CHAPTER 11Gases

11.1 Gases and Pressures

Define pressure, give units of pressure, and describe how pressure is measured.

State the standard conditions of temperature and pressure and convert units of pressure.

Use Dalton’s law of partial pressures to calculate partial pressures and total pressures.

Pressure (P): the force per unit area on a surface.

What causes pressure? collisions of the gas molecules with each other and with

surfaces with which they come into contact.

depends on volume, temperature, and the number of molecules present.

Pressure

Pressure Video

Equation for Pressure

Pressure = Force Area

where P = Pressure, F = Force & A = Area

The SI unit for force is the Newton, (N):the force that will increase the speed of a one-kilogram mass by one meter per second each second that the force is applied.

Calculating Force

Consider a person with a mass of 51 kg. At Earth’s surface, gravity has an acceleration of 9.8 m/s2. What is the value of force?

Force = mass x acceleration

Force = 51 kg × 9.8 m/s2 = 500 kg • m/s2 = 500 N

Calculating Pressure

Pressure is force per unit area, so the pressure of a 500 N person on an area of the floor that is 325 cm2 is:

500 N ÷ 325 cm2 = 1.5 N/cm2

The greater the force on a given area, the greater the pressure.

The smaller the area is on which a given force acts, the greater the pressure.

Relationship Between Pressure, Force and Area

Measuring Pressure barometer: device used to measure

atmospheric pressure

Units for Measuring Pressure

millimeters of mercury (mm Hg)A pressure of 1 mm Hg is also called 1 torr in honor of

Torricelli for his invention of the barometer. torr atmosphere of pressure (atm) bar pounds per square inch (psi) Pascal (Pa) – SI Unit pressure exerted by a

force of 1 N acting on an area of one square meterkiloPascal (kPa)

1 atm = 101.3 kPa = 760 mmHg = 760 Torr

Review- Units of Pressure

Pressure Conversions

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

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

Given: atmospheric pressure = 0.830 atm

Unknown: a. pressure in mm Hg b. pressure in kPa

Pressure Conversions Answers

A)

B)

760 mm Hg

atm mm Hg; atm mm Hgatm

101.325 kPa

atm kPa; atm kPaatm

760 mm Hg

0.830 atm a

631 tm

mm Hg

101.325 kPa

0.830 atm at

84.1m

kPa

Dalton’s Law of Partial Pressures

The pressure of each gas in a mixture is called the partial pressure.

John Dalton discovered that the pressure exerted by each gas in a mixture is independent of that exerted by other gases present.

Dalton’s law of partial pressures: the total pressure of a gas mixture is the sum of the partial pressures of each gas.

Dalton’s Law of Partial Pressures

Dalton derived the following equation:

PT = P1 + P2 + P3 + …

Total Pressure = sum of pressures of each individual gas

Partial Pressure Video

Gases Collected by Water Displacement

Water molecules at the liquid surface evaporate and mix with the gas molecules. Water vapor, like other gases, exerts a pressure known as vapor pressure.

Gases produced in the laboratory are often collected over water. The gas produced by the reaction displaces the water in the reaction bottle.

Particle Model for a Gas Collected Over Water

Gases Collected by Water Displacement (ctd)

Step 1: Raise bottle until water level inside matches the water level outside. (Ptot = Patm)

Step 2: Dalton’s Law of Partial Pressures states:

Patm = Pgas + PH2O

To get Patm, record atmospheric pressure.

Step 3: look up the value of PH2O at the temperature of the experiment in a table, you can then calculate Pgas.

Dalton’s Law of Partial Pressures Sample Problem

KClO3 decomposes and the oxygen gas was collected by water displacement. The barometric pressure and the temperature during the experiment were 731.0 torr and 20.0°C. respectively. What was the partial pressure of the oxygen collected?

Given:PT = Patm = 731.0 torrPH2O = 17.5 torr (vapor pressure of water at

20.0°C, from table A-8 in your book)

Patm = PO2 + PH2O

Unknown: PO2 in torr

Dalton’s Law Sample Problem Solution

Solution:

Patm = PO2 + PH2O

PO2 = Patm - PH2O

substitute the given values of Patm and

into the equation: PO2 =731.0 torr – 17.5 torr = 713.5 torr

Mole Fractions (X)

Mole fraction of a gas(XA) = Moles of gas A (nA)

Total number of moles of a gas(ntot)

mole fraction: ratio of the number of moles of one component of a mixture to the total number of moles

Go To: Page 2 of Packet

Calculating Partial Pressure

PA = XA PT

Partial pressures can be determined from mole fractions using the following equation:

Go To: Page 3 of Packet

11.2 The Gas Laws

Use the kinetic-molecular theory to explain the relationships between gas volume, temperature and pressure.

