i. behavior of gases (read p. 66-69) topic 4- gases s.panzarella
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I. Behavior of Gases
(Read p. 66-69)
I. Behavior of Gases
(Read p. 66-69)
Topic 4- GasesTopic 4- Gases
S.Panzarella
Review of Review of Characteristics of GasesCharacteristics of Gases
Review of Review of Characteristics of GasesCharacteristics of Gases
Gases expand to fill any container.• random motion, no attraction
Gases have very low densities.• no volume = lots of empty space
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Gases can be compressed.• no volume = lots
of empty space
Gases undergo diffusion & effusion.• random motion;
entropy
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I. Kinetic Molecular Theory I. Kinetic Molecular Theory –– (under ideal circumstances) - explains how an “ideal” gas acts
I. Kinetic Molecular Theory I. Kinetic Molecular Theory –– (under ideal circumstances) - explains how an “ideal” gas acts
A. 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 avg. KE directly related to Kelvin temperature (KE = ½ MV2)
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B. Gas behavior is most ideal… at low pressures
at high temperatures
in nonpolar atoms/molecules of low molecular mass ie. H2 and He
think about the conditions for a beach vacation (high temps and low pressure)
C. Real Gases Particles….
Real gases deviate from an ideal gas at low temperatures and high pressures.• Have their own volume (due
to increased pressure)• Attract each other (due to
increased temps
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Homework Homework Homework Homework
Read page 66, 68-69pg 4-5 of guide, #’s 1-11
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III. The Gas LawsIII. The Gas LawsIII. The Gas LawsIII. The Gas Laws Shows the relationships
between TEMPERATURE, PRESSURE, VOLUME and number of moles of a gas
Used to determine what effect changing one of those variables will have on any of the others.
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Normal conditions……Normal conditions……
• Standard Temperature & Pressure - Standard Temperature & Pressure - *STPSTP
• 273 K (or 0◦C) and 101.3 kPa or 1 ATM
• Temperature MUST be (KELVIN) when working with gases. K = ºC + 273
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*Found in Reference Table A
1 Atm = 760 torr =760 mmHg = 101.3 kPa = 14.7 lb/in2
Bill Nye VideoBill Nye VideoBill Nye VideoBill Nye Video
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1. Combined Gas law1. Combined Gas law1. Combined Gas law1. Combined Gas lawCombines 3 gas laws# moles are held constant P, T and V changeFormula: found on Table T
P1V1
T1
=P2V2
T2S.Panzarella
Combined Gas Law – Combined Gas Law – Table TTable TCombined Gas Law – Combined Gas Law – Table TTable T
P1V1
T1
=P2V2
T2
P1V1T2 = P2V2T1
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GIVEN:V1 = 2.00 L
P1 = 80.0 kPa
T1 = 300 K
V2 = 1.00 L
P2 = 240. kPa
T2 = ?
WORK:P1V1T2 = P2V2T1
(80.0 kPa)(2.00 L)(T2) = (240.kPa)(1.00 L)(300 K)
T2 = 450 K
COMBINED COMBINED Gas Law Problem Gas Law Problem #1#1COMBINED COMBINED Gas Law Problem Gas Law Problem #1#1 2.00 L sample of gas at 300. K and a pressure of
80.0 kPa is placed into a 1.00 L container at a pressure of 240. kPa. What is the new temperature of the gas?
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GIVEN:V1 = 7.84 cm3
P1 = 71.8 kPa
T1 = 25°C = 298 K
V2 = ?
P2 = 101.3 kPa
T2 = 273 K
WORK:P1V1T2 = P2V2T1
(71.8 kPa)(7.84 cm3)(273 K)
= (101.3 kPa) V2 (298 K)
V2 = 5.09 cm3
COMBINED COMBINED Gas Law Problem Gas Law Problem #2 #2 COMBINED COMBINED Gas Law Problem Gas Law Problem #2 #2 A gas occupies 7.84 cm3 at 71.8 kPa &
25°C. Find its volume at STP.
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2. Boyle’s Law 2. Boyle’s Law 2. Boyle’s Law 2. Boyle’s Law Video: Let’s watch (1:01): http://www.youtube.
com/watch?feature=player_embedded&v=DcnuQoEy6wA
Pressure vs. Volume (Constant Temperature): (think: squeezing a balloon)
As pressure is INCREASED,
volume is DECREASED
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Boyle’s Law con’t.Boyle’s Law con’t.Boyle’s Law con’t.Boyle’s Law con’t.
INVERSE relationship:
PV = K
formula
P1V1= P2V2
P
V
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GIVEN:
V1 = 100. mL
P1 = 150. kPa
V2 = ?
P2 = 200. kPa
WORK:
P1V1T2 = P2V2T1
BOYLE’S BOYLE’S Law ProblemsLaw ProblemsBOYLE’S BOYLE’S Law ProblemsLaw ProblemsA gas occupies 100. mL at 150.
kPa. Find its volume at 200. kPa.
