GAS LAWS Kinetic Molecular Theory Particles in an ideal gas… –have no volume. –have elastic...
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Transcript of GAS LAWS Kinetic Molecular Theory Particles in an ideal gas… –have no volume. –have elastic...
GAS LAWSGAS LAWS
Kinetic Molecular TheoryKinetic Molecular Theory
• Particles in an ideal gas…– have no volume.– have elastic collisions. – are in constant, random, straight-line motion.– don’t attract or repel each other.– have an average KE directly related to Kelvin
temperature.
Real GasesReal Gases• Particles in a REAL gas…
– have their own volume– attract and repel each other
• Gas behavior is most ideal…– at low pressures– at high temperatures
***Most real gases act like ideal gases except under high pressure and low temperature.
Characteristics of GasesCharacteristics of Gases• Gases expand to fill any container.
– Take the shape and volume of their container.
• Gases are fluids (like liquids).– Little to no attraction between the particles
• Gases have very low densities.= lots of empty space between the particles
Characteristics of GasesCharacteristics of Gases• Gases can be compressed.
– lots of empty space between the particles– Indefinite density
• Gases undergo diffusion.– random motion – scatter in all directions
PressurePressure
area
forcepressure
Which shoes create the most pressure?
Pressure- how much a gas is pushing
on a container.
• Atmospheric pressure- atmospheric gases push on everything on Earth
• UNITS AT SEA LEVEL
1 atm =101.3 kPa (kilopascal)= 760 mmHg =760 torr
PressurePressure• Barometer
– measures atmospheric pressure
Mercury Barometer
Aneroid Barometer
PressurePressure• Manometer
– measures contained gas pressure
C. JohannessonU-tube Manometer Bourdon-tube gauge
C. Johannesson
Temperature= how fast the molecules are moving
ºF
ºC
K
-459 32 212
-273 0 100
0 273 373
K = ºC + 273
• Always use absolute temperature (Kelvin) when working with gases.
Standard Temperature & PressureStandard Temperature & Pressure
0°C 273 K
1 atm 101.3 kPa
760 mm Hg
-OR-
STP
-OR-
Volume = how much space a gas occupies
Units– L, mL, cm3
• 1000 mL = 1 L • 1 mL = 1 cm3
BASIC GAS LAWS
Charles’ LawCharles’ Law
• T V (temperature is directly proportional to volume)
• T ↑ V↑ & T↓ V↓
• V1 = V2
T1 T2 T is always in K
– K = °C + 273
– P and n = constant
V
T
CharlesLaw.exe
Timberlake, Chemistry 7th Edition, page 259
(Pressure is held constant)
T1 T2
V1 V2=
Charles’ Law
Charles’ Law
Timberlake, Chemistry 7th Edition, page 254
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
Charles’ LawCharles’ LawThe egg out of the bottle
• Mrs. Rodriguez inflates a balloon for a party. She is in an air-conditioned room at 27.0oC, and the balloon has a volume of 4.0 L. Because she is a curious and intrepid chemistry teacher, she heats the balloon to a temperature of 57.0oC. What is the new volume of the balloon if the pressure remains constant?
Charles’ Law ProblemCharles’ Law Problem
Given Unkown Equation
Substitute and Solve
T1 = 27.0oC +273= 300 K V1 = 4.0 L T2 = 57.0oC +273= 330 K
V2 = ? L P1V1 = P2V2
T1 V1T2
4.0 L = V2 =300 K 330K
4.4 L
• A 25 L balloon is released into the air on a warm afternoon (42º C). The next morning the balloon is recovered on the ground. It is a very cold morning and the balloon has shrunk to 22 L. What is the temperature in º C?
Charles’ Law Learning CheckCharles’ Law Learning Check
Given Unkown Equation
Substitute and Solve
V1 = 25 L T1 = 42 oC +273= 315 K V2 = 22 L
T2 = ? ºC P1V1 = P2V2
T1 V1T2
25 L = 22 L =315 K T2
277.2 K -273 = 4.2 ºC
Boyle’s LawBoyle’s Law
• P↓ V ↑ & P↑ V ↓• P 1/V (pressure is inversely proportional to volume)
• P1V1 = P2V2– T and n = constant
P
V
Boyle'sLaw.exe
Timberlake, Chemistry 7th Edition, page 253
P1V1 = P2V2
(Temperature is held constant)
Boyle’s LawBoyle’s Law
Timberlake, Chemistry 7th Edition, page 254
Marshmallows in a vacuum
Boyle’s LawBoyle’s Law
Mechanics of Breathing
Timberlake, Chemistry 7th Edition, page 254
Boyle’s LawBoyle’s Law
A balloon is filled with 30.L of helium gas at 1.00 atm. What is the volume when the balloon rises to an altitude where the pressure is only 0.25 atm?
