I. Behavior of Gases

(Read p. 66-69)

I. Behavior of Gases

(Read p. 66-69)

Topic 4- GasesTopic 4- Gases

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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