1 chapter 7 gases. 2 particles of a gas move rapidly in straight lines and are in constant motion....

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Chapter 7 GasesChapter 7 Gases

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Particles of a gasParticles of a gas Move rapidly in straight lines and Move rapidly in straight lines and

are in constant motion. are in constant motion. Have kinetic energy that increases Have kinetic energy that increases

with an increase in temperature.with an increase in temperature. Are very far apart.Are very far apart. Have essentially no attractive (or Have essentially no attractive (or

repulsive) forces. repulsive) forces. Have very small volumes Have very small volumes

compared to the volume of the compared to the volume of the container they occupy. container they occupy.

Kinetic Theory of GasesKinetic Theory of Gases

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Gases are described in terms of four Gases are described in terms of four properties: pressure (P), volume (V), properties: pressure (P), volume (V), temperature (T), and amount (n).temperature (T), and amount (n).

Properties of GasesProperties of Gases

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A barometer A barometer measures the measures the pressure exerted pressure exerted by the gases in by the gases in the atmosphere.the atmosphere.

The atmospheric The atmospheric pressure is pressure is measured as the measured as the height in mm of height in mm of the mercury the mercury column.column.

BarometerBarometer

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A. The downward pressure of the Hg in a A. The downward pressure of the Hg in a barometer is _____ than/as the weight of barometer is _____ than/as the weight of the atmosphere.the atmosphere.

1) greater 1) greater 2) less 2) less 3) the same 3) the same

B. A water barometer is 13.6 times taller B. A water barometer is 13.6 times taller than a Hg barometer (Dthan a Hg barometer (DHgHg = 13.6 g/mL) = 13.6 g/mL) becausebecause

1) H1) H22O is less dense O is less dense

2) H2) H22O is heavierO is heavier

3) air is more dense than H3) air is more dense than H22OO

Learning CheckLearning Check

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A.The downward pressure of the Hg A.The downward pressure of the Hg in a barometer is in a barometer is 3) the same3) the same as as the weight of the atmosphere.the weight of the atmosphere.

B. A water barometer is 13.6 times B. A water barometer is 13.6 times taller than a Hg barometer (Dtaller than a Hg barometer (DHgHg = =

13.6 g/mL) because13.6 g/mL) because

1) H1) H22O is less denseO is less dense

SolutionSolution

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A gas exerts pressure, which is A gas exerts pressure, which is defined as a force acting on a defined as a force acting on a specific area. specific area.

Pressure (P) = Pressure (P) = ForceForce Area Area

One atmosphere (1 atm) is 760 mm One atmosphere (1 atm) is 760 mm Hg.Hg.

1 mm Hg = 1 torr1 mm Hg = 1 torr1.00 atm = 760 mm Hg = 760 1.00 atm = 760 mm Hg = 760 torrtorr

PressurePressure

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In science, pressure is stated in In science, pressure is stated in atmospheres (atm), millimeters of atmospheres (atm), millimeters of mercury (mm Hg), and Pascals (Pa).mercury (mm Hg), and Pascals (Pa).

Units of PressureUnits of Pressure

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A. What is 475 mm Hg expressed in atm?A. What is 475 mm Hg expressed in atm?1) 475 atm1) 475 atm2) 0.638 atm2) 0.638 atm3) 3.61 x 103) 3.61 x 1055 atm atm

B. The pressure in a tire is 2.00 atm. B. The pressure in a tire is 2.00 atm. What is this pressure in mm Hg?What is this pressure in mm Hg?1) 1) 2.00 mm Hg2.00 mm Hg2) 2) 1520 mm Hg1520 mm Hg3)3) 22,300 mm Hg22,300 mm Hg


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A. What is 475 mm Hg expressed in atm?A. What is 475 mm Hg expressed in atm?2) 0.638 atm2) 0.638 atm485 mm Hg x 485 mm Hg x 1 atm 1 atm = 0.638 = 0.638 atm atm

760 mm Hg760 mm HgB. The pressure of a tire is measured as 2.00 B. The pressure of a tire is measured as 2.00

atm. What is this pressure in mm Hg?atm. What is this pressure in mm Hg?2) 2) 1520 mm Hg1520 mm Hg2.00 atm x 2.00 atm x 760 mm Hg760 mm Hg = 1520 mm = 1520 mm HgHg

1 atm1 atm

SolutionSolution

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The pressure The pressure of a gas is of a gas is inversely inversely related to its related to its volume when volume when T and n are T and n are constant.constant.

