# Chapter 6 Gases 6.1 Properties of Gases.

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Chapter 6 Gases 6.1 Properties of Gases

Kinetic Molecular Theory
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 with an increase in temperature

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

Pressure Gas particles are very small and have lots of energy.
Pressure is measure of collisions with sides of container. Gas particles in air exert pressure on us called atmospheric pressure.

Volume The volume of a gas is the same as the volume of
container it occupies. More collisions with sides of container will increase its volume.

Temperature Temperature of gas is measured in Kelvin (K)
temperature scale. Temperature of gas relates to the average kinetic energy of the molecules. Decrease temperature, molecules move less. Increase temperature, molecules move more.

Chapter 6 Gases 6.2 Gas Pressure

Gas Pressure Gas pressure is a force acting on a specific area
Pressure (P) = force area has units of atm, mmHg, torr, lb/in.2, or kilopascals (kPa) 1 atm = mmHg (exact) 1 atm = 760 torr 1 atm = lb/in.2 1 atm = Pa 1 atm = kPa

Altitude and Atmospheric Pressure

Learning Check 1. What is 475 mmHg expressed in atm? A. 475 atm
B atm C  105 atm 2. The pressure in a tire is 2.00 atm. What is this pressure in mmHg? A mmHg B mmHg C mmHg

Solution 1. What is 475 mmHg expressed in atm?
The answer is B, atm. 475 mmHg  atm = atm 760 mmHg 2. The pressure in a tire is 2.00 atm. What is this pressure in mmHg? The answer is B, 1520 mmHg. 2.00 atm  760 mmHg = mmHg 1 atm

Barometer Measures Pressure
A barometer measures the pressure exerted by the gases in the atmosphere indicates atmospheric pressure as the height in mm of the mercury column 760 mmHg = 1 atm

Atmospheric Pressure Atmospheric pressure
is the pressure exerted by a column of air from the top of the atmosphere to the surface of Earth decreases as altitude increases is 1 atm at sea level is higher on rainy day

Learning Check 1. The downward pressure of the Hg in a barometer is _____ the pressure of the atmosphere. A. greater than B. less than C. the same as 2. A water barometer is 13.6 times taller than a Hg barometer (dHg = 13.6 g/mL) because A. H2O is less dense than mercury B. H2O is heavier than mercury C. air is more dense than H2O

Solution 1. The downward pressure of the Hg in a barometer is _____ the pressure of the atmosphere. The answer is C, the same as. 2. A water barometer is 13.6 times taller than a Hg barometer (dHg = 13.6 g/mL) because The answer is A, H2O is less dense than mercury.

Chapter 6 Gases 6.3 Pressure and Volume Boyle’s Law

Boyle’s Law Boyle’s law states that
the pressure of a gas is inversely related to its volume when T and n are constant the product P  V is constant when temperature and moles are held constant if volume decreases, the pressure increases P1V1 = P2V2

Boyle’s Law: PV = Constant
P1V1 = 8.0 atm  2.0 L = 16 atm L P2V2 = 4.0 atm  4.0 L = 16 atm L P3V3 = 2.0 atm  8.0 L = 16 atm L Boyle’s law can be stated as P1V1 = P2V2 (T, n constant)

Boyle’s Law and Breathing
During an inhalation, the lungs expand the pressure in the lungs decreases air flows toward the lower pressure in the lungs

Boyle’s 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

Solving for a Gas Law Factor
The equation for Boyle’s law can be rearranged to solve for any factor. P1V1 = P2V Boyle’s law To solve for V2 , divide both sides by P2. P1V1 = P2V2 P P2 V1 x P1 = V2 P2

Barometer Measures Pressure
A barometer measures the pressure exerted by the gases in the atmosphere indicates atmospheric pressure as the height in mm of the mercury column 760 mmHg = 1 atm

Atmospheric Pressure Atmospheric pressure is the pressure exerted by
a column of air from the top of the atmosphere to the surface of Earth decreases as altitude increases is 1 atm at sea level is higher on a rainy day

Guide to Using Gas Laws

Calculations Using Boyle’s Law
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? Step 1 Organize the data in a table of initial and final conditions. Conditions Initial Final P1 = mmHg P2 = mmHg V1 = L V2 = ?

