Kinetic Theory of Gases. 6.1 Properties of Gases 6.2 Gas Pressure. Properties That Describe a Gas. Gas Pressure. Learning Check.

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Chapter 6 Gases

6.1 Properties of Gases

6.2 Gas Pressure

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Kinetic Theory of Gases

A gas consists of small particles that

• move rapidly in straight lines.

• have essentially no attractive (or repulsive) forces.

• are very far apart.

• have very small volumes compared to the volume of the container they occupy.

• have kinetic energies that increase

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Properties That Describe a Gas

Gases are described in terms of four properties:

pressure (P), volume (V), temperature (T), and amount (n).

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Gas pressure

• is a force acting on a specific area.

Pressure (P) = force area

• has units of atm, mmHg, torr, lb/in.², and kilopascals(kPa).

1 atm = 760 mm Hg (exact)

1 atm = 760 torr

1 atm = 14.7 lb/in.²

1 atm = 101 325 Pa

1 atm = 101.325 kPa

Gas Pressure

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A. What is 475 mmHg expressed in atm?

1) 475 atm 2) 0.625 atm 3) 3.61 x 10⁵atm

B. The pressure in a tire is 2.00 atm. What is this pressure in mmHg?

1) 2.00 mmHg 2) 1520 mmHg 3) 22 300 mmHg

Learning Check

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Solution

A. What is 475 mmHg expressed in atm?

2) 0.625 atm

475 mmHg x 1 atm = 0.625 atm 760 mmHg

B. The pressure in a tire is 2.00 atm. What is this pressure in mmHg?

2) 1520 mmHg

2.00 atm x 760 mmHg = 1520 mmHg 1 atm

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Atmospheric Pressure

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

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Altitude and Atmospheric Pressure

Atmospheric pressure

• is about 1 atmosphere at sea level.

• depends on the altitude and the weather.

• is lower at higher altitudes, where the density of air is less.

• is higher on a rainy day

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Barometer

A barometer

• measures the pressure exerted by the gases in the atmosphere.

• indicates atmospheric pressure as the height in mm of the mercury column.

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

1) greater 2) less 3) the same B. A water barometer is 13.6 times taller than a Hg

barometer (D_Hg= 13.6 g/mL) because 1) H₂O is less dense than mercury.

2) H₂O is heavier than mercury.

3) air is more dense than H₂O.

Learning Check

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

B. A water barometer is 13.6 times taller than a Hg barometer (D_Hg= 13.6 g/mL) because

1) H₂O is less dense than mercury.

Solution Chapter 6 Gases

6.3 Pressure and Volume (Boyle’s Law)

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Boyle’s Law

Boyle’s lawstates that

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

• if volume decreases, the pressure increases.

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In Boyle’s law, the product P x V is constant as long as T and n do not change.

P₁V₁= 8.0 atm x 2.0 L = 16 atm L P₂V₂= 4.0 atm x 4.0 L = 16 atm L P₃V₃= 2.0 atm x 8.0 L = 16 atm L Boyle’s law can be stated as

P₁V₁ = P₂V₂ (T, n constant)

PV Constant in Boyle’s Law

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Solving for a Gas Law Factor

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

P₁V₁ = P₂V2 Boyle’s law To solve for V₂, divide both sides by P₂.

P₁V₁ = P₂V₂

P₂ P₂

V₁ x P₁ = V₂ P₂

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Boyles’ Law and Breathing

During an inhalation,

• the lungs expand.

• the pressure in the lungs decreases.

• air flows towards the lower pressure in the lungs.

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Boyles’ Law and Breathing

During an exhalation,

• lung volume decreases.

• pressure within the lungs increases.

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

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Calculations with Boyle’s Law

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Freon-12, CCl₂F₂, is used in refrigeration systems.

What is the new volume (L) of a 8.0 L sample of Freon gas initially at 550 mmHg after its pressure is changed to 2200 mmHg at constant T and n?

1. Set up a data table:

Conditions 1 Conditions 2 P₁ = 550 mmHg P₂= 2200 mmHg V₁ = 8.0 L V₂ =

Calculation with Boyle’s Law

?

