22
PROPERTIES OF GASES Chapter 7 §7.1 Common Gases 2 Of the elements on the periodic table, some exist as a gas at room temperature, these include the Noble Gases, Group 8A (18), H 2 , N 2 , O 2 , F 2 , Cl 2 , and many oxides of nonmetals such as CO, CO 2 , NO, NO 2 , SO 2 , and SO 3 . Gases and Environmental Concerns 3 Some gases are responsible for environmental and health concerns including methane, CH 4 (green house gas, flammable, respiratory concerns) chlorofluorocarbons (CFCs) (destroys the ozone layer) nitrogen oxides found in smog (respiratory concerns) volatile organic compounds (VOCs), such as compounds found in paint thinners (just plain old really bad stuff) Kinetic Theory of Gases A gas consists of small particles that move randomly with high velocities. have essentially no attractive (or repulsive) forces toward each other. have a very small volume compared to the volume of the containers they occupy. are in constant motion. have kinetic energies that increase with an increase in temperature. 4

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Page 1: Gases and Environmental Concerns Kinetic Theory of Gasesthemalloryfamily.net/teaching_resources/chemistry/chemistry51/han… · Kinetic Theory of Gases A gasconsists of small particles

PROPERTIES OF GASESChapter 7 §7.1

Common Gases

2

Of the elements on the periodic table, some exist as a gas at room temperature, these include

the Noble Gases, Group 8A (18),

H2, N2, O2, F2, Cl2, and

many oxides of nonmetals such as CO, CO2, NO, NO2, SO2, and SO3.

Gases and Environmental Concerns

3

Some gases are responsible for environmental and health concerns including

methane, CH4 (green house gas, flammable, respiratory concerns)

chlorofluorocarbons (CFCs) (destroys the ozone layer)

nitrogen oxides found in smog (respiratory concerns)

volatile organic compounds (VOCs), such as compounds found in paint thinners (just plain old really bad stuff)

Kinetic Theory of Gases

A gas consists of small particles that

move randomly with high velocities.

have essentially no attractive (or repulsive) forces toward each other.

have a very small volume compared to the volume of the containers they occupy.

are in constant motion.

have kinetic energies that increase with an increase in temperature.

4

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Properties of GasesGases are described in terms of four properties:

1. pressure (P)

2. volume (V)

3. temperature (T)

4. amount (n)

5

GAS PRESSUREChapter 7 §7.2

Gas pressure

is the force acting on a specific area.

𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑃

has units of atm, mmHg, torr, lb/in.2, and kilopascals(kPa).1 atm = 760 mmHg (exact)1 atm = 760 torr (exact)1 atm = 14.7 lb/in.2

1 atm = 101.325 kPa

Gas Pressure

7

Atmospheric pressure is the pressure exerted by a column of air from the top of the atmosphere to the surface of the Earth.

8

Atmospheric Pressure

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Atmospheric Pressure

Atmospheric pressure

is about 1 atmosphere at sea level.

depends on the altitude and the weather.

is lower at high altitudes where the density of air is less.

is higher on a rainy day than on a sunny day.

9

Altitude and Atmospheric Pressure

10

Barometer

A barometer measures the pressure exerted by the gases in the atmosphere by the height of mercury in the column (mmHg).

A barometer: The pressure exerted by the gases in the atmosphere is equal to the downward pressure of a mercurycolumn in a closed glass tube.

11

Barometer

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MaximumDepth forWater is29 feet

PRESSURE AND VOLUME (BOYLE’S LAW)Chapter 7 §7.3

15

Robert William Boyle (1627–91)Born 25 January 1627 Lismore, County Waterford, IrelandDied 31 December 1691 (aged 64) London, EnglandNationality IrishFields Physics, chemistryEducation Eton College

BOYLE’S LAW

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Boyle’s Law

Boyle’s law states that

the pressure of a gas is inversely related to its volume when temperature (T) and amount of gas (n) are constant.

if the pressure (P) increases, then the volume (V) decreases.

17

In Boyle’s law, the product P × V is constant as long as T and n do

not change.P1V1 = 8.0 atm x 2.0 L = 16 atm*LP2V2 = 4.0 atm x 4.0 L = 16 atm*L

it can be stated that since P × V is a constant.P1V1 = P2V2 = constant (T, n constant)

PV Constant in Boyle’s Law

18

Boyle’s Law and Breathing: Inhalation

During inhalation,

the lungs expand,

the pressure in the lungs decreases, and

air flows towards the lower pressure in the lungs.

