The Gas The Gas Laws Laws
Mar 31, 2015
The Gas LawsThe Gas Laws
The Gas LawsThe Gas LawsBoyle’s LawBoyle’s LawAmonton’s LawAmonton’s LawCharles’s LawCharles’s LawCombined Gas Combined Gas
LawLaw
Gay-Lussac’s Gay-Lussac’s LawLaw
Avogadro’s LawAvogadro’s LawDalton’s LawDalton’s Law
Boyle’s LawBoyle’s Law
At a constant temperature, pressure is At a constant temperature, pressure is inverselyinversely proportional to volume. proportional to volume.
Pre
ssur
eP
ress
ure
VolumeVolume
Boyle’s LawBoyle’s Law
At a constant temperature, pressure is At a constant temperature, pressure is inverselyinversely proportional to volume. proportional to volume.
1/P
ress
ure
1/P
ress
ure
VolumeVolume
Boyle’s LawBoyle’s Law
At a constant temperature, pressure At a constant temperature, pressure is is inverselyinversely proportional to proportional to volume.volume.
PP11VV11 = P = P22VV22
PV = kPV = kP P 11
VV
Year: 1662Year: 1662
Amonton’s LawAmonton’s Law““Air thermometer”, 1695.Air thermometer”, 1695.
This is This is notnot Gay-Lussac’s law. Gay-Lussac’s law.
P P TT
PP11
TT11==
PP22
TT22
Diagram from http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch4/gaslaws3.htmlDiagram from http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch4/gaslaws3.html
Amonton’s LawAmonton’s Law
P P TT
PP11
TT11==
PP22
TT22
This is why you measure This is why you measure your tire pressure when your tire pressure when the tire is cold. Tire the tire is cold. Tire pressures vary pressures vary with temperature.with temperature.
Measure the pressures of a gas Measure the pressures of a gas at various temperatures at a at various temperatures at a
constant volume.constant volume.
Amonton’s air thermometer Amonton’s air thermometer was used to find the value of was used to find the value of absolute zero.absolute zero.
Amonton’s LawAmonton’s Law
Finding Absolute ZeroFinding Absolute ZeroP
ress
ure
Pre
ssur
e
Temperature (C)Temperature (C)-300 -150 0 100 200 300 -300 -150 0 100 200 300
-273 C-273 C
Extrapolate to the x-axisExtrapolate to the x-axis
Charles’s LawCharles’s Law
At constant pressure, volume is At constant pressure, volume is directly proportional to temp.directly proportional to temp.
Vol
ume
Vol
ume
TemperatureTemperature
Charles’s LawCharles’s LawAt constant pressure, volume is At constant pressure, volume is directly proportional to directly proportional to temperature.temperature.
V V TT
VVTT = k= kVV11
TT11==
VV22
TT22
Charles’s LawCharles’s Law• Studied gases during 1780’s.Studied gases during 1780’s.• Hydrogen balloon assents, 3000 m, Hydrogen balloon assents, 3000 m,
in 1783.in 1783.• Collaborated with the Montgolfier Collaborated with the Montgolfier
brothers on hot air balloons, 1783.brothers on hot air balloons, 1783.• Charles’s gas studies published by Charles’s gas studies published by
Gay-Lussac in1802.Gay-Lussac in1802.
Charles’s Charles’s LawLaw
Hydrogen Hydrogen balloon assent, balloon assent, 3000 m, 1783.3000 m, 1783.
Combined Gas LawCombined Gas Law
givesgives P P TTVV
P P 11VV
P P TT
Combined Gas LawCombined Gas Law
P P TTVV
Next we convert Next we convert
equation by adding a constant. equation by adding a constant.
to anto an
P P kTkTVV
Combined Gas LawCombined Gas Law
Rearranging the equation gives:Rearranging the equation gives:
= k= kPVPVTT
P P kTkTVV
Combined Gas LawCombined Gas Law
Combining the Combining the laws of Boyle, laws of Boyle, Amonton and Amonton and Charles produces Charles produces the combined gas the combined gas law.law.