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2. When pressure increases, volume decreases.

Solve Boyle’s law for V₂: P₁V₁ = P₂V₂

V₂ = V₁x P₁ P₂

V₂ = 8.0 L x 550 mmHg = 2.0 L 2200 mmHg pressure ratio decreases volume

Calculation with Boyle’s Law

(continued)

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Learning Check

For a cylinder containing helium gas, indicate if cylinder A or cylinder B represents the new volume for the following changes (n and T are constant).

1) pressure decreases 2) pressure increases

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Solution

For a cylinder containing helium gas, indicate if cylinder A or cylinder B represents the new volume for the following changes (n and T are constant).

1) pressure decreases B 2) pressure increases A

Learning Check

If a sample of helium gas has a volume of 120 mL and a pressure of 850 mmHg, what is the new volume if the pressure is changed to 425 mmHg?

1) 60 mL 2) 120 mL 3) 240 mL

3) 240 mL

P₁ = 850 mmHg P₂ = 425 mmHg

V₁ = 120 mL V₂ = ??

V₂ = V₁x P₁ = 120 mL x 850 mmHg = 240 mL

P₂ 425 mmHg

Pressure ratio increases volume

Solution

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Learning Check

A sample of helium gas in a balloon has a volume of 6.4 L at a pressure of 0.70 atm. At 1.40 atm (T is constant), is the new volume represented by A, B, or C?

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Solution

A sample of helium gas in a balloon has a volume of 6.4 L at a pressure of 0.70 atm. At a higher pressure (T constant), the new volume is represented by the smaller balloon A.

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

A) 3.2 L B) 6.4 L C) 12.8 L

Learning Check

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Solution

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

A) 3.2 L

V₂ = V₁x P₁ P₂

V₂ = 6.4 L x 0.70 atm = 3.2 L 1.40 atm

Volume decreases when there is an increase in the pressure (temperature is constant.)

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Learning Check

A sample of oxygen gas has a volume of 12.0 L at 600. mmHg.

What is the new pressure when the volume changes to 36.0 L?

(T and n constant).

1) 200. mmHg 2) 400. mmHg 3) 1200 mmHg

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Solution

1) 200. mmHg Data Table

Conditions 1 Conditions 2 P₁ = 600. mmHg P₂ = ???

V₁ = 12.0 L V₂ = 36.0 L P₂ = P₁ x V₁

V₂

600. mmHg x 12.0 L = 200. mmHg 36.0 L

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If the sample of nitrogen (gas) has a volume of 360 mL at a pressure of 720 mmHg, what is the new volume when the pressure is increased to 1.20 atm (T constant)?

A) 284 mL B) 456 mL C) 2160 mL

Learning Check

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Solution

We need to make the units for initial and final pressure the same:

1.20 atm x 760 mmHg = 912 mmHg 1 atm

V₂ = V₁x P₁ P₂

V₂ = 360 mL x 720 mmHg = 284 mL (A) 912 mmHg

Volume decreases when there is an increase in the pressure (temperature is constant.)

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Chapter 6 Gases

6.4 Temperature and Volume

(Charles’s Law)

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Charles’s Law

In Charles’s Law,

• the Kelvin temperature of a gas is directly related to the volume.

• P and n are constant.

• when the temperature of a gas increases, its volume increases.

Charles’s Law: V and T

• For two conditions, Charles’s law is written V₁ = V₂(P and n constant) T₁ T₂

• Rearranging Charles’s law to solve for V₂: T₂x V₁ = V₂x T₁

T₁ T₁

V₂ = V₁ x T₂ T₁

Learning Check

Solve Charles’s law expression for T_2.

V₁ = V₂ T₁ T₂

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Solution

V₁ = V₂ T₁ T₂

Cross-multiply to give:

V₁T₂ = V₂T₁

Isolate T₂by dividing through by V₁: V₁T₂ = V₂T₁

V₁ V₁

T₂ = T₁x V₂ V₁

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A balloon has a volume of 785 mL at 21 °C. If the temperature drops to 0 °C, what is the new volume of the balloon (P constant)?