19

Boyle’s Law and Breathing: Exhalation

During exhalation,

lung volume decreases,

pressure within the lungs increases, and

air flows from the higher pressure in the lungs to the outside.

20

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7.75 At a restaurant, a customer chokes on a piece of food. You put your arms around the person’s waist and use your fists to push up on the person’s abdomen, an action called the Heimlich maneuver.

a. How would this action change the volume of the chest and lungs?

b. Why does it cause the person to expel the food item from the airway? TEMPERATURE AND VOLUME (CHARLES’ LAW)

Chapter 7 §7.4

Jacques Alexandre César Charles, 1820Born November 12, 1746 Beaugency, OrléanaisDied April 7, 1823 (aged 76) ParisNationality France

Charles’s Law

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Charles’s LawCharles’s law states that,

the Kelvin temperature of a gas is directly related to the volume of the gas,

P and n are constant, and

as the temperature of a gas increases, the molecules move faster and its volume increases to maintain constant P.

25

= Constant

Charles’s law states that when the temperature of a gas increases, making the molecules move faster, the volume of the gas must increase to maintain constant pressure.

Charles’s Law: V and T

26

TEMPERATURE AND PRESSURE(GAY-LUSSAC’S LAW)

Chapter 7 §7.5

Joseph Louis Gay-LussacBorn 6 December 1778 Saint-Léonard-de-NoblatDied 9 May 1850 (aged 71) Paris, FranceNationality FrenchFields Chemistry

Gay-Lussac’s Law

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Gay-Lussac’s Law: P and T

Gay-Lussac’s law states that the pressure exerted by a gas is

directly related to the Kelvin temperature.

V and n are constant. an increase in temperature

increases the pressure of a gas.

30

= Constant

7.79 A fire extinguisher has a pressure of 10. atm at 25 °C. What is the pressure, in atmospheres, if the fire extinguisher is used at a temperature of 75 °C and V and n remain constant?

𝑃𝑇

𝑃𝑇

10 𝑎𝑡𝑚273 25 𝐾

𝑃273 75 𝐾

348 𝐾298 𝐾 ∗ 10 𝑎𝑡𝑚 𝑃

𝑃 12 𝑎𝑡𝑚

7.81 A sample of hydrogen (H2) gas at 127 °C has a pressure of 2.00 atm. At what temperature (°C) will the pressure of the H2 decrease to 0.25 atm, if V and n remain constant?

𝑃𝑇

𝑃𝑇

2.00 𝑎𝑡𝑚273 127 𝐾

0.25 𝑎𝑡𝑚𝑇

0.25 𝑎𝑡𝑚2.00 𝑎𝑡𝑚 ∗ 400 𝐾 𝑇

𝑇 50 𝐾 220 ℃

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7.82 A sample of nitrogen (N2) gas has a pressure of 745 mmHg at 30. °C. What is the pressure when the temperature rises to 125 °C?

𝑃𝑇

𝑃𝑇

745 𝑚𝑚 𝐻𝑔273 30 𝐾

𝑃

273 125 𝐾398 𝐾303 𝐾 ∗ 745 𝑚𝑚 𝐻𝑔 𝑃

𝑃 979 𝑚𝑚 𝐻𝑔

Vapor Pressure

In a closed container, the vapor above a liquid accumulates and creates pressure called vapor pressure.

Each liquid exerts its own vapor pressure at a given temperature. As temperature increases, more vapor forms, and vapor pressure increases.

36

Vapor Pressure and Boiling Point

A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure.

37

Boiling Point of Water

The boiling point of water depends on the vapor pressure.

is lower at higher altitudes.

is increased by using an autoclave to increase external pressure.

38

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THE COMBINED GAS LAWChapter 7 §7.6

Combined Gas Law

The combined gas law uses

Boyle’s Law

Charles’s Law

Gay-Lussac’s Law

P1V1 = P2V2 = constant

= constant

= constant

𝑃 𝑉𝑇

𝑃 𝑉𝑇

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Combined Gas Law

The combined gas law:

= constant

Combined Gas Law

By using the combined gas law, we can derive any of the gas laws by omitting those properties that do not change.