If volume If volume decreases, decreases, the pressure the pressure increases.increases.

Boyle’s LawBoyle’s Law

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The product P x V remains constant as The product P x V remains constant as long as T and long as T and nn do not change. do not change.PP11VV1 1 = 8.0 atm x 2.0 L = 16 atm L= 8.0 atm x 2.0 L = 16 atm LPP22VV2 2 = 4.0 atm x 4.0 L = 16 atm L= 4.0 atm x 4.0 L = 16 atm LPP33VV3 3 = 2.0 atm x 8.0 L = 16 atm L= 2.0 atm x 8.0 L = 16 atm L

Boyle’s Law can be stated asBoyle’s Law can be stated as PP11VV11 = P = P22VV22 (T, n constant)(T, n constant)

PV Constant in Boyle’s LawPV Constant in Boyle’s Law

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Solving for a Gas Law Solving for a Gas Law FactorFactor

The equation for Boyle’s Law can be The equation for Boyle’s Law can be rearranged to solve for any factor.rearranged to solve for any factor.

To solve for VTo solve for V22, divide both sides by P, divide both sides by P22..PP11VV11 = = PP22VV22 Boyle’s LawBoyle’s Law PP2 2 PP22

PP11VV11 = = VV22

PP22 Solving for PSolving for P22

PP22 = = P P11 V V11

VV22

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PV in Breathing MechanicsPV in Breathing Mechanics

When the lungs When the lungs expand, the expand, the pressure in the pressure in the lungs decreases.lungs decreases.

Inhalation Inhalation occurs as air occurs as air flows towards flows towards the lower the lower pressure in the pressure in the lungs.lungs.

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PV in Breathing MechanicsPV in Breathing Mechanics

When the lung When the lung volume decreases, volume decreases, pressure within pressure within the lungs the lungs increases.increases.

Exhalation occurs Exhalation occurs as air flows from as air flows from the higher the higher pressure in the pressure in the lungs to the lungs to the outside.outside.

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Freon-12, CClFreon-12, CCl22FF22, is used in , is used in refrigeration systems. What is the refrigeration systems. What is the new volume (L) of a 8 L sample of new volume (L) of a 8 L sample of Freon gas initially at 50 mm Hg after Freon gas initially at 50 mm Hg after its pressure is changed to 200 mm Hg its pressure is changed to 200 mm Hg at constant T?at constant T?

1. Set up a data table1. Set up a data tableConditions 1Conditions 1 Conditions 2Conditions 2PP11 = 50 mm Hg= 50 mm HgPP22 = 200 mm Hg= 200 mm Hg

VV11 = 8 L= 8 L VV22 ==

Calculation with Boyle’s LawCalculation with Boyle’s Law

?

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2.2. When pressure increases, volume When pressure increases, volume decreases.decreases.

Solve Boyle’s Law for VSolve Boyle’s Law for V22: :

PP11VV11 = P = P22VV22

VV22 = = VV11PP11

PP22 VV22 = 8 L x = 8 L x 50 mm Hg 50 mm Hg = 2 L = 2 L

200 mm Hg200 mm Hgpressure ratio decreases volumepressure ratio decreases volume

Calculation with Boyle’s Calculation with Boyle’s Law (continued)Law (continued)

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The helium in a cylinder The helium in a cylinder has a volume of 120 mL has a volume of 120 mL and a pressure of 840 mm and a pressure of 840 mm Hg. A change in the volume Hg. A change in the volume results in a lower pressure results in a lower pressure inside the cylinder. Does inside the cylinder. Does cylinder A or B represent the final cylinder A or B represent the final volume?volume?

At a new pressure of 420 mm Hg, what is At a new pressure of 420 mm Hg, what is the new volume?the new volume? 1) 60 mL 1) 60 mL 2) 120 mL2) 120 mL 3) 240 mL 3) 240 mL


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The helium in a cylinder has a volume of The helium in a cylinder has a volume of 120 mL and a pressure of 840 mm Hg. A 120 mL and a pressure of 840 mm Hg. A change in the volume results in a lower change in the volume results in a lower pressure inside the cylinder. Does cylinder pressure inside the cylinder. Does cylinder A or B represent the final volume?A or B represent the final volume?B) If P decreases, V increases. B) If P decreases, V increases.