P V
(150.kPa)(100.mL)=(200.kPa)V2
V2 = 75.0 mL
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3. Charles’ Law3. Charles’ Law3. Charles’ Law3. Charles’ LawLet’s Watch Video (33 sec) : http://www.youtube.com/watch?v=XHiYKfAmTMc&feature=player_embedded
Volume vs. Temperature (Constant Pressure) (think: hot air balloon)
As temperature is INCREASED, volume is
INCREASED
• As the temperature of the water increases, the volume of the balloon increases.
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Charles’ Law con’t.Charles’ Law con’t.Charles’ Law con’t.Charles’ Law con’t.
DIRECT relationship:
V/T = K
Formula
V1 T2 = V2 T1
V
T
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GIVEN:
V1 = 5.00 L
T1 = 36°C = 309K
V2 = ?
T2 = 94°C = 367K
WORK:
P1V1T2 = P2V2T1
CHARLES’ CHARLES’ Law ProblemsLaw Problems CHARLES’ CHARLES’ Law ProblemsLaw ProblemsA gas occupies 5.00L at 36°C. Find
its volume at 94°C.
T V
(5.00 L)(367 K)=V2(309 K)
V2 = 5.94 LS.Panzarella
4. Gay-Lussac’s 4. Gay-Lussac’s LawLaw4. Gay-Lussac’s 4. Gay-Lussac’s LawLaw Temperature vs. Pressure (Constant
Volume)- (think: car tires or pressure cooker)
As temperature is INCREASED,
pressure is INCREASED
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Gay-Lussac’s Law con’tGay-Lussac’s Law con’tGay-Lussac’s Law con’tGay-Lussac’s Law con’t
DIRECT relationship: P/T = K Formula:
P1T2 = P2T1
P
T
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GIVEN:
P1 = 1.00 atm
T1 = 200 K
P2 = ?
T2 = 800 K
WORK:
P1V1T2 = P2V2T1
GAY-LUSSAC’S Law GAY-LUSSAC’S Law ProblemProblemGAY-LUSSAC’S Law GAY-LUSSAC’S Law ProblemProblem A 10.0 L sample of gas in a rigid
container at 1.00 atm and 200. K is heated to 800. K. Assuming that the volume remains constant, what is the new pressure of the gas?
P T
(1.00 atm) (800 K) = P2(200 K)
P2 = 4.00 atm
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HomeworkHomeworkHomeworkHomework
See guide pg 5-6
• Part A (use your RB)
• Part B #12-21
• Quiz on Thursday
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2 more Gas laws2 more Gas laws
(TEXT BOOKp. 350-353)Read these pages first!
(TEXT BOOKp. 350-353)Read these pages first!
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5. Graham’s 5. Graham’s LawLaw5. Graham’s 5. Graham’s LawLaw
States that the rate of effusion (diffusion) of a gas is inversely proportional to the square root of the gas’s molar mass.
Helium effuses (and diffuses) nearly three times faster than nitrogen at the same temperature.
S.BarryS.Panzarella
Graham’s Law cont.Graham’s Law cont.Graham’s Law cont.Graham’s Law cont.
Gases of SMALL MASS (or DENSITY) diffuse faster than gases of higher molar mass.
ex. At STP, which gas will diffuse more rapidly? Use Table S
a) He b) Ar c) Kr d) Xe
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6. Dalton’s Law6. Dalton’s Law6. Dalton’s Law6. Dalton’s Law At constant volume and temperature, the total pressure exerted by a mixture of
gases is equal to the sum of the partial pressures of the component gases. Formula
Three gases are combined in container T
Ptotal = P1 + P2 + P3 ...
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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
Dalton’s Law example #1Dalton’s Law example #1Dalton’s Law example #1Dalton’s Law example #1 Hydrogen gas is collected over water at
22.5°C and 2.72 kPa. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa.
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GIVEN:
PO2 = ?
PN2 = 79 kPa
PCO2 = 0.034 kPa Pothers = 0.95 kPa Ptotal = 101.3 kPa
WORK:Ptotal = PO2 + PN2 + PCO2 + Poth
101.3 kPa = PO2 + 79 kPa + 0.034 kPa + 0.95 kPa
101.3 kPa = PO2 + 79.984 kPa
21.316 kPa = PO2
Ex. Air contains oxygen, nitrogen, and carbon dioxide and other gases. What is the partial pressure of O2 at 101.3 kPa of pressure if the PN2 = 79 kPa , the PCO2 = 0.034 kPa and Pothers = 0.95 kPa?