Solution Step 2 When pressure increases, volume decreases.
Solve Boyle’s law for V2: P1V1 = P2V V2 = V1 P1 P2 Step 3 Substitute values into the gas law equation and calculate. V2 = L  550 mmHg = 2.0 L 2200 mmHg pressure ratio decreases volume

Learning Check 27 A sample of oxygen gas has a volume of 12.0 L at mmHg. What is the new pressure when the volume changes to 36.0 L (T and n constant)? A mmHg B mmHg C mmHg

Solution A sample of oxygen gas has a volume of 12.0 L at
28 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)? Step 1 Organize the data in a table of initial and final conditions. Data Table Conditions 1 Conditions 2 P1 = mmHg P2 = ? V1 = L V2 = L

Solution A sample of oxygen gas has a volume of 12.0 L at
29 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)? Step 2 Rearrange the gas law equation to solve for the unknown quantity. P2 = P1  V1 V2 Step mmHg  12.0 L = mmHg The answer is A L The answer is A, 200. mmHg.

Learning Check 30 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

Learning Check 31 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 cylinder B 2. pressure increases cylinder A

Learning Check If a sample of helium gas has a volume of 120 mL and a
32 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? A. 60 mL B mL C mL

Solution If a sample of helium gas has a volume of 120 mL and a
33 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? Step 1 Organize the data in a table of initial and final conditions. Data Table Conditions Conditions 2 P1 = mmHg P2 = mm Hg V1 = mL V2 = ? Step 2 Rearrange the gas law equation to solve for the unknown quantity. V2 = V1  P1 P2

Solution If a sample of helium gas has a volume of 120 mL and a
34 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? Step 3 Substitute values into the gas law equation and calculate. V2 = V1  P1 = mL  850 mmHg = 240 mL P mmHg Pressure ratio increases volume The answer is C, 240 mL.

Learning Check 35 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?

Solution 36 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.

Temperature and Volume
Chapter 6 Gases 6.4 Temperature and Volume Charles’s Law

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 V1 = V2 T T2

Charles’s Law: V and T For two conditions, Charles’s law is written
V1 = V (P and n constant) T T2 Rearranging Charles’s law to solve for V2: T2  V1 = V2  T1 T T1 V2 = V1  T2 T1

Learning Check Solve Charles’s law expression for T2. V1 = V2 T T2

Solution V1 = V2 T1 T2 Cross-multiply to give: V1T2 = V2T1
Isolate T2 by dividing through by V1: V1T = V2T1 V V1 T2 = T1 x V2 V1

Calculations Using Charles’s Law
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)? Step 1 Organize the data into a table of initial and final conditions. Conditions Initial Final 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
Step 2 Rearrange to solve for unknown quantity: V2 V1 = V V2 = V1  T2 T T T1 Step 3 Substitute the values into the gas law equation and calculate. V2 = 785 mL  273 K = 729 mL 294 K

Learning Check 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)? A C B C C. −82 C

Solution 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)? Step 1 Organize the data into a table of initial and final conditions. Conditions Initial Final V1 = 420 mL V2 = 640 mL T1 = 18 C = 291 K T2 = ?

Solution 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)? Step 2 Rearrange to solve for unknown quantity: T2 V1 = V2 T2 = V2  T1 T T V1 Step 3 Substitute the values into the gas law equation and calculate. T2 = 291 K  640 mL = 443 K 420 mL = 443 K – 273 = 170 C. The answer is B.

Learning Check Use the gas laws to complete each sentence with increases or 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.

Solution Use the gas laws to complete each sentence with increases or decreases. A. Pressure increases when V decreases. B. When T decreases, V decreases. C. Pressure decreases when V changes from 12 L to 24 L. D. Volume increases when T changes from 15 C to 45 C.

Temperature and Pressure
Chapter 6 Gases 6.5 Temperature and Pressure Gay-Lussac’s Law

Gay-Lussac’s Law 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 T T2

Learning Check Solve Gay-Lussac’s law for P2. P1 = P2 T T2

Solution Solve Gay-Lussac’s law for P2. P1 = P2 T1 T2
Multiply both sides by T2 and cancel: P1  T2 = P2  T2 T T2 P2 = P1  T2 T1

Calculations Using Gay-Lussac’s Law
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)? Step 1 Organize the data into a table of initial and final conditions. Conditions Initial Final P1 = 2.0 atm P2 = ? T1 = 18 C T2 = 62 C + 273 = 291 K = 335 K Be sure to use the Kelvin (K) temperature in gas calculations.