At a new pressure of 420 mm At a new pressure of 420 mm Hg, what is the new volume Hg, what is the new volume of the cylinder?of the cylinder?

3) 240 mL3) 240 mL

SolutionSolution

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A sample of helium gas A sample of helium gas has a volume of 6.4 L at has a volume of 6.4 L at a pressure of 0.70 atm. a pressure of 0.70 atm. What is the new volume What is the new volume when the pressure is when the pressure is increased to 1.40 atm increased to 1.40 atm (T constant)?(T constant)?

A) 3.2 LA) 3.2 L B) 6.4 LB) 6.4 L C) 12.8 C) 12.8 LL


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A) 3.2 LA) 3.2 L

Solve for VSolve for V22: P: P11VV11 = P = P22VV22

VV22 = = VV11PP11

PP22 VV22 = 6.4 L x = 6.4 L x 0.70 atm 0.70 atm = 3.2 L = 3.2 L

1.40 atm1.40 atmVolume decreases when there is an Volume decreases when there is an increase in the pressure (Temperature is increase in the pressure (Temperature is constant).constant).

SolutionSolution

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A sample of oxygen gas has a volume of A sample of oxygen gas has a volume of 12.0 L at 600. mm Hg. What is the new 12.0 L at 600. mm Hg. What is the new pressure when the volume changes to pressure when the volume changes to 36.0 L? (T and n constant.)36.0 L? (T and n constant.)

1) 200. mm Hg 1) 200. mm Hg

2) 400. mm Hg 2) 400. mm Hg

3) 1200 mm Hg3) 1200 mm Hg


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1) 200. mm Hg 1) 200. mm Hg Data table Data table Conditions 1Conditions 1 Conditions 2Conditions 2PP11 = 600. mm Hg = 600. mm Hg PP22 = ??? = ???VV11 = 12.0 L = 12.0 L VV22 = 36.0 L = 36.0 L

PP2 = 2 = PP11 V V11

VV22

600. mm Hg x 600. mm Hg x 12.0 L12.0 L = 200. mm Hg = 200. mm Hg 36.0 L 36.0 L

SolutionSolution

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The Kelvin The Kelvin temperature of temperature of a gas is directly a gas is directly related to the related to the volume (P and n volume (P and n are constant).are constant).

When the When the temperature of temperature of a gas increases, a gas increases, its volume its volume increases.increases.

Charles’ LawCharles’ Law

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For two conditions, Charles’ Law is For two conditions, Charles’ Law is writtenwritten

VV11 = = VV2 2 (P and n constant)(P and n constant)

TT11 T T22

Rearranging Charles’ Law to solve for Rearranging Charles’ Law to solve for VV22

VV2 2 = = VV11TT22

TT11

Charles’ Law V and TCharles’ Law V and T

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Solve Charles’ Law expression for TSolve Charles’ Law expression for T2.2.

VV11 = = VV22

TT11 T T22

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SolutionSolution

VV11 = = VV22

TT11 T T22

Cross multiply to giveCross multiply to give

VV11TT2 2 == VV22TT11

Isolate TIsolate T2 2 by dividing through by Vby dividing through by V11

VV11TT2 2 == VV22TT11

VV11 V V11

TT22 = = VV22TT11

VV11

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A balloon has a volume of 785 mL at 21°C. IfA balloon has a volume of 785 mL at 21°C. IfThe temperature drops to 0°C, what is the The temperature drops to 0°C, what is the

new new volume of the balloon (P constant)? volume of the balloon (P constant)?

1.1. Set up data table:Set up data table:Conditions 1Conditions 1 Conditions 2Conditions 2VV11 = 785 mL = 785 mL VV22 = ? = ?TT11 = 21°C = 294 K = 21°C = 294 K TT22 = 0°C = 273 K = 0°C = 273 KBe sure that you always use the Kelvin (K)Be sure that you always use the Kelvin (K)temperature in gas calculations.temperature in gas calculations.