Given Unkown Equation
• Substitute and Solve
Boyle’s Law ProblemBoyle’s Law Problem
V2 0.25 atm = 30 L x 1.0 atm = 120 L
V1 = 30 LP1 = 1 atmP2 = .25atm
V2 = ? L P1V1 = P2V2
T1 T2
A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa.
Given Unkown Equation
•Substitute and Solve
Boyle’s Law Learning CheckBoyle’s Law Learning Check
V1 = 100. mL = 0.100 LP1 = 150. kPaP2 = 200. kPa
V2 = ? L P1V1 = P2V2
T1 T2
AVOGADRO’S LAWAVOGADRO’S LAW• Vn Vn • V n (direct)
• V1 = V2
n1 n2
– T & P Constant
Avogadro'sLaw.exe
n
V
A 3.0 liter sample of gas contains 7.0 moles. How much gas will there be, in order for the sample to be 2.3 liters? P & T do not change
Given Unkown Equation
• Substitute and Solve
Avogadro’s Law ProblemAvogadro’s Law Problem
V1 = 3.0 Ln1 = 7.0 molV2 = 2.3 L
n2 = ? mol P1V1 = P2V2
n1T1 n2T2
3.0 L = 2.3 L =7.0 mol n2 mol
5.4 mol
Gay-Lussac’s LawGay-Lussac’s Law
• P1 = P2
T1 T2
– V & n constant
• Direct relationship
• PT PT
P
T
Gay-Lussac'sLaw.exe
Gay-Lussac LawGay-Lussac Law• Collapsing Barrel
Gay-LussacGay-Lussac LawLaw• Tank car implosion
COMBINED IDEAL GAS LAWCOMBINED IDEAL GAS LAW
• P1V1 = P2V2
n1T1 n2T2
• If P, V, n, or T are constant then they cancel out of the equation.
• n usually constant (unless you add or remove gas), so
• P1V1 = P2V2
T1 T2
Combined Gas Law ProblemCombined Gas Law Problem• Ms. Evans travels to work in a hot air balloon from the Rocky
Mountains. At her launch site, the temperature is 5.00 °C, the atmospheric pressure is 0.801 atm, and the volume of the air in the balloon is 120.0 L. When she lands in Plano, the temperature is 28.0 °C and the atmospheric pressure is 101.3 kPa. What is the new volume of the air in the balloon?
Given Unkown Equation
Substitute and Solve V2 x 1 atm = 120.0 L x 0.801 atm = 104 L 301K 278 K
T1 = 5.0oC +273= 278 KP1 = 0.801 atmV1 = 120.0 LT2 = 28.0oC +273= 301 KP2 = 101.3 kPa = 1 atm
V2 = ? L V1 x P1 = V 2 x P2 T 1 T 2
Combined Gas Law Learning CheckCombined Gas Law Learning Check• Nitrogen gas is in a 7.51 L container at 5.C and 0.58 atm.
What is the new volume of the gas at STP?
Given Unkown Equation
Substitute and Solve
V2 x 1.0 atm = 7.51L x 0.58 atm = 4.3 L 273 K 278 K
T1 = 5.0oC +273= 278 KP1 = 0.58 atmV1 = 7.51 LT2 = 273 KP2 = 1 atm
V2 = ? L V1 x P1 = V 2 x P2 T 1 T 2
Ideal Gas Law (“Pivnert”)Ideal Gas Law (“Pivnert”)PV=nRT R = The Ideal Gas Constant R = 0.0821 (L*atm) R = 62.4 (L*mm Hg) (mol*K) (mol*K) R = 8.31 (L*kPa) (mol*K)•V has to be in Liters, n in Moles, T in Kelvin, •P can be in atm, kPa or mmHg•* Choose which R to used based on the units of your pressure.
P V = n R T (atm) (L) = (moles) (L*atm/mol*K) (K) (kPa) (L) = (moles) (L*kPa/mol*K) (K) mm Hg (L) = (moles) (L*mmHg/mol*K) (K)
Ideal Gas Law ProblemIdeal Gas Law Problem• A rigid steel cylinder with a volume of 20.0 L is filled with nitrogen
gas to a final pressure of 200.0 atm at 27.0 oC. How many moles of gas does the cylinder hold?
Given Unkown Equation
Substitute and Solve
n 0821 atm L/K Mole x 300 K = 200.0 atm x 20.0L= 162 moles
V = 20.0 LP = 200.0 atmT =27.0oC +273= 300 K
moles of nitrogen?