Calculations Using Gay-Lussac’s Law
Step 2 Rearrange to solve for unknown quantity: P2 Solve Gay-Lussac’s law for P2: P1 = P2 P2 = P1  T2 T T T1 Step 3 Substitute the values into the gas law equation and calculate. P2 = 2.0 atm  335 K = 2.3 atm 291 K Temperature ratio increases pressure

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)?

Solution 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)? Step 1 Organize the data into a table of initial and final conditions. Conditions Initial Final P1 = 645 torr P2 = 824 torr T1 = 128 C T2 = K – 273 = ? C = 401 K

Solution 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)? Step 2 Rearrange the gas law equation to solve for the unknown quantity. Solve Gay-Lussac’s law for T2: P1 = P2 T2 = T1  P2 T T P1

Solution 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)? Step 3 Substitute values into the gas law equation and calculate. T2 = 401 K  824 torr = 512 K − 273 = 239 C 645 torr Pressure ratio increases temperature

Vapor Pressure and Boiling Point
is the pressure that accumulates when molecules of a liquid collect over the surface of a liquid in a closed container is specific for a given temperature increases when the temperature increases When the vapor pressure of a liquid equals the external pressure, the liquid reaches its boiling point.

Temperature and Vapor Pressure of Water

Pressure and Boiling Point of Water
People who live at high altitudes often use pressure cookers to obtain higher temperatures when preparing food. In a pressure cooker, water is heated in a closed container so that pressures above 1 atm are obtained, raising the boiling point of water.

Learning Check Explain why water boils at a lower temperature in the mountains than at sea level.

Solution Explain why water boils at a lower temperature in the mountains than at sea level. Atmospheric pressure in the mountains is less than at sea level. The vapor pressure of the water reaches the atmospheric pressure at a lower temperature.

Temperature and Pressure
Chapter 6 Gases 6.5 Temperature and Pressure Gay-Lussac’s Law

Gay-Lussac’s Law 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 T T2

Learning Check Solve Gay-Lussac’s law for P2. P1 = P2 T T2

Solution Solve Gay-Lussac’s law for P2. P1 = P2 T1 T2
Multiply both sides by T2 and cancel: P1  T2 = P2  T2 T T2 P2 = P1  T2 T1

Calculations Using Gay-Lussac’s Law
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)? Step 1 Organize the data into a table of initial and final conditions. Conditions Initial Final P1 = 2.0 atm P2 = ? T1 = 18 C T2 = 62 C + 273 = 291 K = 335 K Be sure to use the Kelvin (K) temperature in gas calculations.

Calculations Using Gay-Lussac’s Law
Step 2 Rearrange to solve for unknown quantity: P2 Solve Gay-Lussac’s law for P2: P1 = P2 P2 = P1  T2 T T T1 Step 3 Substitute the values into the gas law equation and calculate. P2 = 2.0 atm  335 K = 2.3 atm 291 K Temperature ratio increases pressure

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)?

Solution 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)? Step 1 Organize the data into a table of initial and final conditions. Conditions Initial Final P1 = 645 torr P2 = 824 torr T1 = 128 C T2 = K – 273 = ? C = 401 K

Solution 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)? Step 2 Rearrange the gas law equation to solve for the unknown quantity. Solve Gay-Lussac’s law for T2: P1 = P2 T2 = T1  P2 T T P1

Solution 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)? Step 3 Substitute values into the gas law equation and calculate. T2 = 401 K  824 torr = 512 K − 273 = 239 C 645 torr Pressure ratio increases temperature

Vapor Pressure and Boiling Point
is the pressure that accumulates when molecules of a liquid collect over the surface of a liquid in a closed container is specific for a given temperature increases when the temperature increases When the vapor pressure of a liquid equals the external pressure, the liquid reaches its boiling point.

Temperature and Vapor Pressure of Water

Pressure and Boiling Point of Water
People who live at high altitudes often use pressure cookers to obtain higher temperatures when preparing food. In a pressure cooker, water is heated in a closed container so that pressures above 1 atm are obtained, raising the boiling point of water.

Learning Check Explain why water boils at a lower temperature in the mountains than at sea level.

Solution Explain why water boils at a lower temperature in the mountains than at sea level. Atmospheric pressure in the mountains is less than at sea level. The vapor pressure of the water reaches the atmospheric pressure at a lower temperature.