Calculations Using Charles’ Calculations Using Charles’ LawLaw

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Calculations Using Charles’ Calculations Using Charles’ Law (continued)Law (continued)

2. Solve Charles’ law for V2

V1 = V2

T1 T2

V2 = V1 T2

T1

V2 = 785 mL x 273 K = 729 mL 294 K

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A sample of oxygen gas has a A sample of oxygen gas has a volume of 420 mL at a temperature volume of 420 mL at a temperature of 18°C. At what temperature (in of 18°C. At what temperature (in °C) will the volume of the oxygen °C) will the volume of the oxygen be 640 mL (P and n constant)?be 640 mL (P and n constant)?

1) 443°C1) 443°C

2) 170°C 2) 170°C

3) – 82°C3) – 82°C


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170°C170°C

TT2 2 = = TT11VV22

VV11

TT22 = 291 K x = 291 K x 640 mL640 mL = = 443 K443 K

420 mL420 mL

= 443 K – 273 K = 443 K – 273 K = = 170°C170°C

SolutionSolution

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The pressure The pressure exerted by a gas exerted by a gas is directly is directly related to the related to the Kelvin Kelvin temperature of temperature of the gas at the gas at constant V and constant V and n.n.

PP11 = = PP22

TT11 TT22

Gay-Lussac’s Law: P and TGay-Lussac’s Law: P and T

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A gas has a pressure at 2.0 atm at 18°C. A gas has a pressure at 2.0 atm at 18°C. WhatWhat

is the new pressure when the temperature is is the new pressure when the temperature is 62°C? (V and n constant)62°C? (V and n constant)1.1. Set up a data table.Set up a data table.

Conditions 1Conditions 1 Conditions 2Conditions 2 PP11 = 2.0 atm= 2.0 atm PP22 = = TT11 = 18°C + 273 = 18°C + 273 TT22 = 62°C + 273= 62°C + 273

= 291 K= 291 K = 335 K= 335 K

Calculation with Gay-Calculation with Gay-Lussac’s LawLussac’s Law

?

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Calculation with Gay-Calculation with Gay-Lussac’s Law (continued)Lussac’s Law (continued)

2. Solve Gay-Lussac’s Law for P2. Solve Gay-Lussac’s Law for P22

PP11 = = PP22

TT11 T T22

PP22 = = PP11 T T22

TT11

PP22 = 2.0 atm x = 2.0 atm x 335 K335 K = 2.3 atm = 2.3 atm

291 K291 K

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Use the gas laws to complete with Use the gas laws to complete with 1)1) Increases 2) Decreases Increases 2) Decreases 2)2) A. Pressure _______ when V decreases. A. Pressure _______ when V decreases.

B. When T decreases, V _______. B. When T decreases, V _______.

C. Pressure _______ when V changesC. Pressure _______ when V changes

from 12.0 L to 24.0 L.from 12.0 L to 24.0 L.

D. Volume _______when T changes fromD. Volume _______when T changes from

15.0 °C to 45.0°C.15.0 °C to 45.0°C.


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Use the gas laws to complete with Use the gas laws to complete with 1) Increases 2) Decreases1) Increases 2) Decreases

A. Pressure A. Pressure 1) Increases1) Increases, when V , when V decreases.decreases.

B. When T decreases, V B. When T decreases, V 2) Decreases.2) Decreases.

C. Pressure C. Pressure 2) Decreases2) Decreases when V changes when V changes

from 12.0 L to 24.0 Lfrom 12.0 L to 24.0 LD. Volume D. Volume 1) Increases1) Increases when T changes when T changes

fromfrom 15.0 °C to 45.0°C15.0 °C to 45.0°C

SolutionSolution

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Next TimeNext Time

We complete Chapter 7We complete Chapter 7

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The combined gas law uses Boyle’s The combined gas law uses Boyle’s Law, Charles’ Law, and Gay-Lussac’s Law, Charles’ Law, and Gay-Lussac’s Law (n is constant).Law (n is constant).

PP11 V V11 = = PP22 V V22

TT11 T T22

Combined Gas LawCombined Gas Law

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A sample of helium gas has a volume of A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. At what temperature of 29°C. At what temperature (°C) will the helium have a temperature (°C) will the helium have a volume of 90.0 mL and a pressure of 3.20 volume of 90.0 mL and a pressure of 3.20 atm (n constant)?atm (n constant)?