44

4545

Review Gas Laws

Boyles Law Constant Temperature PV = constant

Charles Law Constant Pressure V/T = constant

Gay Lussacs Law Constant Volume P/T = constant

Combined Gas Law PV/T = constant

Ideal Gas Law PV=nRT

45

Constant Amount of Gas 𝑃 𝑉𝑇

𝑃 𝑉𝑇

Remember 1‐atm = 760 mmHg745 760⁄ 𝑎𝑡𝑚 · 425 𝑚𝑙

273 24 𝐾 0.115 𝑎𝑡𝑚 · 𝑉

273 95 𝐾0.980 𝑎𝑡𝑚0.115 𝑎𝑡𝑚 ∗

178 𝐾297 𝐾 ∗ 425 𝑚𝑙 𝑉

𝑉 2170 𝑚𝑙

7.78 Your spaceship has docked at a space station above Mars. The temperature inside the space station is a carefully controlled 24 °C at a pressure of 745 mmHg. A balloon with a volume of 425 mL drifts into the airlock where the temperature is ‐95 °C and the pressure is 0.115 atm. 

What is the new volume, in milliliters, of the balloon if n remains constant and the balloon is very elastic? 

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7.80 A weather balloon has a volume of 750 L when filled with helium at 8 °C at a pressure of 380 torr. What is the new volume of the balloon when the pressure is 0.20 atm, the temperature is ‐45 °C, and n remains constant?

𝑃 𝑉𝑇

𝑃 𝑉𝑇

Remember 1‐atm = 760 mmHg380 𝑡𝑜𝑟𝑟 · 750 𝐿

273 8 𝐾 0.20 ∗ 760 𝑡𝑜𝑟𝑟 𝑉

273 45 𝐾380 𝑡𝑜𝑟𝑟150 𝑡𝑜𝑟𝑟 ∗

228 𝐾281 𝐾 ∗ 750𝐿 𝑉

𝑉 1,600𝐿

VOLUME AND MOLES(AVOGADRO’S LAW)

Chapter 7 §7.7

Avogadro’s Law: Volume and Moles

Avogadro’s law states that the volume of a gas is directly

related to the number of moles (n) of gas.

T and P are constant.

49

= constant

Amedeo AvogadroBorn 9 August 1776 Turin, Piedmont-SardiniaDied 9 July 1856 (aged 79) Turin, Piedmont-SardiniaNationality Italian

Avogadro’s Law

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To make comparisons between different gases, we use arbitrary conditions called standard temperature (273 K) and standard pressure (1 atm). Standard temperature and pressure is abbreviated as STP.

Standard temperature (T) = 0 C or 273 K

Standard pressure (P) = 1 atm (760 mmHg)

STP

51

Molar VolumeThe molar volume of a gas measured at STP (standard temperature and pressure) is 22.4 L for 1 mole of any gas.

Avogadro’s law indicates that 1 mole of any gas at STP has a volume of 22.4 L.52

Molar Volume as a Conversion Factor

The molar volume at STP has about the same volume as 3 basketballs. can be used to form 2 conversion factors.

53

22.4 𝐿1 𝑚𝑜𝑙𝑒 𝑎𝑛𝑑

1 𝑚𝑜𝑙𝑒22.4 𝐿

Gases in Chemical Reactions

The volume or amount of a gas at STP in a chemicalreaction can be calculated from

STP conditions.

mole−mole factors from the balanced equation.

54

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7.87 When heated, calcium carbonate decomposes to give calcium oxide and carbon dioxide gas. If 56.0 g of CaCO3react, how many liters of CO2 gas are produced at STP?

CaCO3(s) → CaO(s)+ CO2(g)

𝐶𝑎𝐶𝑂 ∆→ 𝐶𝑎𝑂 𝐶𝑂

𝑛 56.0 𝑔𝐶𝑎𝐶𝑂 ∗ 1 𝑚𝑜𝑙𝑒 𝐶𝑎𝐶𝑂100 𝑔 𝐶𝑎𝐶𝑂

0.560 𝑚𝑜𝑙𝑒𝑠 𝐶𝑎𝐶𝑂

𝑚𝑜𝑙𝑒𝑠 𝐶𝑎𝐶𝑂 𝑚𝑜𝑙𝑒𝑠 𝐶𝑂 0.560 𝑚𝑜𝑙𝑒𝑠

0.560 𝑚𝑜𝑙𝑒𝑠 𝐶𝑂 ∗ 22.4 𝐿1 𝑚𝑜𝑙𝑒 12.5 𝐿 𝐶𝑂

7.88 Magnesium reacts with oxygen to form magnesium oxide. How many liters of oxygen gas at STP are needed to react completely with 8.0 g of magnesium?