1. Set up data table:

Conditions 1 Conditions 2

V₁ = 785 mL V₂ = ?

T₁ = 21 °C = 294 K T₂ = 0 °C = 273 K Be sure to use the Kelvin (K) temperature in gas calculations.

Calculations Using Charles’s Law

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Calculations Using Charles’s Law

(continued)

2. Solve Charles’s law for V₂:

V₁ = V₂ T₁ T₂ V₂ = V₁ x T₂

T₁

V₂ = 785 mL x 273 K = 729 mL 294 K

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

1) 443 °C 2) 170 °C 3) - 82 °C

Learning Check

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2) 170 °C T₂ = T₁x V₂

V₁

T₂ = 291 K x 640 mL = 443 K 420 mL

= 443 K – 273 = 170 °C

Solution

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Use the gas laws to complete each sentence with 1) increases or 2) decreases.

A. Pressure _______ when V decreases.

B. When T decreases, V _______.

C. Pressure _______ when V changes from 12 L to 24 L.

D. Volume _______when T changes from 15 °C to 45 °C.

Learning Check

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Use the gas laws to complete each sentence with 1) increases or 2) decreases.

A. Pressure 1) increaseswhen V decreases.

B. When T decreases, V2) decreases.

C. Pressure 2) decreaseswhen V changes from 12 L to 24 L.

D. Volume 1) increaseswhen T changes from 15 °C to 45 °C.

Solution

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Chapter 6 Gases

6.5 Temperature and Pressure

(Gay-Lussac’s Law)

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Gay-Lussac’s Law: P and T

In Gay-Lussac’s law

• the pressure exerted by a gas is directly related to the Kelvin temperature.

• V and n are constant.

P₁ = P₂ T₁ T₂

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Learning Check

Solve Gay-Lussac’s law for P₂. P₁ = P₂

T₁ T₂

Solution

Solve Gay-Lussac’s law for P₂. P₁ = P₂

T₁ T₂

Multiply both sides by T₂and cancel:

P₁x T₂ = P₂x T₁

T₁ T₁

P₂ = P₁ x T₂ T₁

A gas has a pressure at 2.0 atm at 18 °C. What is the new pressure when the temperature is 62 °C?

(V and n constant) 1. Set up a data table:

P₁ = 2.0 atm P₂ =

T₁ = 18 °C + 273 T₂ = 62 °C + 273

= 291 K = 335 K

Calculation with Gay-Lussac’s

Law

?

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Calculation with Gay-Lussac’s

Law (continued)

2. Solve Gay-Lussac’s law for P₂: P₁ = P₂

T₁ T₂ P₂ = P₁ x T₂

T₁

P₂ = 2.0 atm x 335 K = 2.3 atm 291 K

temperature ratio increases pressure

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Learning Check

A gas has a pressure of 645 torr at 128 °C. What is the temperature in Celsius if the pressure increases to 824 torr? (n and V remain constant)

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Solution

A gas has a pressure of 645 torr at 128 °C. What is the temperature in Celsius if the pressure increases to 1.50 atm? (n and V remain constant)

1. Set up a data table:

P₁= 645 torr P₂= 1.50 atm x 760 torr = 1140 torr 1 atm

T₁= 128 °C + 273 T₂ = K – 273 = ? °C

= 401 K

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Solution

2. Solve Gay-Lussac’s law for T₂: P₁ = P₂

T₁ T₂ T₂ = T₁ x P₂

P₁

T₂ = 401 K x 1140 torr = 709 K - 273 = 436 °C 645 torr

pressure ratio increases temperature

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Chapter 6 Gases

6.6 The Combined Gas Law

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The combined gas law uses Boyle’s

law, Charles’s law, and Gay-Lussac’s

law (n is constant).

P

₁

V

₁

= P

₂

V

₂

T

₁

T

₂

Combined Gas Law

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A gas has a volume of 675 mL at 35 °C

and 0.850

atm pressure. What is the volume (mL)

of the gas at

-95 °C and a pressure of 802 mmHg? (n

constant)

Learning Check

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Solution

Data Table

T₁= 308 K T₂= -95 °C + 273 = 178 K

V₁= 675 mL V₂= ???

