Chapter 13 Gas Laws

Pressure and Force

• Pressure is the force per unit area on a surface.

Pressure = Force

Area

Chapter 13 – Gases and Pressure

Gases in the Atmosphere

• The atmosphere of Earth is a layer of gases surrounding the planet that is retained by Earth's gravity.

• By volume, dry air is 78% nitrogen, 21% oxygen, 0.9% argon, 0.04% CO2, and small amounts of other gases.

Chapter 13 – Gases and Pressure

Atmospheric Pressure

• Atmospheric pressure is the force per unit area exerted on a surface by the weight of the gases that make up the atmosphere above it.

Chapter 13 – Gases and Pressure

Measuring Pressure

• A common unit of pressure is millimeters of mercury (mm Hg).

• 1 mm Hg is also called 1 torr in honor of Evangelista Torricelli who invented the barometer (used to measure atmospheric pressure).

• The average atmospheric pressure at sea level at 0°C is 760 mm Hg, so one atmosphere (atm) of pressure is 760 mm Hg.

Chapter 13 – Gases and Pressure

Measuring Pressure (continued)

•Pressure can also be measured in pascals (Pa): 1 Pa = 1 N/m2.

•One pascal is very small, so usually kilopascals (kPa) are used instead.

•One atm is equal to 101.3 kPa.

1 atm = 760 mm Hg (Torr) = 101.3 kPa

Chapter 13 – Gases and Pressure

Units of Pressure

Chapter 13 – Gases and Pressure

Converting Pressure Sample Problem

The average atmospheric pressure in Denver, CO is 0.830 atm. Express this pressure in:

a. millimeters of mercury (mm Hg) b. kilopascals (kPa)

0.830 atm atm

mm Hg

1

760 x =

0.830 atm atm

kPa

1

101.3 x =

631 mm Hg

84.1 kPa

Chapter 13 – Gases and Pressure

Dalton’s Law of Partial Pressures

• Dalton’s law of partial pressures - the total pressure of a gas mixture is the sum of the partial pressures of the component gases.

PT = P1 + P2 + P3 …

Chapter 13 – Gases and Pressure

Dalton’s Law of Partial Pressures Sample Problem

A container holds a mixture of gases A, B & C. Gas A has a pressure of 0.5 atm, Gas B has a pressure of 0.7 atm, and Gas C has a pressure of 1.2 atm.

a. What is the total pressure of this system?

b. What is the total pressure in mm Hg?

PT = P1 + P2 + P3 …

PT = 0.5 atm + 0.7 atm + 1.2 atm = 2.4 atm

2.4 atm atm

mm Hg

1

760 x = 1800 mm Hg

Chapter 13 – Gases and Pressure

#3 Gases and Pressure WS Explain how to calculate the partial pressure of a dry gas that is collected over water when the total pressure is atmospheric pressure.

Chapter 13 – Gases and Pressure

Dalton’s Law of Partial Pressures Sample Problem

• Because water molecules at a

liquid surface evaporates…gas

collected is not pure!

• The gas collected is a mixture of

dry gas and water vapor

• Use Dalton’s Law to solve for

partial pressure:

pdry gas= Patm – pwater

Gases and Pressure

• Gas pressure is caused by collisions of the gas molecules with each other and with the walls of their container.

• The greater the number of collisions, the higher the pressure will be.

Chapter 13 –The Gas Laws

Pressure – Volume Relationship

• When the volume of a gas is decreased, more collisions will occur.

• Pressure is caused by collisions.

• Therefore, pressure will increase.

• This relationship between pressure and volume is inversely proportional.

Chapter 13 –The Gas Laws

Boyle’s Law

• Boyle’s Law – The volume of a fixed mass of gas varies inversely with the pressure at a constant temperature.

• P1 and V1 represent

initial conditions, and P2 and V2 represent another set of conditions.

P1V1 = P2V2

Chapter 13 –The Gas Laws

Boyle’s Law Sample Problem

A sample of oxygen gas has a volume of 150.0 mL when its pressure is 0.947 atm. What will the volume of the gas be at a pressure of 0.987 atm if the temperature remains constant?

Solution:

P1V1 = P2V2 (0.947 atm) (150.0 mL) = (0.987 atm) V2

V2 = (0.947 atm) (150.0 mL)

(0.987 atm) = 144 mL

Chapter 13 –The Gas Laws

Volume – Temperature Relationship

• the pressure of gas inside and outside the balloon are the same.

• at low temperatures, the gas molecules don’t move as much – therefore the volume is small.

• at high temperatures, the gas molecules move more – causing the volume to become larger.

Chapter 13 –The Gas Laws

• Charles’s Law – The volume of a fixed mass of gas at constant pressure varies directly with the Kelvin temperature.

Charles’s Law

V1 V2 =

T1 T2

• V1 and T1 represent initial conditions, and V2 and T2 represent another set of conditions.

Chapter 13 –The Gas Laws

The Kelvin Temperature Scale

• Absolute zero – The theoretical lowest possible temperature where all molecular motion stops.

• The Kelvin temperature scale starts at absolute zero (-273oC.)

• This gives the following relationship between the two temperature scales:

K = oC + 273

Chapter 13 –The Gas Laws

Charles’s Law Sample Problem

A sample of neon gas occupies a volume of 752 mL at 25°C. What volume will the gas occupy at 50°C if the pressure remains constant?