Chapter 6 Gases 6.6 The Combined Gas Law

The Combined Gas Law The combined gas law uses Boyle’s law, Charles’s law, and Gay-Lussac’s law (n is constant). P1 V = P2 V2 T1 T2

Summary of Gas Laws

Calculations Using Combined Gas Law
A gas has a volume of 675 mL at 35 C and atm pressure. What is the volume (mL) of the gas at −95 C and a pressure of 802 mmHg (n constant)? Step 1 Organize the data into a table of initial and final conditions Conditions Initial Final T1 = 35 C T2 = −95 C + 273 = 308 K = 178 K V1 = 675 mL V2 = ? P1 = 646 mmHg P2 = 802 mmHg

Calculations Using Combined Gas Law
A gas has a volume of 675 mL at 35 C and atm pressure. What is the volume (mL) of the gas at −95 C and a pressure of 802 mmHg (n constant)? Step 2 Rearrange to solve for unknown quantity: V2 P1 V1 = P2 V2 T T2 V2 = V1  P1  T2 P2  T1

Calculations Using Combined Gas Law
A gas has a volume of 675 mL at 35 C and atm pressure. What is the volume (mL) of the gas at −95 C and a pressure of 802 mmHg (n constant)? Step 3 Substitute the values into the gas law equation and calculate. V2 = 675 mL  646 mmHg  178 K = 314 mL 802 mmHg  308 K

Learning Check A sample of helium gas has a volume of 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)?

Solution Step 1 Organize the data into a table of initial and final conditions. Conditions Initial Final P1 = atm P2 = atm V1 = L V2 = mL (180 mL) T1 = 29 C T2 = ? = 302 K Be sure to use the Kelvin (K) temperature in gas calculations.

Solution Step 2 Rearrange to solve for unknown quantity: T2
P1 V1 = P2 V and T2 = T1  P2  V2 T T P1  V1 Step 3 Substitute the values into the gas law equation and calculate. T2 = 302 K  3.20 atm  90.0 mL = 604 K 0.800 atm  mL T2 = 604 K − 273 = 331 C

Chapter 6 Gases 6.7 Volume and Moles Avogadro’s Law

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 n n2

If 0.75 mole of helium gas occupies a volume of 1.5 L, what volume (L) will 1.2 moles of helium occupy at the same temperature and pressure? A L B. 1.8 L C. 2.4 L

If 0.75 mole of helium gas occupies a volume of 1.5 L, what volume (L) will 1.2 moles of helium occupy at the same temperature and pressure? Step 1 Organize the data into a table of initial and final conditions. Conditions Initial Final V1 = 1.5 L V2 = ? n1 = 0.75 mol n2 = 1.2 moles

Calculations Using Combined Gas Law
If 0.75 mole of helium gas occupies a volume of 1.5 L, what volume (L) will 1.2 moles of helium occupy at the same temperature and pressure? Step 2 Rearrange to solve for unknown quantity: V2 V2 = V1  n2 n1 Step 3 Substitute the values into the gas law equation and calculate. V2 = 1.5 L  1.2 moles = 2.4 L The answer is C. 0.75 mole

STP: Standard Temperature and Pressure
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)

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. Use this equality as a conversion factor: At STP, 22.4 L = 1 mole 22.4 L and mole 1 mole L

Molar Volume

Guide to Using Molar Volume

Calculations Using Molar Volume
What is the volume occupied by 2.75 moles of N2 gas at STP? Step 1 State the given and needed quantities. Given: moles N2 Needed: Volume of N2 gas at STP Step 2 Write a plan. The molar volume is used to convert moles to liters. Molar Moles N2 Volume Volume N2

Calculations Using Molar Volume
What is the volume occupied by 2.75 moles of N2 gas at STP? Step 3 Write conversion factors including 22.4 L/mole at STP. At STP, 22.4 L = 1 mole 22.4 L and 1 mole 1 mole L Step 4 Set up problem with factors to cancel units. 2.75 moles N2  L = L 1 mole N2

Learning Check 1. What is the volume at STP of 4.00 g of CH4?
A L B L C L 2. How many grams of He are present in 8.00 L of gas at STP? A g B g C g

Solution Step 1 State given and needed quantities.
1. What is the volume at STP of 4.00 g of CH4? Given: grams CH4 Needed: Volume of CH4 gas at STP 2. How many grams of He are present in 8.00 L of gas at STP? Given: L of He at STP Needed: grams of He