P₁= 646 mmHg P₂= 802 mmHg Solve for V₂:

V₂= V₁ x P₁x T₂ P2 T₁

V₂= 675 mL x 646 mmHg x 178 K = 314 mL 802 mmHg x 308 K

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A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm, and a temperature of 29 °C.

At what temperature (°C) will the helium have a volume of 90.0 mL and a pressure of 3.20 atm? (n is constant)

1. Set up data table.

Conditions 1 Conditions 2 P₁= 0.800 atm P₂ = 3.20 atm V₁= 0.180 L (180 mL) V₂= 90.0 mL T₁= 29 °C + 273 = 302 K T₂ = ??

Combined Gas Law Calculation

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Combined Gas Law Calculation

(continued)

2. Solve for T₂: P₁V₁= P₂V₂ T₁ T₂ T₂ = T₁x P₂ x V₂

P1 V₁

T₂= 302 K x 3.20 atm x 90.0 mL = 604 K 0.800 atm 180.0 mL

T₂ = 604 K - 273 = 331 °C

Chapter 6 Gases

6.7 Volume and Moles (Avogadro’s Law)

Avogadro's Law: Volume and

Moles

In Avogadro’s law

• the volume of a gas is directly related to the number of moles (n) of gas.

• T and P are constant.

V₁ = V₂ n₁ n₂

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Learning Check

If 0.75 mole of helium gas

occupies a volume of 1.5

L, what volume will 1.2

moles of helium occupy

at the same temperature

and pressure?

1) 0.94 L

2) 1.8 L

3) 2.4 L

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Solution

3) 2.4 L

STEP 1: Conditions 1 Conditions 2

V₁ = 1.5 L V₂= ???

n₁ = 0.75 mole of He n₂= 1.2 moles of He

STEP 2: Solve for unknown V₂. V₂ = V₁x n₂

n₁

STEP 3: Substitute values and solve for V₂. V₂ = 1.5 L x 1.2 moles He = 2.4 L

0.75 mole He

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The volumes of gases can be compared at

STP, Standard Temperature and Pressure,

when they have

•the same temperature.

standard temperature (T) 0 °C or 273 K

•the same pressure.

standard pressure (P) 1 atm (760 mmHg)

STP

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Molar Volume

At standard temperature and pressure

(STP), 1 mole of a gas occupies a

volume of 22.4 L, which is called its

molar volume.

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Molar Volume as a Conversion

Factor

The molar volume at STP can be used

to write conversion factors.

22.4 L and 1 mole

1 mole 22.4 L

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Using Molar Volume

What is the volume occupied by 2.75 moles of N₂gas at STP?

The molar volume is used to convert moles to liters.

2.75 moles N₂ x 22.4 L = 61.6 L 1 mole

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Guide to Using Molar Volume

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A. What is the volume at STP of 4.00 g of CH₄? 1) 5.60 L 2) 11.2 L 3) 44.8 L

B. How many g of He are present in 8.00 L of gas at STP?

1) 25.6 g 2) 0.357 g 3) 1.43 g

Learning Check

69

A. 1) 5.60 L

4.00 g CH₄x 1 mole CH₄ x 22.4 L (STP) = 5.60 L 16.0 g CH₄ 1 mole CH₄

B. 3) 1.43 g

8.00 L x 1 mole He x 4.00 g He = 1.43 g He 22.4 L 1 mole He

Solution

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Gases in Equations

The volume or amount of a gas at STP

in a chemical

reaction can be calculated from

• STP conditions.

• mole factors from the balanced

equation.

Guide to Using Molar Volume

for Reactions

STP and Gas Equations

What volume (L) of O₂gas at STP is needed to completely react with 15.0 g of aluminum?

4Al(s) + 3O₂(g) 2Al₂O₃(s)

Plan: g Al mole Al mole O₂ L O₂(STP)

15.0 g Al x 1 mole Al x 3 moles O₂ x 22.4 L (STP) 27.0 g Al 4 moles Al 1 mole O₂

= 9.33 L of O₂at STP

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What mass of Fe will react with 5.50 L of

O

₂

at STP?