1. 1. Set up Data TableSet up Data Table

Conditions 1Conditions 1 Conditions 2Conditions 2PP1 1 = 0.800 atm = 0.800 atm PP22 = 3.20 atm = 3.20 atmVV11 = 0.180 L (180 mL) = 0.180 L (180 mL) VV2 2 = 90.0 mL= 90.0 mLTT11 = 29°C + 273 = 302 K = 29°C + 273 = 302 K TT2 2 = ??= ??

Combined Gas Law Combined Gas Law CalculationCalculation

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2. Solve for T2. Solve for T2 2 PP11 V V11 = = PP22 V V22

TT11 T T22

TT22 = = TT11 P P22VV22

PP11VV11

TT22 = 302 K x = 302 K x 3.20 atm 3.20 atm x x 90.0 mL 90.0 mL = = 604 K604 K

0.800 atm 180.0 mL0.800 atm 180.0 mL

TT22 = 604 K – 273 = 331 °C = 604 K – 273 = 331 °C

Combined Gas Law Combined Gas Law Calculation (continued)Calculation (continued)

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A gas has a volume of 675 mL at A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. 35°C and 0.850 atm pressure. What is the volume(mL) of the gas What is the volume(mL) of the gas at –95°C and a pressure of 802 mm at –95°C and a pressure of 802 mm Hg (n constant)?Hg (n constant)?


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Data TableData TableTT11 = 308 K = 308 K T T22 = -95°C + 273 = = -95°C + 273 =

178K178KVV11 = 675 mL = 675 mL V V22 = ??? = ???PP11 = 646 mm Hg = 646 mm Hg P P22 = 802 mm Hg = 802 mm Hg Solve for TSolve for T22

VV22 = = VV11 P P11 T T22

PP22TT11

VV22 = = 675 mL x 646 mm Hg x 178K675 mL x 646 mm Hg x 178K = = 314 mL314 mL 802 mm Hg x 308K 802 mm Hg x 308K

SolutionSolution

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The volume The volume of a gas is of a gas is directly directly related to the related to the number of number of moles of gas moles of gas when T and P when T and P are constant.are constant.VV11 = = VV22 nn11 n n22

Avogadro's Law: Volume Avogadro's Law: Volume and Molesand Moles

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If 0.75 mole of helium gas If 0.75 mole of helium gas occupies a volume of 1.5 L, what occupies a volume of 1.5 L, what volume will 1.2 moles of helium volume will 1.2 moles of helium occupy at the same temperature occupy at the same temperature and pressure?and pressure?

1) 1) 0.94 L0.94 L

2)2) 1.8 L1.8 L

3) 3) 2.4 L2.4 L

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SolutionSolution

3) 3) 2.4 L2.4 LConditions 1Conditions 1 Conditions 2Conditions 2VV11 = 1.5 L = 1.5 L VV22 = ??? = ???nn11 = 0.75 mole He = 0.75 mole He nn22 = 1.2 = 1.2 moles Hemoles HeVV22 = = VV11nn22

nn11

VV22 = 1.5 L x = 1.5 L x 1.2 moles He1.2 moles He = 2.4 L = 2.4 L 0.75 mole He0.75 mole He

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The volumes of gases can be compared The volumes of gases can be compared when they have the same conditions of when they have the same conditions of temperature and pressure (STP).temperature and pressure (STP).

Standard temperature (T) Standard temperature (T) 0°C or 273 K0°C or 273 K

Standard pressure (P)Standard pressure (P)

1 atm (760 mm Hg)1 atm (760 mm Hg)

STPSTP

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At STP, 1 At STP, 1 mole of a gas mole of a gas occupies a occupies a volume of volume of 22.4 L.22.4 L.

The volume of The volume of one mole of a one mole of a gas is called gas is called the the molar molar volume.volume.

Molar VolumeMolar Volume

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The molar volume at STP can be The molar volume at STP can be used to form conversion factors.used to form conversion factors.

22.4 L22.4 L andand 1 mole 1 mole 1 mole 22.4 L1 mole 22.4 L

Molar Volume as a Molar Volume as a Conversion FactorConversion Factor

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A. What is the volume at STP of 4.00 g A. What is the volume at STP of 4.00 g

of CHof CH44??