Solution Step 2 Write a plan.
1. What is the volume at STP of 4.00 g of CH4? The molar volume is used to convert moles to liters. mass CH moles CH volume CH4 2. How many grams of He are present in 8.00 L of gas at STP? volume He moles He mass He

Solution Step 3 Write conversion factors including 22.4 L/mole at STP.
1. What is the volume at STP of 4.00 g of CH4? At STP, 22.4 L = 1 mole L and 1 mole 1 mole L 1 mole of CH4 = g of CH4 1 mole CH4 and g CH4 16.05 g CH mole CH4

Solution Step 3 Write conversion factors including 22.4 L/mole at STP.
2. How many grams of He are present in 8.00 L of gas at STP? At STP, 22.4 L = 1 mole L and 1 mole 1 mole L 1 mole of He = 4.00 g He 1 mole He and g He 4.00 g He mole He

Solution Step 4 Set up problem with factors to cancel units.
1. What is the volume at STP of 4.00 g of CH4? 4.00 g CH4  1 mole CH4  L (STP) = 5.60 L 16.0 g CH mole CH4 The answer is A, 5.60 L. 2. How many grams of He are present in 8.00 L of gas at STP? 8.00 L  1 mole He  g He = 1.43 g He 22.4 L (STP) mole He The answer is C, 1.43 g.

Chapter 6 Gases 6.8 Partial Pressure Dalton’s Law

Partial Pressure The partial pressure of a gas is the pressure that each gas in a mixture would exert if it were by itself in the container.

Dalton’s Law of Partial Pressures
Dalton’s law of partial pressures indicates 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 PT = P1 + P2 + P

Total Pressure 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. Gas mixtures 0.4 mole O2 0.6 mole He 1.0 mole 0.5 mole O2 0.3 mole He 0.2 mole Ar 1.0 mole 1.0 mole N2 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 N2 will form bubbles in the blood, a dangerous and painful condition called “the bends.” Helium, which does not dissolve in the blood, is mixed with O2 to prepare breathing mixtures for deep descents.

Air is a Gas Mixture The air we breathe is a gas mixture
contains mostly N2 and O2, and small amounts of other gases

Guide to Solving for Partial Pressure

Guide to Solving for Partial Pressure
A scuba tank contains O2 with a pressure of atm and He at 855 mmHg. What is the total pressure in mmHg in the tank? Step 1 Write the equation for the sum of the partial pressures. Ptotal = PO2 + PHe Step 2 Solve for the unknown pressure Ptotal = PO2 + PHe Convert the pressure in atm to mmHg. 0.450 atm  760 mmHg = 342 mmHg = PO atm

Guide to Solving for Partial Pressure
Step 3 Substitute known pressures and calculate the unknown partial pressure. Ptotal = PO2 + PHe Ptotal = 342 mmHg mmHg = mmHg

Learning Check For a deep dive, a scuba diver uses a mixture of helium and oxygen with a pressure of atm. If the oxygen has a partial pressure of 1280 mmHg, what is the partial pressure of the helium? A mmHg B mmHg C mmHg

Solution For a deep dive, a scuba diver uses a mixture of helium and
oxygen with a pressure of atm. If the oxygen has a partial pressure of 1280 mmHg, what is the partial pressure of the helium? Step 1 Write the equation for the sum of the partial pressures. Ptotal = PO2 + PHe Step 2 Solve for the unknown pressure. PHe = Ptotal − PO2 Convert the pressure in atm to mmHg. Ptotal = 8.00 atm  760 mmHg = mmHg 1 atm

Solution For a deep dive, a scuba diver uses a mixture of helium and
oxygen with a pressure of atm. If the oxygen has a partial pressure of 1280 mmHg, what is the partial pressure of the helium? Step 3 Substitute known pressures and calculate the unknown partial pressure. PHe = mmHg – 1280 mmHg = mmHg The answer is C, 4800 mm Hg.

Blood Gases In the lungs, O2 enters the blood, while CO2 from the blood is released. In the tissues, O2 enters the cells, which releases 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 blood

Partial Pressures in Blood
Partial Pressures in Blood and Tissue Gas Oxygenated Blood Deoxygenated Blood Tissues O2 100 40 30 or less CO2 46 50 or greater

Gas Exchange During Breathing

Gas Law Concepts