= k= kPVPVTT
Combined Gas LawCombined Gas Law
PP11VV11
TT11
== kkPP22VV22
TT22
== kk
Consider a confined gas at two sets Consider a confined gas at two sets of conditions. Since the number of of conditions. Since the number of molecules is the same and the molecules is the same and the values of k are the same, then we values of k are the same, then we can combine the two equations.can combine the two equations.
Combined Gas LawCombined Gas Law
PP11VV11
TT11
==PP22VV22
TT22
PP11VV11
TT11
== kkPP22VV22
TT22
== kk
Combined Gas LawCombined Gas Law
PP11VV11
TT11
==PP22VV22
TT22
Use the combined gas law Use the combined gas law whenever you are asked to find a whenever you are asked to find a new P, V or T after changes to a new P, V or T after changes to a confined gas.confined gas.
Sample Combined Gas Law ProblemSample Combined Gas Law Problem
Consider a confined gas in a Consider a confined gas in a cylinder with a movable cylinder with a movable piston. The pressure is 0.950 piston. The pressure is 0.950 atm. Find the new pressure atm. Find the new pressure when the volume is reduced when the volume is reduced from 100.0 mL to 65.0 mL, from 100.0 mL to 65.0 mL, while the temperature while the temperature remains constant?remains constant?
100 mL
65 mL
Sample Combined Gas Law ProblemSample Combined Gas Law Problem
Start with the equation for Start with the equation for the combined gas law.the combined gas law.
100 mL
65 mL
Sample Combined Gas Law ProblemSample Combined Gas Law Problem
100 mL
65 mL
PP11VV11
TT11
==PP22VV22
TT22
Since the temperature is Since the temperature is constant, we can cancel constant, we can cancel out Tout T11 and T and T22..
Sample Combined Gas Law ProblemSample Combined Gas Law Problem
100 mL
65 mL
PP11VV11 == PP22VV22
Next, solve Next, solve for Pfor P22..
This becomes This becomes Boyle’s LawBoyle’s Law
PP11VV11==VV22
PP22
Sample Combined Gas Law ProblemSample Combined Gas Law Problem
100 mL
65 mL
PP11VV11==VV22
PP22
(0.950 atm)(100.0 mL)(0.950 atm)(100.0 mL)==
65.0 mL65.0 mLPP22
==PP22 1.46 atm1.46 atm
Combined Gas Law ProblemsCombined Gas Law Problems
1. A sample of neon gas has a 1. A sample of neon gas has a volume of 2.00 L at 20.0 C and volume of 2.00 L at 20.0 C and 0.900 atm. What is the new 0.900 atm. What is the new pressure when the volume is pressure when the volume is reduced to 0.750 L and the reduced to 0.750 L and the temperature increases to 24.0 C?temperature increases to 24.0 C?
The answer isThe answer is 2.43 atm2.43 atm
Combined Gas Law ProblemsCombined Gas Law Problems
2. Some “left over” propane gas in a 2. Some “left over” propane gas in a rigid steel cylinder has a pressure of rigid steel cylinder has a pressure of 24.6 atm at a temperature of 20.C. 24.6 atm at a temperature of 20.C. When thrown into a campfire the When thrown into a campfire the temperature in the cylinder rises to temperature in the cylinder rises to 313C. What will be the pressure of 313C. What will be the pressure of the propane?the propane?
The answer isThe answer is 49.2 atm49.2 atm
Combined Gas Law ProblemsCombined Gas Law Problems
3. Consider the fuel mixture in the 3. Consider the fuel mixture in the cylinder of a diesel engine. At its cylinder of a diesel engine. At its maximum, the volume is 816 cc. maximum, the volume is 816 cc. The mixture comes in at 0.988 atm The mixture comes in at 0.988 atm and 31 C. What will be the and 31 C. What will be the temperature (in C) when the gas is temperature (in C) when the gas is compressed to 132 cc and 42.4 atm?compressed to 132 cc and 42.4 atm?
The answer isThe answer is 1837 C1837 C
Gay-Lussac’s LawGay-Lussac’s Law
Year: 1802Year: 1802
At a given temperature and pressure, At a given temperature and pressure, the volumes of reacting gases are in the volumes of reacting gases are in a ratio of small, whole numbers.a ratio of small, whole numbers.