Use Boyle’s law to calculate volume-pressure changes at constant temperature.

Use Charles’s law to calculate volume-temperature changes at constant pressure.

Use Gay-Lussac’s law to calculate pressure-temperature changes at constant volume.

Use the combined gas law to calculate volume-temperature-pressure changes.

Boyle’s Law

Constant: temperature, amount of gas If you decrease the volume, what

happens to the pressure? If you increase the volume, what happens

the pressure?

Pressure and volume are _____________ related.

Boyle’s Law Video

Boyle’s Law

Boyle’s Law

P1V1 = P2V2

Boyle’s Law 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?

P1 = 0.947 atm P2 = 0.987 atmV1 = 150.0 mL V2 = ?

Boyle’s Law Problem Solution

1 12

2

PVV

P 22(0.947 atm)(150.0 mL O )

0.987 at144 mL O

m

Charles’ Law

Constant: pressure, amount of gas If you increase the temperature of a gas,

what will happen to the volume? If you decrease the temperature of gas,

what will happen to the volume?

Volume and temperature are ______________ related.

Charles’ Law Video

Charles’ Law

Temperature

Units: Farenheit, Celsius, and Kelvin

absolute zero: when all motion stops

K = 273 + °C.

Charles’ Law

1 2

1 2

V V

T T

Charles’ Law 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?

Temperature must be in KELVIN!!!

V1 = 752 mL V2 = ?T1 = 25°C T2 = 50°C

Charles’ Law Sample Problem Solution

1 22

1

VTV

T

(752 mL Ne)(323 K)

298 K815 mL Ne1 2

21

VTV

T

Gay-Lussac’s Law

Constant: volume, amount of gas If you increase the temperature of a gas

what will happen to the pressure? If you decrease the temperature of gas

what will happen to the pressure?

Pressure and temperature are _____________ related.

GL Law Video

Gay-Lussac’s Law

Gay-Lussac’s Law

1 2

1 2

P P

T T

Gay-Lussac’s Law Problem

The gas in a container is at a pressure of 3.00 atm at 25°C. Directions on the container warn the user not to keep it in a place where the temperature exceeds 52°C. What would the gas pressure in the container be at 52°C?

Temperature must be in KELVIN!!!

P1 = 3.00 atm P2 = ?T1 = 25°C T2 = 52°C

Gay-Lussac’s Law Problem Solution

P2 = P1T2 = (3.00 atm) (325 K) = 3.27 atm

T1 298 K

1 22

1

PTP

T

Summary of the Basic Gas Laws

The Combined Gas Law

Constant: amount of gas combined gas law: used when pressure,

temperature, and volume change within a system

1 1 2 2

1 2

PV PV

T T

NOTE: P & V are directly related to T, while P is inversely related to V

Combined Gas Law Problem

A helium-filled balloon has a volume of 50.0 L at 25.0°C and 1.08 atm. What volume will it have at 0.855 atm and 10.0°C?

Temperature must be in KELVIN!!

P1 = 1.08 atm P2 = 0.855 atmV1 = 50.0 L V2 = ?T1 = 25.0°C T2 = 10.0°C

Combined Gas Law Problem Solution

1 1 22

2 1

PVTV

PT

(1.08 atm)(50.0 L He)(283 K)

(0.855 atm)(298 K)60.0 L He1 1 2

22 1

PVTV

PT

11.3 Gas Volumes and the Ideal Gas Law

State the law of combining volumes.

State Avogadro’s law and explain its significance.

Define standard molar volume of a gas and use it to calculate gas masses and volumes.

State the ideal gas law.

Using the ideal gas law, calculate pressure, volume, temperature, or amount of gas when the other three quantities are known.

Reacting Volumes

In the early 1800s, Joseph Gay-Lussac observed that 2 L of hydrogen can react with 1 L of oxygen to form 2 L of water vapor.

hydrogen gas + oxygen gas → water vapor2 L (2 volumes) 1 L (1 volume) 2 L (2 volumes)

The reaction shows a simple 2:1:2 ratio in the volumes of reactants and products:

2 mL, 1 mL, and 2 mL

Volume Stoich Video

Avogadro’s Law

Avogadro’s law: states that equal volumes of gases at constant temperature and pressure contain equal numbers of molecules.

