
TT11 = 400. K = 400. K TT22 = 600. K = 600. K
PP↑↑V↓V↓ T↑ T↑V↑V↑ 10. L 10. L x x 200.torr200.torr x x 600. K600. K = = 1,200,00L1,200,00L = = 6.7 L6.7 L 1 450.torr 400. K 1 450.torr 400. K 180,000180,000 (2 s.f. in 1 data) (2 s.f. in 1 data)
10. L of Ar gas with a pressure of 200.torr & a 10. L of Ar gas with a pressure of 200.torr & a temperature of 400.K was altered to 450. torr temperature of 400.K was altered to 450. torr and 600.K. What’s the new volume?and 600.K. What’s the new volume?
VV11 = 10.L = 10.L VV22 = = ??
PP11 = 200.torr = 200.torr PP22 = 450.torr = 450.torr
TT11 = 400. K = 400. K TT22 = 600. K = 600. K
PP↑↑V↓V↓ T↑ T↑V↑V↑ 10. L 10. L x x 200.torr200.torr x x 600. K600. K = = 1,200,00L1,200,00L = = 6.7 L6.7 L 1 450.torr 400. K 1 450.torr 400. K 180,000180,000 (2 s.f. in 1 data) (2 s.f. in 1 data)
Make 2 separate fractions, follow the arrow for the basic gas law that governs the unknown variable & that known.
Homework:Homework:
Questions on back of notes: #4-12
READ about Dalton’s Law in workbook, try to do questions #13-15.
KEEP IN MIND: THE GAS LAWS TEST IS THIS WEDNESDAY, 10/19!
Tomorrow we’ll finish notes & do practice test in class. Start studying!
Questions on back of notes: #4-12
READ about Dalton’s Law in workbook, try to do questions #13-15.
KEEP IN MIND: THE GAS LAWS TEST IS THIS WEDNESDAY, 10/19!
Tomorrow we’ll finish notes & do practice test in class. Start studying!
DALTON’S LAWDALTON’S LAW
Sometimes, the best way to isolate a gas Sometimes, the best way to isolate a gas is by using water. This, however, can is by using water. This, however, can influence the gas’ pressure.influence the gas’ pressure.
Look for words like “Look for words like “over waterover water” or “” or “wetwet””To account for that, To account for that, we must subtract the we must subtract the
water vapor pressurewater vapor pressure (using a chart). (using a chart).Then we can just solve it as a combined Then we can just solve it as a combined
gas law problem! gas law problem!
Sometimes, the best way to isolate a gas Sometimes, the best way to isolate a gas is by using water. This, however, can is by using water. This, however, can influence the gas’ pressure.influence the gas’ pressure.
Look for words like “Look for words like “over waterover water” or “” or “wetwet””To account for that, To account for that, we must subtract the we must subtract the
water vapor pressurewater vapor pressure (using a chart). (using a chart).Then we can just solve it as a combined Then we can just solve it as a combined
gas law problem! gas law problem!
QuickTime™ and a decompressor
are needed to see this picture.
Water Vapor Pressure Chart(pg.21 in little white booklet)Water Vapor Pressure Chart(pg.21 in little white booklet)
QuickTime™ and a decompressor
are needed to see this picture.
DALTON’S LAW practiceDALTON’S LAW practice 15 L of Cl15 L of Cl2 2 gas was collected gas was collected over water over water at a at a
temperature of 14.0˚C and a pressure of 752mmtemperature of 14.0˚C and a pressure of 752mmHgHg. . What would the What would the drydry volume be at STP? volume be at STP? PP11 = = 752mm752mmHgHg P P22 = =stpstp= mm= mmHgHg
VV1(wet)1(wet)= = 15L 15L - - = = V V2(dry)2(dry)= = ??
TT11 = 14.0˚C = 14.0˚C + = K T T22 = =stpstp= K= K
PP↑↑V__V__ T↓ T↓V__V__
___L ___L x x _______mm_______mmHg Hg x x _____ K_____ K = = 14 L14 L
1 1 mmmmHgHg K K (2 s.f. in 1 data)(2 s.f. in 1 data)
15 L of Cl15 L of Cl2 2 gas was collected gas was collected over water over water at a at a temperature of 14.0˚C and a pressure of 752mmtemperature of 14.0˚C and a pressure of 752mmHgHg. . What would the What would the drydry volume be at STP? volume be at STP? PP11 = = 752mm752mmHgHg P P22 = =stpstp= mm= mmHgHg
VV1(wet)1(wet)= = 15L 15L - - = = V V2(dry)2(dry)= = ??
TT11 = 14.0˚C = 14.0˚C + = K T T22 = =stpstp= K= K
PP↑↑V__V__ T↓ T↓V__V__
___L ___L x x _______mm_______mmHg Hg x x _____ K_____ K = = 14 L14 L
1 1 mmmmHgHg K K (2 s.f. in 1 data)(2 s.f. in 1 data)
DALTON’S LAW practiceDALTON’S LAW practice 15 L of Cl15 L of Cl2 2 gas was collected gas was collected over water over water at a at a
temperature of 14.0˚C and a pressure of 752mmtemperature of 14.0˚C and a pressure of 752mmHgHg. . What would the What would the drydry volume be at STP? volume be at STP? PP11 = = 752mm752mmHgHg PP22 = =stpstp= 760mm= 760mmHgHg
VV1(wet)1(wet)= = 15L 15L - - 11.9911.99 (chart)(chart)= = 3L3L VV2(dry)2(dry)= = ??
TT11 = 14.0˚C = 14.0˚C +273= 287K TT22 = =stpstp= 273 K= 273 K
PP↑↑V↓V↓ T↓ T↓V↓V↓
3 L 3 L x x 752mm752mmHg Hg x x 273. K273. K = = 2.8 L2.8 L
1 760mm1 760mmHgHg 287. K 287. K (2 s.f. in 1 data)(2 s.f. in 1 data)
15 L of Cl15 L of Cl2 2 gas was collected gas was collected over water over water at a at a temperature of 14.0˚C and a pressure of 752mmtemperature of 14.0˚C and a pressure of 752mmHgHg. . What would the What would the drydry volume be at STP? volume be at STP? PP11 = = 752mm752mmHgHg PP22 = =stpstp= 760mm= 760mmHgHg
VV1(wet)1(wet)= = 15L 15L - - 11.9911.99 (chart)(chart)= = 3L3L VV2(dry)2(dry)= = ??
TT11 = 14.0˚C = 14.0˚C +273= 287K TT22 = =stpstp= 273 K= 273 K
PP↑↑V↓V↓ T↓ T↓V↓V↓
3 L 3 L x x 752mm752mmHg Hg x x 273. K273. K = = 2.8 L2.8 L
1 760mm1 760mmHgHg 287. K 287. K (2 s.f. in 1 data)(2 s.f. in 1 data)
DO NOW!DO NOW!
E13 GAS LAWS PRACTICE TEST: ALL Workbook page 69
Use booklet page 20 &21 for reference. Show work!!! Keep units straight so you can cancel out. CIRCLE FINAL ANSWERS CHECK answers at front desk now or online tonight
Mhschemistry.wordpress.com
E13 GAS LAWS TEST = TOMORROW!!! E13 problems & pract test will be collected
E13 GAS LAWS PRACTICE TEST: ALL Workbook page 69
Use booklet page 20 &21 for reference. Show work!!! Keep units straight so you can cancel out. CIRCLE FINAL ANSWERS CHECK answers at front desk now or online tonight
Mhschemistry.wordpress.com
E13 GAS LAWS TEST = TOMORROW!!! E13 problems & pract test will be collected