Also known as the Also known as the law of combining volumeslaw of combining volumes..
Gay-Lussac’s LawGay-Lussac’s Law
Gay-Lussac found that the volumes Gay-Lussac found that the volumes of gases in a reaction were in ratios of gases in a reaction were in ratios of small, whole numbers.of small, whole numbers.
2 H2 H22(g) + O(g) + O22(g) (g) 2 H 2 H22O(g)O(g)
200 mL of hydrogen reacts 200 mL of hydrogen reacts with 100 mL of oxygen.with 100 mL of oxygen. 2:12:1
Gay-Lussac’s LawGay-Lussac’s Law
2 H2 H22(g) + O(g) + O22(g) (g) 2 H 2 H22O(g)O(g)
The ratio of volumes of The ratio of volumes of gases come from the ratios gases come from the ratios
of the coefficients in the of the coefficients in the balanced equation.balanced equation.
Avogadro’s LawAvogadro’s Law
Equal volumes of gases at Equal volumes of gases at the same temperature and the same temperature and pressure have equal pressure have equal numbers of molecules.numbers of molecules.
VVnn = k= k
Avogadro’s law followed Dalton’s Avogadro’s law followed Dalton’s atomic theory and Gay-Lussac’s law.atomic theory and Gay-Lussac’s law.Year: 1811.Year: 1811. V V nn
VV11
nn11==
VV22
nn22
Dalton’s LawDalton’s LawDalton’s law of partial pressures Dalton’s law of partial pressures
deals with mixtures of gases. deals with mixtures of gases.
PPtotaltotal = P = P11 + P + P22 + P + P33 … …Use when dealing with the pressure of Use when dealing with the pressure of HH22O(g) when collecting a gas over water.O(g) when collecting a gas over water.
The total pressure is the sum of the The total pressure is the sum of the partial pressures:partial pressures:
Dalton’s LawDalton’s Law
Inverted gas Inverted gas collecting bottlecollecting bottle
Flask with Flask with metal and metal and
HC lHC l
Rubber tubing Rubber tubing carrying Hcarrying H22 gas gas
Water in troughWater in trough
““Collecting a gas over water”Collecting a gas over water”
Dalton’s LawDalton’s Law
Inverted gas Inverted gas collecting bottlecollecting bottle
Flask with Flask with metal and metal and
HC lHC l
Rubber tubing Rubber tubing carrying Hcarrying H22 gas gas
Water in troughWater in trough
““Collecting a gas over water”Collecting a gas over water”
Dalton’s LawDalton’s Law
Flask with Flask with metal and metal and
HC lHC lWater in troughWater in trough
Water is displaced through the mouth Water is displaced through the mouth of the bottle as Hof the bottle as H22 gas bubbles in. gas bubbles in.
HH22 gasgas
Dalton’s LawDalton’s Law
Flask with Flask with metal and metal and
HC lHC lWater in troughWater in trough
A mixture of hydrogen gas and water A mixture of hydrogen gas and water vapor is in the collection bottle.vapor is in the collection bottle.
HH22 gasgas
Dalton’s LawDalton’s Law
Flask with Flask with metal and metal and
HC lHC lWater in troughWater in trough
The pressure of the mixture is the sum The pressure of the mixture is the sum of the pressures of Hof the pressures of H22O gas and HO gas and H22 gas. gas.
HH22 gasgas
Dalton’s LawDalton’s Law
Flask with Flask with metal and metal and
HC lHC lWater in troughWater in trough
PPtotaltotal = P = PHH22OO + P + PHH22
HH22 gasgas
Sample ProblemSample Problem
Flask with Flask with metal and metal and
HC lHC lWater in troughWater in trough
The total pressure in the bottle is The total pressure in the bottle is 712.7 torr. The temperature is 19 C. 712.7 torr. The temperature is 19 C. What is the HWhat is the H22 pressure? pressure?