According to Avogadro’s law, one mole of any gas will occupy the same volume as one mole of any other gas at the same conditions, despite mass differences.

standard molar volume of a gas:22.41410 L (rounded to 22.4 L)

Molar Volume Video

Gas Stoichiometry

Gay-Lussac’s law of combining volumes of gases and Avogadro’s law can be applied in calculating the stoichiometry of reactions involving gases.

The coefficients in chemical equations of gas reactions reflect not only molar ratios, but also volume ratios (assuming conditions remain the same).

example—reaction of carbon dioxide formation: 2CO(g) + O2(g) → 2CO2(g)

2 molecules 1 molecule 2 molecules2 mole 1 mole 2 mol2 volumes 1 volume 2 volumes

Gas Stoichiometry Problem

Number 1 on Practice Sheet What volume of nitrogen at STP would

be required to react with 0.100 mol of hydrogen to produce ammonia?

N2 + 3 H2 2 NH3

Gas Stoichiometry Problem Solution

0.100 mol H2 x 1 mol N2 x 22.4 L N2

3 mol H2 1 mol N2

= 0.747 L N2

Avogadro’s Law

Constant: pressure, temperature If you increase the amount of moles, what

happens to the volume? If you decrease the amount of moles what

happens to the volume?

Amount of moles and volume are ____________ related.

Avogadro’s Law

This equation is NOT in the book, it was calculated during the Gas Simulation Lab

V1 = V2

n1 n2

The Ideal Gas Law

ideal gas law: relates all variables – pressure, volume, moles, temperature

PV = nRT

The Ideal Gas Law

ideal gas constant, R:

Its value depends on the units chosen for pressure, volume, and temperature in the rest of the equation.

Measured values of P, V, T, and n for a gas at near-ideal conditions can be used to calculate R:

PVR

nT

(1 atm)(22.414 10 L) L atm0.082 058

(1 mol)(273.15 K) mol K

Usually rounded to 0.0821 (Latm/molK)

Numerical Values of The Gas Constant “R”

ALWAYS MATCH UP YOUR UNITS!!!!

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?

P = ?V = 10.0 Ln = 0.500 molT = 298 K

Ideal Gas Law Problem Solution

nRTP

V

(0.500 mol)(0.0821 L atm)(298 K)

101

.0 L.22 atmP

Gas Stoich and Ideal Gas Law

Number 2 on Practice Sheet What volume of nitrogen at 215OC and

715 mmHg would be required to react with 0.100 mol of hydrogen to produce ammonia?

N2 + 3 H2 2 NH3

Note: This system is NOT at STP!!

Gas Stoichiometry Problem Solution

0.100 mol H2 x 1 mol N2 = 0.033 mol N2

3 mol H2

P = 715 mmHgV = ?n = 0.333 mol N2

R = 62.4 LmmHg/molKT = 25OC + 273 = 488 K

11.4 Diffusion and Effusion

Describe the process of diffusion.

State Graham’s law of effusion.

State the relationship between the average molecular velocities of two gases and their molar masses.

Diffusion and Effusion

REMEMBER:

DIFFUSION: the gradual mixing of two or more gases due to their spontaneous, random motion

EFFUSION: process when the molecules of a gas confined in a container randomly pass through a tiny opening in the container

Effusion Rate Video

Graham’s Law Of Effusion

Graham’s law of effusion: the rates of effusion of gases at the same temperature and pressure are inversely proportional to the square roots of their molar masses.

B

A

MA

B M

rate of effusion of

rate of effusion of

Graham’s Law- Visual Problem

Graham’s Law Problem

Compare the rates of effusion of hydrogen and oxygen at the same temperature and pressure.

Given: identities of two gases, H2 and O2

Unknown: relative rates of effusion

Hydrogen = Compound AOxygen = Compound B

Graham’s Law ProblemHINT: Always put the substance with the

larger molar mass on top as compound B.

1. Calculate:

2. Rearrange the equation:rate of effusion of A = 3.98 rate of effusion

of B

3. Write a sentence:Hydrogen diffuses 3.98 times faster than Oxygen

32.00 g/molrate of effusion of 32.00 g/mol3.98

rate of effusion of 2.02 g/mol2.02 g/molB

A

MA

B M