2Mg(s)+ O2(g) → 2MgO(s)

2𝑀𝑔 𝑂 ∆→ 2𝑀𝑔𝑂

𝑛 8.0 𝑔 𝑀𝑔 ∗ 1 𝑚𝑜𝑙𝑒 𝑀𝑔24.3 𝑔 𝑀𝑔 0.33 𝑚𝑜𝑙𝑒𝑠 𝑀𝑔

2 𝑚𝑜𝑙𝑒𝑠 𝑀𝑔 1 𝑚𝑜𝑙𝑒𝑠 𝑂 0.16 𝑚𝑜𝑙𝑒𝑠 𝑂

0.16 𝑚𝑜𝑙𝑒𝑠 𝑂 ∗22.4 𝐿1 𝑚𝑜𝑙𝑒 3.8 𝐿 𝑂

7.89 In the Haber process, H2 and N2 react to produce ammonia (NH3). How many grams of N2 are needed to produce 150 L of ammonia at STP?

3H2(g)+ N2(g) → 2NH3(g)

3𝐻 𝑁 → 2𝑁𝐻

150 𝐿 𝑁𝐻 ∗ 1 𝑚𝑜𝑙𝑒

22.4 𝐿𝑖𝑡𝑒𝑟𝑠 6.7 𝑚𝑜𝑙𝑒𝑠 𝑁𝐻

6.7 𝑚𝑜𝑙𝑒𝑠 𝑁𝐻 ∗ 1 𝑚𝑜𝑙𝑒𝑠 𝑁

2 𝑚𝑜𝑙𝑒𝑠 𝑁𝐻 3.3 𝑚𝑜𝑙𝑒𝑠 𝑁

3.3 𝑚𝑜𝑙𝑒𝑠 𝑁 ∗ 28.0 𝑔𝑟𝑎𝑚𝑠 𝑁

1 𝑚𝑜𝑙𝑒 𝑁 94 𝑔𝑟𝑎𝑚𝑠 𝑁

7.90 How many liters of H2 gas at STP can be produced from the reaction of 2.45 g of Al with excess HCl?

2Al(s)+ 6HCl(aq) → 2AlCl3(aq)+ 3H2(g)

2𝐴𝑙 6𝐻𝐶𝑙 → 2𝐴𝑙𝐶𝑙 3𝐻

2.45 𝑔 𝐴𝑙 ∗ 1 𝑚𝑜𝑙𝑒 𝐴𝑙27.0 𝑔 𝐴𝑙 0.0907 𝑚𝑜𝑙𝑒𝑠 𝐴𝑙

0.0907 𝑚𝑜𝑙𝑒𝑠 𝐴𝑙 ∗ 3 𝑚𝑜𝑙𝑒𝑠 𝐻2 𝑚𝑜𝑙𝑒𝑠 𝐴𝑙 0.136 𝑚𝑜𝑙𝑒𝑠 𝐻

0.136 𝑚𝑜𝑙𝑒𝑠 𝐻 ∗22.4 𝐿𝑖𝑡𝑒𝑟𝑠

1 𝑚𝑜𝑙𝑒 3.05 𝐿 𝐻

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7.91 Aluminum oxide can be formed from its elements. What volume, in liters, of oxygen at STP is needed to completely react 5.4 g of Al?

4Al(s)+ 3O2(g) → 2Al2O3(s)

4𝐴𝑙 3𝑂∆→2𝐴𝑙 𝑂

5.4 𝑔 𝐴𝑙 ∗ 1 𝑚𝑜𝑙𝑒 𝐴𝑙27.0 𝑔 𝐴𝑙 ∗

3 𝑚𝑜𝑙𝑒𝑠 𝑂4 𝑚𝑜𝑙𝑒𝑠 𝐴𝑙 ∗

22.4 𝐿 𝑂1 𝑚𝑜𝑙𝑒 𝑂 3.4 𝐿 𝑂

7.92 Glucose, C6H12O6, is metabolized in living systems to CO2 and H2O. How many grams of water can be produced from 12.5 L of O2 at STP?