PV=nRT R= .0821 atm L/K Mole
Ideal Gas Law Learning CheckIdeal Gas Law Learning Check• A balloon contains 2.00 mol of nitrogen at a pressure of 0.980
atm and a temperature of 37C. What is the volume of the balloon?
Given Unkown Equation
Substitute and Solve
0.980 atm x V= 2.00 mol x .0821 atm L/K Mole x 310 K = 51.9 L
n = 2.00 molP = 0.980 atmT =37.0oC +273= 310 K
V in L? PV=nRT R= .0821 atm L/K Mole
Dalton’s Law of Partial PressureDalton’s Law of Partial Pressure• The total pressure of a mixture of gases is
equal to the sum of the partial pressures of the component gases.
• Ptotal = Pgas 1 + Pgas 2 + P gas 3 + …
A metal container holds a mixture of 2.00 atm of nitrogen, 1.50 atm of oxygen and 3.00 atm of helium. What is the total pressure in the canister?
6.5 atm
Welcome to Mole Welcome to Mole IslandIsland
Welcome to Mole Welcome to Mole IslandIsland
Welcome to Mole Welcome to Mole IslandIsland
Gas StoichiometryGas StoichiometryMoles Moles Liters of a Gas: Liters of a Gas:
– 22C4H10 (g) + 13O2(g) O8CO2(g) + 10H2O(g)
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
2 mol + 13 mol ? 8 mol + 10 mol
Avogadro’s principle states that one mole of any gas occupies 22.4 L at STP.
Thus when gases are involved, the coefficients in a balanced chemical equation represent not only molar amounts but also relatives volumes
2 L + 13 L ? 8 L + 10 LRecall:The coefficients in a chemical reaction represent molar amounts of substances taking part in the reaction.
Gas Stoichiometry ProblemGas Stoichiometry ProblemIn the following combustion reaction, what volume of methane (CH4) is needed to produce 26 L of
water vapor?
– CH4 (g) + 2O2(g) O CO2(g) + 2H2O(g)
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
x L ? 26 L
1 L ? 2 L
x L = 26 L 1L 2L
x= 13 L
1 mol ? 2 mol
Gas StoichiometryGas Stoichiometry use ideal gas lawuse ideal gas law
– Looking for grams or moles of gas? • Step 1: start with ideal gas law to find moles of gas• Step 2: 1change to grams of gas
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
PV=nRT
Grams/mol? 1) Use Ideal Gas Law 2) Do stoichiometry calculations
Example 1Example 1How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C? PV=nRT
4 Al(s) + 3 O2(g) 2 Al2O3(s)
Given liters: Start with Ideal Gas Law and calculate moles of O2.
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
Given Unkown V O2 = 15.0 L O2 grams of Al2O3? R= .0821 atm L/K MoleP O2 = 97.3 kPa= 0.9605 atm T O2 =21oC +273= 294 K
Step 1: Calculate moles of O2 n = PV = 0.9605 atm x 15.0 L = 0.5969 mol O2
RT 0.0821 atm L/K Mole 294 K
Step 2: Calculate mass of Al2O3
0.5969 mol O2 = X mol Al2O3 = 0.3979 mol Al2O3 3moleO2 2 mole Al2O3
Use stoich to convert moles of O2 to grams Al2O3.
41 g Al2O3 0.3979 mol Al2O3 x 101.96 g Al2O3=
1 mol Al2O3
Gas StoichiometryGas Stoichiometry use ideal gas lawuse ideal gas law
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
PV=nRT
• Looking for volume of gas?•Step 1: start with stoichiometry conversion to find moles of gas•Step 2: use ideal gas law to find the volume
Liters ? 1) Do stoichiometry calculations 2) Use Ideal Gas Law
What volume of CO2 forms from 5.25 g of CaCO3 at 101.3 kPa & 25ºC?
CaCO3 CaO + CO2
Looking for liters: Start with stoich and calculate moles of CO2.
Plug this into the Ideal Gas Law to find volume.
Given Unkown PV=nRTm = 5.25 g CaCO3 volume of CO2? R= .0821 atm L/K MoleP = 101.3 kPa = 1 atm T =25.0oC +273= 298 K
Step 1: Calculate moles of CO2 5.25 g CaCO3 x 1 mole CaCO3 = 0.0525 mol CaCO3
100 g CaCO3
1 mole CO2 = 1mole CaCO3 ; 0.0525 mol CO2
Step 2: Calculate volume of CO2
V = nRT = 0.0525 mol CO2 x .0821 atm L/K Mole x 298 K = 1.28 L P 1 atm
Example 2Example 2