1) 5.60 L1) 5.60 L 2) 11.2 L2) 11.2 L 3) 44.8 3) 44.8

LL

B. How many grams of He are present in B. How many grams of He are present in

8.00 L of gas at STP? 8.00 L of gas at STP?

1) 25.6 g1) 25.6 g 2) 0.357 g2) 0.357 g 3) 3)

1.43 g1.43 g


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A.A. 1) 5.60 L1) 5.60 L

4.00 g CH4.00 g CH44 x x 1 mole CH1 mole CH44 x x 22.4 L (STP)22.4 L (STP) = = 5.60 L 5.60 L

16.0 g CH16.0 g CH44 1 mole CH 1 mole CH44

B. 3) 1.43 gB. 3) 1.43 g8.00 L x 8.00 L x 1 mole He 1 mole He x x 4.00 g He4.00 g He = =

1.43 g He1.43 g He 22.4 L 1 mole He22.4 L 1 mole He

SolutionSolution

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The relationship between the four The relationship between the four properties (P, V, n, and T) of gases properties (P, V, n, and T) of gases can be written equal to a constant can be written equal to a constant R.R.

PVPV = R = RnTnT

Rearranging this expression gives Rearranging this expression gives the expression called the the expression called the ideal gas ideal gas law.law.

PV = nRTPV = nRT

Ideal Gas LawIdeal Gas Law

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The universal gas constant, R, can be The universal gas constant, R, can be calculated using the molar volume of a gas calculated using the molar volume of a gas at STP. at STP.

At STP (273 K and 1.00 atm), 1 mole of a At STP (273 K and 1.00 atm), 1 mole of a gas occupies 22.4 L.gas occupies 22.4 L.

P VP V R = R = PVPV = = (1.00 atm)(22.4 L)(1.00 atm)(22.4 L)

nTnT (1 mole) (273K) (1 mole) (273K) n Tn T

= = 0.0821 0.0821 L atm L atm mole Kmole K

Note there are four units associated with R.Note there are four units associated with R.

Universal Gas Constant, RUniversal Gas Constant, R

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Another value for the universal gas Another value for the universal gas constant is obtained using mm Hg for constant is obtained using mm Hg for the STP pressure. What is the value of R the STP pressure. What is the value of R when a pressure of 760 mm Hg is placed when a pressure of 760 mm Hg is placed in the R value expression? in the R value expression?


54

What is the value of R when the STP What is the value of R when the STP value for P is 760 mmHg? value for P is 760 mmHg?

R = R = PV PV = = (760 mm Hg) (760 mm Hg) (22.4 L) (22.4 L)

nT (1 mole) nT (1 mole) (273K)(273K)

= 62.4 = 62.4 L mm HgL mm Hg mole K mole K

SolutionSolution

55

Dinitrogen oxide (NDinitrogen oxide (N22O), laughing O), laughing gas, is used by dentists as an gas, is used by dentists as an anesthetic. If a 20.0 L tank of anesthetic. If a 20.0 L tank of laughing gas contains 2.8 moles laughing gas contains 2.8 moles NN22O at 23°C, what is the pressure O at 23°C, what is the pressure (mm Hg) in the tank? (mm Hg) in the tank?


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1. Adjust the units of the given 1. Adjust the units of the given properties to match the units of R. properties to match the units of R.

V = 20.0 L, T = 296 K, n = 2.8 moles, V = 20.0 L, T = 296 K, n = 2.8 moles, P = ?P = ?

2. Rearrange the ideal gas law for P.2. Rearrange the ideal gas law for P.P = P = nRTnRT VVP = (P = (2.8 moles)(62.4 L mm Hg)(296 K)2.8 moles)(62.4 L mm Hg)(296 K)

(20.0 L) (mole K) (20.0 L) (mole K)

= 2.6 x 10= 2.6 x 1033 mm Hg mm Hg

SolutionSolution

57

A cylinder contains 5.0 L A cylinder contains 5.0 L of Oof O22 at 20.0°C and 0.85 at 20.0°C and 0.85 atm. How many grams of atm. How many grams of oxygen are in the oxygen are in the cylinder?cylinder?