C6H12O6(s) + 6O2(g) → 2CO2(g)+ 6H2O(l)

𝐶 𝐻 𝑂 6𝑂 → 2𝐶𝑂 6𝐻 𝑂

12.5 𝐿 𝑂 ∗ 1 𝑚𝑜𝑙𝑒 𝑂22.4 𝐿 𝑂 ∗

6 𝑚𝑜𝑙𝑒𝑠 𝐻 𝑂6 𝑚𝑜𝑙𝑒𝑠 𝑂 ∗

18.0 𝑔 𝐻 𝑂1 𝑚𝑜𝑙𝑒 𝐻 𝑂 10.0 𝑔

7.94 What is the molar mass, g/mole, of a gas if 1.15 g of the gas has a volume of 225 mL at STP?

0.225 𝐿 ∗ 1 𝑚𝑜𝑙𝑒22.4 𝐿 0.0100 𝑚𝑜𝑙𝑒𝑠

𝑀𝑜𝑙𝑎𝑟 𝑀𝑎𝑠𝑠 𝐺𝑟𝑎𝑚𝑠𝑀𝑜𝑙𝑒

1.15 𝑔𝑟𝑎𝑚𝑠0.0100 𝑚𝑜𝑙𝑒𝑠 114 𝑔𝑟𝑎𝑚𝑠

𝑚𝑜𝑙𝑒

7.95 Nitrogen dioxide reacts with water to produce oxygen and ammonia. How many liters of O2 are produced when 0.42 mole of NO2 reacts at STP?

4NO2(g) + 6H2O(g) → 7O2(g) + 4NH3(g)

4𝑁𝑂 6𝐻 𝑂 ∆→ 7𝑂 4𝑁𝐻

0.42 𝑚𝑜𝑙𝑒 𝑁𝑂 ∗ 7 𝑚𝑜𝑙𝑒𝑠 𝑂

4 𝑚𝑜𝑙𝑒𝑠 𝑁𝑂 ∗22.4 𝐿 𝑂𝑚𝑜𝑙𝑒 𝑂 16 𝐿 𝑂

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7.96 What is the volume, in liters, of H2 gas produced at STP from the reaction of 25.0 g of Al?

2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2(g)

2𝐴𝑙 3𝐻 𝑆𝑂 → 𝐴𝑙 𝑆𝑂 3𝐻

25.0 𝑔 𝐴𝑙 ∗ 1 𝑚𝑜𝑙𝑒 𝐴𝑙

27.0 𝑔𝑟𝑎𝑚𝑠 𝐴𝑙 ∗ 3 𝑚𝑜𝑙𝑒𝑠 𝐻2 𝑚𝑜𝑙𝑒𝑠 𝐴𝑙 ∗

22.4 𝐿 𝐻1 𝑚𝑜𝑙𝑒 𝐻 31.1 𝐿 𝐻

7.97 A weather balloon is partially filled with helium to allow for expansion at high altitudes. At STP, a weather balloon is filled with enough helium to give a volume of 25.0 L. 

How many grams of helium were added to the balloon?

25.0 𝐿 𝐻𝑒 ∗ 1 𝑚𝑜𝑙𝑒 𝐻𝑒22.4 𝐿 𝐻𝑒 ∗

4.00 𝑔 𝐻𝑒1 𝑚𝑜𝑙𝑒 𝐻𝑒 4.46 𝑔 𝐻𝑒

THE IDEAL GAS LAWChapter 7 §7.8

The four properties used in the measurement of a gas, pressure (P), volume (V), temperature (T), and amount (n),can be combined to give a single expression called theideal gas law. PV = nRT

Ideal Gas Law

70

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Rearranging the ideal gas law equation shows that the four gas properties equal a constant, R.

To calculate the value of R, we substitute the STP conditions for molar volume into the expression:

R, Ideal Gas Constant

71

𝑷𝑽 𝒏𝑹𝑻

𝑷𝑽𝒏𝑻

= R

𝑹 𝟏. 𝟎𝟎 𝒂𝒕𝒎 𝟐𝟐. 𝟒 𝑳

𝟏 𝒎𝒐𝒍𝒆 𝟐𝟕𝟑 𝑲 𝟎. 𝟎𝟖𝟐𝟏 𝑳 · 𝒂𝒕𝒎

𝒎𝒐𝒍𝒆 · 𝑲

Unit Summary for R, the Ideal Gas Constant

72

7.77 In 1783, Jacques Charles launched his first balloon filled with hydrogen gas, which he chose because it was lighter than air. The balloon had a volume of 31,000 L when it reached an altitude of 1000 m, where the pressure was 658 mmHg and the temperature was   ‐8 °C. 