Solution:

V1 V2 =

T1 T2 752 mL V2

= 298 K 323 K

K = oC + 273 T1 = 25 + 273 = 298 T2 = 50 + 273 = 323

752 mL V2 =

298 K 323 K x = 815 mL

Chapter 13 –The Gas Laws

Pressure – Temperature Relationship

• Increasing temperature means increasing kinetic energy of the particles.

• The energy and frequency of collisions depend on the average kinetic energy of the molecules.

• Therefore, if volume is kept constant, the pressure of a gas increases with increasing temperature.

Chapter 13 –The Gas Laws

Gay-Lussac’s Law

• Gay-Lussac’s Law – The pressure of a fixed mass of gas varies directly with the Kelvin temperature.

• P1 and T1 represent

initial conditions. P2 and T2 represent another set of conditions.

P1 P2 =

T1 T2

Chapter 13 –The Gas Laws

The Combined Gas Law

• The combined gas law is written as follows:

• Each of the other gas laws can be obtained from the combined gas law when the proper variable is kept constant.

P1 P2 =

T1 T2

V1 V2

Chapter 13 –The Gas Laws

The Combined Gas Law Sample Problem

A helium-filled balloon has a volume of 50.0 L at 25°C and 1.08 atm. What volume will it have at 0.855 atm and 10.0°C? Solution:

(1.08 atm) (0.855 atm) =

298 K 283 K

K = oC + 273 T1 = 25 + 273 = 298 T2 = 10 + 273 = 283

(1.08 atm) V2 =

(298 K)

(283 K) = 60.0 L

P1 P2 =

T1 T2

V1 V2

(50.0 L) V2

(50.0 L)

(0.855 atm)

Chapter 13 –The Gas Laws

Avogadro’s Law

• In 1811, Amedeo Avogadro discovered that the volume of a gas is proportional to the number of molecules (or number of moles.)

• Avogadro’s Law - equal volumes of gases at the same temperature and pressure contain equal numbers of molecules, or:

V1 V2 =

n1 n2

Chapter 13 –The Gas Laws

Standard Molar Volume

• Standard Temperature and Pressure (STP) is 0oC and 1 atm.

• The Standard Molar Volume of a gas is the volume occupied by one mole of a gas at STP. It has been found to be 22.4 L.

Chapter 13 –Gas Volumes and the Ideal Gas Law

Molar Volume Conversion Factor

• Standard Molar Volume can be used as a conversion factor to convert from the number of moles of a gas at STP to volume (L), or vice versa.

Chapter 13 –Gas Volumes and the Ideal Gas Law

Molar Volume Conversion Sample Problem

a. What quantity of gas, in moles, is contained in 5.00 L at STP?

b. What volume does 0.768 moles of a gas occupy at STP?

5.00 L L

mol

22.4

1 x = 0.223 mol

0.768 mol mol

L

1

22.4 x = 17.2 L

Chapter 13 –Gas Volumes and the Ideal Gas Law

Volume Ratios

• You can use the volume ratios as conversion factors just like you would use mole ratios.

2 CO(g) + O2(g) → 2 CO2(g) 2 molecules 1 molecule 2 molecules 2 mole 1 mole 2 mol 2 volumes 1 volume 2 volumes

• Example: What volume of O2 is needed to react completely with 0.626 L of CO to form CO2?

0.626 L CO

L CO

L O2

2

1 x = 0.313 L O2

Chapter 13 –Gas Volumes and the Ideal Gas Law

The Mole Map

• You can now convert between number of particles, mass (g), and volume (L) by going through moles.

Chapter 13 –Gas Volumes and the Ideal Gas Law

Gas Stoichiometry Sample Problem

Assume that 5.61 L H2 at STP reacts with excess CuO according to the following equation:

CuO(s) + H2(g) → Cu(s) + H2O(g)

a. How many moles of H2 react?

b. How many grams of Cu are produced?

5.61 L H2 L H2

mol H2

22.4

1 x = 0.250 mol H2

5.61 L H2 L H2

mol H2

22.4

1 x mol H2

mol Cu

1

1 x mol Cu

g Cu

1

63.5 x = 15.9 g Cu

Chapter 13 –Gas Volumes and the Ideal Gas Law

The Ideal Gas Law

• All of the gas laws you have learned so far can be combined into a single equation, the ideal gas law:

• R represents the ideal gas constant which has a value of 0.0821 (L•atm)/(mol•K).

PV = nRT

Chapter 13 –Gas Volumes and the Ideal Gas Law

The Ideal Gas Law Sample Problem

What is the pressure in atmospheres exerted by a 0.500 mol sample of nitrogen gas in a 10.0 L container at 298 K?

Solution:

PV = nRT

P (10.0 L) = (0.500 mol) (0.0821 L•atm/mol•K)

P = (0.500 mol) (298 K)

(10.0 L) = 1.22

atm

(298 K)

(0.0821 L•atm/mol•K)

Chapter 13 –Gas Volumes and the Ideal Gas Law

Diffusion and Effusion

• Diffusion is the gradual mixing of two or more gases due to their spontaneous, random motion.

• Effusion is the process whereby the molecules of a gas confined in a container randomly pass through a tiny opening in the container.

Chapter 13 –Diffusion and Effusion

Graham’s Law of Effusion

• Light molecules move faster than heavy ones. • Graham’s

law of effusion says the greater the molar mass of a gas, the slower it will effuse.

Chapter 13 –Diffusion and Effusion