58

1. Determine the given properties.1. Determine the given properties.P = 0.85 atm, V = 5.0 L, T = 293 K, n (or g =?)P = 0.85 atm, V = 5.0 L, T = 293 K, n (or g =?)2. Rearrange the ideal gas law for n (moles).2. Rearrange the ideal gas law for n (moles).

n = n = PVPV RTRT = = (0.85 atm)(5.0 L)(mole K)(0.85 atm)(5.0 L)(mole K) = 0.18 mole = 0.18 mole OO22

(0.0821atm L)(293 K)(0.0821atm L)(293 K) 3. Convert moles to grams using molar mass.3. Convert moles to grams using molar mass.

= 0. 18 mole O= 0. 18 mole O2 2 x x 32.0 g O32.0 g O2 2 = 5.8 g O= 5.8 g O22

1 mole O1 mole O22

SolutionSolution

59

What is the molar mass of a gas if 0.250 g of the What is the molar mass of a gas if 0.250 g of the gasgas

occupy 215 mL at 0.813 atm and 30.0°C?occupy 215 mL at 0.813 atm and 30.0°C?1. Solve for the moles (n) of gas.1. Solve for the moles (n) of gas. n = n = PVPV = = (0.813 atm) (0.215 L) (0.813 atm) (0.215 L)

RT (0.0821 L atm/mole K)(303K) RT (0.0821 L atm/mole K)(303K) = 0.00703 mole= 0.00703 mole

2. Set up the molar mass relationship.2. Set up the molar mass relationship.

Molar mass = Molar mass = g g = = 0.250 g 0.250 g = =

35.6g/mole35.6g/mole

mole 0.00703 mole mole 0.00703 mole

Molar Mass of a GasMolar Mass of a Gas

60

Gases in EquationsGases in Equations

The amounts of gases reacted or The amounts of gases reacted or produced in a chemical reaction can be produced in a chemical reaction can be calculated using the ideal gas law and calculated using the ideal gas law and mole factors.mole factors.

Problem:Problem:

What volume (L) of ClWhat volume (L) of Cl22 gas at 1.2 atm gas at 1.2 atm and 27and 27°C is needed to completely react °C is needed to completely react with with 1.5 g of aluminum?1.5 g of aluminum?

2Al(s) + 3Cl2Al(s) + 3Cl22 (g) 2AlCl (g) 2AlCl33(s)(s)

61

Gases in Equations Gases in Equations (continued)(continued)

2Al(s) + 3Cl2Al(s) + 3Cl22 (g) (g) 2AlCl2AlCl33(s)(s) 1.5 g ? L 1.2 atm, 300K1.5 g ? L 1.2 atm, 300K

1.1. Calculate the moles of ClCalculate the moles of Cl22 needed. needed.

1.5 g Al x 1.5 g Al x 1 mole Al1 mole Al x x 3 moles Cl3 moles Cl2 2 = 0.083 mole = 0.083 mole ClCl22

27.0 g Al 2 moles Al27.0 g Al 2 moles Al

2. 2. Place the moles ClPlace the moles Cl22 in the ideal gas equation. in the ideal gas equation.

V = V = nRTnRT = ( = (0.083 mole Cl0.083 mole Cl22)(0.0821 Latm/moleK))(0.0821 Latm/moleK)(300K)(300K)

P 1.2 atmP 1.2 atm= 1.7 L Cl= 1.7 L Cl22

62

What volume (L) of OWhat volume (L) of O22 at at 24°C and 24°C and 0.950 atm are needed to react with 0.950 atm are needed to react with 28.0 g NH28.0 g NH33??

4NH4NH33(g) + 5O(g) + 5O22(g) 4NO(g) + (g) 4NO(g) + 6H 6H22O(g)O(g)


63

1. Calculate the moles of O1. Calculate the moles of O22 needed. needed.

28.0 g NH28.0 g NH33 x x 1 mole NH1 mole NH33 x x 5 mole O5 mole O22

17.0 g NH17.0 g NH33 4 mole NH 4 mole NH33

= 2.06 mole O= 2.06 mole O22

2. Place the moles O2. Place the moles O22 in the ideal gas equation. in the ideal gas equation.

V = V = nRTnRT = = (2.06 moles)(0.0821 L atm/moleK)(2.06 moles)(0.0821 L atm/moleK)(297K)(297K)

PP 0.950 atm0.950 atm

= 52.9 L O= 52.9 L O22

SolutionSolution

64

In a mixture of gases, the In a mixture of gases, the partial pressurepartial pressure of each gas of each gas is the pressure that gas is the pressure that gas would exert if it were by would exert if it were by itself in the container.itself in the container.