How many kilograms of hydrogen were used to fill the balloon at STP?

𝑃𝑉 𝑛𝑅𝑇

𝑛 𝑃𝑉𝑅𝑇

𝑛 658 𝑚𝑚 𝐻𝑔 31,000 𝐿

0.0821 𝑎𝑡𝑚 · 𝐿𝑚𝑜𝑙𝑒 · 𝐾 273 8 𝐾

∗1 𝑎𝑡𝑚

760 𝑚𝑚 𝐻𝑔

1,200 𝑚𝑜𝑙𝑒𝑠

1,200 𝑚𝑜𝑙𝑒𝑠 𝐻 ∗ 2.0 𝑔 𝐻

1 𝑚𝑜𝑙𝑒 𝐻 ∗ 1 𝑘𝑔

1,000 𝑔 2.4 𝐾𝑔 𝐻

7.83 How many moles of CO2 are in 35.0 L of CO2(g) at 1.2 atm and 5 °C?

𝑃𝑉 𝑛𝑅𝑇

𝑛 𝑃𝑉𝑅𝑇

𝑛 1.2 𝑎𝑡𝑚 · 35.0𝐿

0.0821 𝑎𝑡𝑚 · 𝐿𝑚𝑜𝑙𝑒 · 𝐾 · 273 5 𝐾

𝑛 1.8 𝑚𝑜𝑙𝑒𝑠

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7.84 A container is filled with 0.67 mole of O2 at 5 °C and 845 mmHg.  What is the volume, in milliliters, of the container?

𝑃𝑉 𝑛𝑅𝑇

𝑉𝑛𝑅𝑇

𝑃

𝑉0.67 𝑚𝑜𝑙𝑒 · 0.0821 𝑎𝑡𝑚 · 𝐿

𝑚𝑜𝑙𝑒 · 𝐾 · 273 5 𝐾

845 𝑚𝑚 𝐻𝑔760 𝑚𝑚 𝐻𝑔/𝑎𝑡𝑚

𝑉 14 𝐿𝑉 14,000 𝑚𝐿

7.85 A 2.00‐L container is filled with methane gas (CH4) at a pressure of 2500. mmHg and a temperature of 18 °C. How many grams of methane are in the container?

𝑃𝑉 𝑛𝑅𝑇

𝑛 𝑃𝑉𝑅𝑇

𝑛 2500. 𝑚𝑚 𝐻𝑔 2.00 𝐿

0.0821 𝑎𝑡𝑚 · 𝐿𝑚𝑜𝑙𝑒 · 𝐾 273 18 𝐾

∗1 𝑎𝑡𝑚

760 𝑚𝑚 𝐻𝑔 0.275 𝑚𝑜𝑙𝑒𝑠

0.275 𝑚𝑜𝑙𝑒𝑠 𝐶𝐻 ∗16.0 𝑔 𝐶𝐻1 𝑚𝑜𝑙𝑒 𝐶𝐻 ∗

1 𝑘𝑔1,000 𝑔 4.40 𝑔 𝐶𝐻

7.86 A steel cylinder with a volume of 15.0 L is filled with 50.0 g of nitrogen gas at 25 °C. What is the pressure, in atmospheres, of the N2 gas in the cylinder?

𝑃𝑉 𝑛𝑅𝑇

𝑃 𝑛𝑅𝑇

𝑉

𝑛 50.0 𝑔𝑁 ∗ 1 𝑚𝑜𝑙𝑒 𝑁28.0 𝑔 𝑁 1.79 𝑚𝑜𝑙𝑒𝑠

𝑃 1.79 𝑚𝑜𝑙𝑒𝑠 · 0.0821 𝑎𝑡𝑚 · 𝐿

𝑚𝑜𝑙𝑒 · 𝐾 · 273 25 𝐾

15.0 𝐿𝑃 2.91 𝑎𝑡𝑚

7.93 A sample of gas with a mass of 1.62 g has a volume of 941 mL at a pressure of 748 torr and a temperature of 20 °C. 

What is the molar mass, g/mole, of the gas?