4Fe(s) + 3O

₂

(g) 2Fe

₂

O

₃

(s)

Learning Check

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4Fe(s) + 3O₂(g) 2Fe₂O₃(s)

? 5.50 L at STP

5.50 L O₂x 1 mole x 4 moles Fe x 55.9 g Fe = 18.3 g of Fe 22.4 L 3 moles O₂ 1 mole Fe

Solution

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Chapter 6 Gases

6.8 Partial Pressures (Dalton’s Law)

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The partial pressure of a gas

• is the pressure of each gas in a

mixture.

• is the pressure that gas would exert if it

were by itself in the container.

Partial Pressure

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Dalton’s law of partial pressuresindicates that

• pressure depends on the total number of gas particles, not on the types of particles.

• the total pressure exerted by gases in a mixture is the sum of the partial pressures of those gases.

P_T= P₁+ P₂+ P₃+....

Dalton’s Law of Partial Pressures

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Dalton’s Law of Partial Pressures

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For example, at STP, 1 mole of a pure gas in a volume of 22.4 L will exert the same pressure as 1 mole of a gas mixture in 22.4 L.

V = 22.4 L

Gas mixtures

Total Pressure

0.5 mole O₂ 0.3 mole He 0.2 mole Ar 1.0 mole 1.0 mole N₂

0.4 mole O₂ 0.6 mole He 1.0 mole

1.0 atm 1.0 atm 1.0 atm

Scuba Diving

• When a scuba diver dives, the increased pressure causes N₂(g) to dissolve in the blood.

• If a diver rises too fast, the dissolved N₂will form bubbles in the blood, a dangerous and painful condition called "the bends."

• Helium, which does not dissolve in the blood, is mixed with O₂to prepare breathing mixtures for deep descents.

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Learning Check

A scuba tank contains

O

₂

with a pressure of

0.450 atm and He at

855 mmHg. What is the

total pressure in mmHg

in the tank?

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Solution

1. Convert the pressure in atm to mmHg.

0.450 atm x 760 mmHg = 342 mmHg = PO₂ 1 atm

2. Calculate the sum of the partial pressures.

P_total = PO₂+ P_He

P_total = 342 mmHg + 855 mmHg

= 1197 mmHg

For a deep dive, a scuba diver uses a

mixture of helium and oxygen with a

pressure of 8.00 atm. If the oxygen

has a partial pressure of 1280 mmHg,

what is the partial pressure of the

helium?

1) 520 mmHg

2) 2040 mmHg

3) 4800 mmHg

Learning Check

3) 4800 mmHg

P_Total= 8.00 atm x 760 mmHg = 6080 mmHg 1 atm

P_Total= P_O+ P_He

2

P_He = P_Total- PO₂

P_He = 6080 mmHg - 1280 mmHg

= 4800 mmHg

Solution

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Gases We Breathe

The air we

breathe

• is a gas

mixture.

• contains

mostly N

₂

and

O

₂

, and small

amounts of

other gases.

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A. If the atmospheric pressure today is 745 mmHg, what is the partial pressure (mmHg) of O₂in the air?

1) 35.6 2) 156 3) 760

B. At an atmospheric pressure of 714, what is the partial pressure (mmHg) N₂in the air?

1) 557 2) 9.14 3) 0.109

Learning Check

87

A. If the atmospheric pressure today is 745 mmHg, what is the partial pressure (mmHg) of O₂in the air?

2) 156

B. At an atmospheric pressure of 714, what is the partial pressure (mmHg) N₂in the air?

1) 557

Solution

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Blood Gases

• In the lungs, O₂ enters the blood, while CO₂from the blood is released.

• In the tissues, O₂ enters the cells, which releases CO₂into the blood.

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Blood Gases

In the body,

• O₂flows into the tissues because the partial pressure of O₂is higher in blood, and lower in the tissues.

• CO₂flows out of the tissues because the partial pressure of CO₂is higher in the tissues, and lower in the blood.

Partial Pressures in Blood and Tissue Oxygenated Deoxygenated

Gas Blood Blood Tissues

O2 100 40 30 or less

CO₂ 40 46 50 or greater

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