HH22 gasgas
Flask with Flask with metal and metal and
HC lHC lWater in troughWater in trough
At 19C, the vapor pressure of water At 19C, the vapor pressure of water is 16.5 torr. The His 16.5 torr. The H22 pressure is … pressure is …
HH22 gasgas
Sample ProblemSample Problem
Flask with Flask with metal and metal and
HC lHC lWater in troughWater in trough
PPHH22 = P = Ptotaltotal - P - PHH22OO
HH22 gasgas
PPHH22 = 712.7 = 712.7 torrtorr – 16.5 – 16.5 torrtorr = 696.2 = 696.2 torrtorr
Sample ProblemSample Problem
Another ProblemAnother ProblemA pressure sensor is attached to a A pressure sensor is attached to a sealed flask containing 80.0 mL of sealed flask containing 80.0 mL of mercury at 50.0 C. The pressure mercury at 50.0 C. The pressure sensor indicates 693 torr. The sensor indicates 693 torr. The pressure of the air inside the empty pressure of the air inside the empty flask at 50.0 C is 684 torr. flask at 50.0 C is 684 torr. Determine the vapor pressure of Determine the vapor pressure of mercury at 50.0 C.mercury at 50.0 C.
Another ProblemAnother ProblemA pressure sensor is A pressure sensor is attachedattached to a to a sealed flask containing 80.0 mL of sealed flask containing 80.0 mL of mercury at 50.0 C. The pressure mercury at 50.0 C. The pressure sensor indicates 693 torr. The sensor indicates 693 torr. The pressure of the air inside the empty pressure of the air inside the empty flask at 50.0 C is 684 torr. flask at 50.0 C is 684 torr. Determine the vapor pressure of Determine the vapor pressure of mercury at 50.0 C.mercury at 50.0 C.
PPtotaltotal = P = P11 + P + P22
Another ProblemAnother ProblemA pressure sensor is A pressure sensor is attachedattached to a to a sealed flask containing 80.0 mL of sealed flask containing 80.0 mL of mercury at 50.0 C. The pressure mercury at 50.0 C. The pressure sensor indicates 693 torr. The sensor indicates 693 torr. The pressure of the air inside the empty pressure of the air inside the empty flask at 50.0 C is 684 torr. flask at 50.0 C is 684 torr. Determine the vapor pressure of Determine the vapor pressure of mercury at 50.0 C.mercury at 50.0 C.
PPtotaltotal = P = Pairair+ P+ PHgHg
Another ProblemAnother ProblemA pressure sensor is A pressure sensor is attachedattached to a to a sealed flask containing 80.0 mL of sealed flask containing 80.0 mL of mercury at 50.0 C. The pressure mercury at 50.0 C. The pressure sensor indicates 693 torr. The sensor indicates 693 torr. The pressure of the air inside the empty pressure of the air inside the empty flask at 50.0 C is 684 torr. flask at 50.0 C is 684 torr. Determine the vapor pressure of Determine the vapor pressure of mercury at 50.0 C.mercury at 50.0 C.
PPHgHg = P = Ptotal total - P- Pairair
Another ProblemAnother ProblemA pressure sensor is A pressure sensor is attachedattached to a to a sealed flask containing 80.0 mL of sealed flask containing 80.0 mL of mercury at 50.0 C. The pressure mercury at 50.0 C. The pressure sensor indicates 693 torr. The sensor indicates 693 torr. The pressure of the air inside the empty pressure of the air inside the empty flask at 50.0 C is 684 torr. flask at 50.0 C is 684 torr. Determine the vapor pressure of Determine the vapor pressure of mercury at 50.0 C.mercury at 50.0 C.
PPHgHg = = 693 693 torrtorr – 684 – 684 torrtorr
Another ProblemAnother ProblemA pressure sensor is A pressure sensor is attachedattached to a to a sealed flask containing 80.0 mL of sealed flask containing 80.0 mL of mercury at 50.0 C. The pressure mercury at 50.0 C. The pressure sensor indicates 693 torr. The sensor indicates 693 torr. The pressure of the air inside the empty pressure of the air inside the empty flask at 50.0 C is 684 torr. flask at 50.0 C is 684 torr. Determine the vapor pressure of Determine the vapor pressure of mercury at 50.0 C.mercury at 50.0 C.
PPHgHg = = 99 torrtorr
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