𝑃𝑉 𝑛𝑅𝑇

𝑛 𝑃𝑉𝑅𝑇

𝑛 748 𝑡𝑜𝑟𝑟 0.941𝐿

0.0821 𝑎𝑡𝑚 · 𝐿𝑚𝑜𝑙𝑒 · 𝐾 273 20 𝐾

∗1 𝑎𝑡𝑚

760 𝑡𝑜𝑟𝑟 0.0385 𝑚𝑜𝑙𝑒𝑠

𝑀𝑜𝑙𝑎𝑟 𝑀𝑎𝑠𝑠 𝐺𝑟𝑎𝑚𝑠𝑀𝑜𝑙𝑒

1.62 𝑔𝑟𝑎𝑚𝑠0.0385 𝑚𝑜𝑙𝑒𝑠 42.1 𝑔𝑟𝑎𝑚𝑠

𝑚𝑜𝑙𝑒

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7.98 At an altitude of 30.0 km, where the temperature is ‐35 °C, a weather balloon containing 1.75 moles of helium has a volume of 2,460 L. 

What is the pressure, in mmHg, of the helium inside the balloon?

𝑃𝑉 𝑛𝑅𝑇

𝑃 𝑛𝑅𝑇

𝑉

𝑃 1.75 𝑚𝑜𝑙𝑒𝑠 · 0.0821 𝑎𝑡𝑚 · 𝐿

𝑚𝑜𝑙𝑒 · 𝐾 · 273 35 𝐾

2,460 𝐿 0.0139 𝑎𝑡𝑚

0.0139 𝑎𝑡𝑚 ∗ 760 𝑚𝑚 𝐻𝑔

1 𝑎𝑡𝑚 10.6 𝑚𝑚 𝐻𝑔

PARTIAL PRESSURES(DALTON’S LAW)

Chapter 7 §7.9

In a gas mixture, each gas exerts its partial pressure, which is the pressure it would exert if it were the only gas in the container.

Dalton’s law states that the total pressure of a gas mixture is the sum of the partial pressures of the gases in the mixture.

Partial Pressure, Dalton’s Law

81

𝑃 𝑃 𝑃 + 𝑃

Dalton’s Law of Partial Pressures indicates that pressure depends on the total number of gas particles.

Dalton’s Law of Partial Pressures

82

PHe = 2.0 atm PAr = 4.0 atm Ptotal = PHe + PAr = 2.0 atm + 4.0 atm = 6.0 atm

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For example, at STP, one mole of a pure gas willexert the same pressure as one mole of a gasmixture in a 22.4 L container.

V = 22.4 LGas mixtures

Total Pressure

0.5 mole O20.3 mole He0.2 mole Ar1.0 mole

1.0 mole N2

0.4 mole O20.6 mole He1.0 mole

1.0 atm 1.0 atm 1.0 atm

83

Composition of Air

Air is a mixture of gases, including nitrogen, oxygen, carbon dioxide, argon, and water gases.

84

Blood Gases

In the lungs, O2 enters the blood, at the same time that CO2 is released.

In the tissues, O2 enters the cells, which release CO2 into the blood.

Blood Gases

In the body, O2 flows into the tissues because the partial pressure of O2 is

higher in blood and lower in the tissues. CO2 flows out of the tissues because the partial pressure of

CO2 is higher in the tissues and lower in the blood.

Partial Pressures (mmHg) in Blood and TissueGas Oxygenated Deoxygenated Tissues

Blood BloodO2 100 mmHg 40 mmHg 30 mmHg or lessCO2 40 mmHg 46 mmHg 50 mmHg or greater

86

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Changes in Partial Pressures of Blood Gases During Breathing

87

7.76 An airplane is pressurized to 650. mmHg.

a. If air is 21% oxygen, what is the partial pressure of oxygen on the plane?

b. If the partial pressure of oxygen drops below 100. mmHg, passengers become drowsy. If the total cabin pressure is lowered this happens and oxygen masks are released. 

What is the total cabin pressure at which oxygen masks are dropped?

650 𝑚𝑚 𝐻𝑔 𝑥 21% 𝑂 140 𝑚𝑚 𝐻𝑔 is the partial pressure of 𝑂

100 𝑚𝑚 𝑂21% 𝑂 480 𝑚𝑚 𝐻𝑔 𝑡𝑜𝑡𝑎𝑙 𝑐𝑎𝑏𝑖𝑛 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒

7.99 A gas mixture contains oxygen and argon at partial pressures of 0.60 atmand 425 mm Hg. If nitrogen gas added to the sample increases the total pressure to 1,250 torr, what is the partial pressure, in torr, of the nitrogen added?

