Chapter 13 Gas Laws - BOHS CP CHEMISTRY - Homebohscpchemistry.weebly.com/.../cp_chapter_13_2017.pdf · Atmospheric Pressure • Atmospheric pressure is the force per unit area exerted

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.


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.


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.


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


Units of 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


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 …


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


#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.


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.


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


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


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.


• 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.


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


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


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.


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


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


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)


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


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.


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


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


The Mole Map

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


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


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


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)


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

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Chapter 13 Gas Laws - BOHS CP CHEMISTRY - Homebohscpchemistry.weebly.com/.../cp_chapter_13_2017.pdf · Atmospheric Pressure • Atmospheric pressure is the force per unit area exerted