Partial PressurePartial Pressure

65

The total The total pressure exerted pressure exerted by a gas mixture by a gas mixture is the sum of the is the sum of the partial partial pressures of the pressures of the gases in that gases in that mixture.mixture.

PPTT = P = P11 + P + P22

+ .....+ .....

Dalton’s Law of Partial Dalton’s Law of Partial PressuresPressures

66

The total pressure of a gas The total pressure of a gas mixture depends on the mixture depends on the total number of gas total number of gas particles, particles, notnot on the types on the types of particles.of particles.

Partial PressuresPartial Pressures

67

For example, at STP, one mole of gas For example, at STP, one mole of gas particles in a volume of 22.4 L will particles in a volume of 22.4 L will exert the same pressure as one mole exert the same pressure as one mole of a mixture of gas particles in 22.4 L. of a mixture of gas particles in 22.4 L.

V = 22.4 LV = 22.4 L

Total PressureTotal Pressure

0.5 mole O2

0.3 mole He0.2 mole Ar1.0 mole

1.0 mole N2

0.4 mole O2

0.6 mole He1.0 mole

1.0 atm 1.0 atm 1.0 atm

68

A scuba tank contains OA scuba tank contains O2 2 with a with a pressure of 0.450 atm and He at 855 pressure of 0.450 atm and He at 855 mm Hg. What is the total pressure in mm Hg. What is the total pressure in mm Hg in the tank?mm Hg in the tank?


69

1. Convert the pressure in atm to mm Hg1. Convert the pressure in atm to mm Hg 0.450 atm x 0.450 atm x 760 mm Hg760 mm Hg = 342 mm Hg = = 342 mm Hg =

PPOO 1 atm 1 atm 22

2. Calculate the sum of the partial 2. Calculate the sum of the partial pressures.pressures.

PPtotal total = = PPO O + + PPHeHe

22

PPtotaltotal = 342 mm Hg + 855 mm Hg = 342 mm Hg + 855 mm Hg

= 1197 mm Hg= 1197 mm Hg

SolutionSolution

70

Gases We BreatheGases We Breathe

71

When a scuba diver makes a deep dive, When a scuba diver makes a deep dive, the increased pressure causes more Nthe increased pressure causes more N2 2

(g)(g) to dissolve in the blood.to dissolve in the blood. If a diver rises too fast, the dissolved NIf a diver rises too fast, the dissolved N22

will form bubbles in the blood, a will form bubbles in the blood, a dangerous and painful condition called dangerous and painful condition called "the bends." "the bends."

Helium, which does not dissolve in the Helium, which does not dissolve in the blood, is mixed with Oblood, is mixed with O22 to prepare to prepare breathing mixtures for deep descents. breathing mixtures for deep descents.

Health Note: Scuba DivingHealth Note: Scuba Diving

72

For a deep dive, some scuba divers For a deep dive, some scuba divers are using a mixture of helium and are using a mixture of helium and oxygen gases with a pressure of oxygen gases with a pressure of 8.00 atm. If the oxygen has a partial 8.00 atm. If the oxygen has a partial pressure of 1280 mm Hg, what is pressure of 1280 mm Hg, what is the partial pressure of the helium?the partial pressure of the helium?

1) 520 mm Hg1) 520 mm Hg

2) 2040 mm Hg2) 2040 mm Hg

3) 4800 mm Hg3) 4800 mm Hg


73

3)3) 4800 mm Hg4800 mm Hg

PPTotal Total = 8.00 atm x = 8.00 atm x 760 mm Hg760 mm Hg = 6080 = 6080 mm Hgmm Hg

1 atm1 atmPPTotal Total = = PPO O + + PPHe He 22

PPHeHe = = PPTotal Total - - PPO O

22

PPHe He = 6080 mm Hg - 1280 mm Hg = 6080 mm Hg - 1280 mm Hg = 4800 mm Hg = 4800 mm Hg

SolutionSolution

1 chapter 7 gases. 2 particles of a gas move rapidly in straight lines and are in constant motion....

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