Dalton’s Law example #2Dalton’s Law example #2Dalton’s Law example #2Dalton’s Law example #2
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III. Avogadro’s HypothesisIII. Avogadro’s HypothesisIII. Avogadro’s HypothesisIII. Avogadro’s Hypothesis Definition: Equal volume of gases at the same
temperature and pressure contain equal
numbers of particles (molecules) regardless of
their mass
Let’s watch (2:05)http://www.youtube.com/watch?v=fexEvn0ZOpo&feature=player_embedded
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1) Which rigid cylinder contains the same number of gas molecules at STP as a 2.0-liter rigid cylinder containing H2(g) at STP?
(a) 1.0-L cylinder of O2(g)
(b) 2.0-L cylinder of CH4(g)
(c) 1.5-L cylinder of NH3(g)
(d) 4.0-L cylinder of He(g)
2) Which two samples of gas at STP contain the same total 2) Which two samples of gas at STP contain the same total number of molecules?number of molecules?
(a) 1 L of CO(a) 1 L of CO(g)(g) and 0.5 L of N and 0.5 L of N2(g)2(g)
(b) 2 L of CO(b) 2 L of CO(g)(g) and 0.5 L of NH and 0.5 L of NH3(g)3(g)
(c) 1 L of H(c) 1 L of H2(g)2(g) and 2 L of Cl and 2 L of Cl2(g)2(g)
(d) 2 L of H(d) 2 L of H2(g)2(g) and 2 L of Cl and 2 L of Cl2(g)2(g)
3) At the same temperature and pressure, which sample 3) At the same temperature and pressure, which sample contains the same number of moles of particles as 1 liter contains the same number of moles of particles as 1 liter of Oof O22(g)? (g)?
• 1 L Ne(g) 1 L Ne(g) (c) 0.5 L SO (c) 0.5 L SO22(g)(g)• (b) 2 L N(b) 2 L N2(2(g) g) (d) 1 L H (d) 1 L H22O(ℓ)O(ℓ)
You
Try
B. Molar volume:B. Molar volume: B. Molar volume:B. Molar volume: The number of molecules in 22.4 L of
any gas at STP has been chosen as a standard unit called 1 mole
1 mole = 22.4 L of any gas at STP contains 6.02x1023 particles
22.4 L of any gas at STP is = to its molecular mass
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Molar volume cont.Molar volume cont.
Density = molecular mass of gas volume (22.4 L)
ex. What is the density of 1 mole of oxygen gas?
ex. Which gas has a density of 1.70 g/L at STP?
a) F2 b) He c) N2 d) SO2
You
Try
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IV. Ideal Gas LawIV. Ideal Gas Law
PV=nRT
Recall - Avogadro’s law, which is derived from this basic idea, says that the volume of a gas
maintained at constant temperature and pressure is directly proportional to the number of moles of the gas
Video: http://education-portal.com/academy/lesson/the-ideal-gas-law-and-the-gas-constant.html#lesson S.Panzarella
PV
TVn
PVnT
Ideal Gas Law Ideal Gas Law video http://education-portal.com/academy/lesson/using-the-ideal-gas-law-to-predict-video http://education-portal.com/academy/lesson/using-the-ideal-gas-law-to-predict-the-effect-of-changes-to-a-gas.html#lessonthe-effect-of-changes-to-a-gas.html#lesson
Ideal Gas Law Ideal Gas Law video http://education-portal.com/academy/lesson/using-the-ideal-gas-law-to-predict-video http://education-portal.com/academy/lesson/using-the-ideal-gas-law-to-predict-the-effect-of-changes-to-a-gas.html#lessonthe-effect-of-changes-to-a-gas.html#lesson
= kUNIVERSAL GAS
CONSTANTR=0.0821 Latm/molKR=8.315 dm3kPa/molK
= R
You don’t need to memorize these values!
• Merge the Combined Gas Law with Avogadro’s Principle:
n is the number of moles of the gas
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GIVEN:
P = ? atm
n = 0.412 mol
T = 16°C = 289 K
V = 3.25 LR = 0.0821Latm/molK
WORK:
PV = nRT
P(3.25)=(0.412)(0.0821)(289) L mol Latm/molK K
P = 3.01 atm
Ideal Gas Law Problem #1Ideal Gas Law Problem #1Ideal Gas Law Problem #1Ideal Gas Law Problem #1
Calculate the pressure in atmospheres of 0.412 mol of He at 16°C & occupying 3.25 L.
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GIVEN:
V = ?
n = 85 g
T = 25°C = 298 K
P = 104.5 kPaR = 8.315 dm3kPa/molK
Ideal Gas Law Problem #2Ideal Gas Law Problem #2Ideal Gas Law Problem #2Ideal Gas Law Problem #2
Find the volume of 85 g of O2 at 25°C and 104.5 kPa.
= 2.7 mol
WORK:85 g 1 mol = 2.7 mol
32.00 g
PV = nRT(104.5)V=(2.7) (8.315) (298) kPa mol dm3kPa/molK K
V = 64 dm3S.Panzarella