𝑃𝑎𝑟𝑡𝑖𝑎𝑙 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑑𝑢𝑒 𝑡𝑜 𝑂 0.60 𝑎𝑡𝑚 ∗ 760 𝑡𝑜𝑟𝑟

1 𝑎𝑡𝑚 460 𝑡𝑜𝑟𝑟

𝑃𝑎𝑟𝑡𝑖𝑎𝑙 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑑𝑢𝑒 𝑡𝑜 𝐴𝑟 425 𝑚𝑚 𝐻𝑔 ∗ 1 𝑡𝑜𝑟𝑟

1 𝑚𝑚 𝐻𝑔 425 𝑡𝑜𝑟𝑟

𝑃𝑎𝑟𝑡𝑖𝑎𝑙 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑑𝑢𝑒 𝑡𝑜 𝑁 1,250 460 425 𝑡𝑜𝑟𝑟 360 𝑡𝑜𝑟𝑟

7.100 A gas mixture contains helium and oxygen at partial pressures of 255 torr and 0.450 atm. What is the total pressure, in mmHg, of the mixture after it is placed in a container one‐half the volume of the original container?

𝑃𝑎𝑟𝑡𝑖𝑎𝑙 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑑𝑢𝑒 𝑡𝑜 𝐻𝑒 255 𝑡𝑜𝑟𝑟 ∗ 1 𝑚𝑚 𝐻𝑔

1 𝑡𝑜𝑟𝑟 255 𝑚𝑚 𝐻𝑔

𝑃𝑎𝑟𝑡𝑖𝑎𝑙 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑑𝑢𝑒 𝑡𝑜 𝑂 0.450 𝑎𝑡𝑚 ∗ 760 𝑚𝑚 𝐻𝑔

1 𝑎𝑡𝑚 342 𝑚𝑚 𝐻𝑔

𝑇𝑜𝑡𝑎𝑙 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑒𝑟 255 342 𝑚𝑚 𝐻𝑔 597 𝑚𝑚 𝐻𝑔𝑃 𝑉 𝑃 𝑉

𝐼𝑓 𝑡ℎ𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑖𝑠 𝑐𝑢𝑡 𝑖𝑛 ℎ𝑎𝑙𝑓, 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑑𝑜𝑢𝑏𝑙𝑒𝑑𝑇𝑜𝑡𝑎𝑙 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 ℎ𝑎𝑙𝑓 𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑒𝑟 2 ∗ 597 𝑚𝑚 𝐻𝑔 1190 𝑚𝑚 𝐻𝑔

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SLO – Student Learning ObjectivesClass Learning Objectives:

Conceptualize, model and explain chemical processes qualitatively at the molecular level. Extract appropriate information, analyze and synthesize experimental results to reach correct conclusions. Perform laboratory techniques safely and accurately and maintain a laboratory notebook according to

standard scientific guidelines.

College Student Learning Objectives:Written and Oral Communication Problem Solving Math Competency

Assessment Method: Embedded questions in exams. Evaluation of lab reports and lab notebooks.

Regarding the test covering chapters 7, 8 & 10…Stuff that will be provided 1 gram = 0.03527

ounces 1 kg = 2.20 lbs (exact) 2.54 cm = 1 inch (exact) 1 liter = 1.06 quarts 1 ml = 1 cm3 (exact) 1 ml H2O = 1 g (exact) 1 cal = 1 g°C for water

(exact) TF = 1.8Tc + 32 (exact) 1 cal = 4.18400 joules Carbohydrate 4 kcal/g

Fat 9 kcal/g Protein 4 kcal/g 1 atm = 760 mm Hg R = 0.0821 atmꞏL/moleꞏK 22.4L = 1 mole at STP A periodic Table Solubility Rules

Someone needs to email me.

All the paper you will need

Stuff you need to bring

A writing instrument A calculator

Did I miss anything?

And Finally…

• Please get sleep the night before the final.

• Please eat decent food before the test.

• Please relax and take you time during the final.

• Grades will be sent out in approximately two weeks.

Thank you each and every one on you. You have greatly enriched my life by attending these classes. I hope that you continue your education and are successful in life.

Sincerely,

Prof.M.

The final is on Thursday May 31st

in this classroom at 5:30 PM