chemistry ii n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Chemistry II PowerPoint Presentation
Download Presentation
Chemistry II

Loading in 2 Seconds...

play fullscreen
1 / 45

Chemistry II - PowerPoint PPT Presentation


  • 163 Views
  • Uploaded on

Chemistry II . Unit 1 Gases. The Nature of Gases. Objectives: Describe the assumption of the kinetic theory as it applies to gases. Interpret gas pressure in terms of kinetic theory Define the relationship between Kelvin temperature and average kinetic energy.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Chemistry II' - betty


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
chemistry ii

Chemistry II

Unit 1 Gases

the nature of gases
The Nature of Gases

Objectives:

  • Describe the assumption of the kinetic theory as it applies to gases.
  • Interpret gas pressure in terms of kinetic theory
  • Define the relationship between Kelvin temperature and average kinetic energy.
  • Explain why gases are easier to compress than solids or liquids.
  • Describe the 3 factors that affect gas pressure.
slide4

Expand to fill their container

  • take the shape of their container
  • low density
  • Compressible
    • Compressibility measures how much the volume of matter decreases under pressure.
  • mixtures of gases are
  • always homogeneous
  • Fluids (flow)
  • Properties of Gases
slide5

Gas pressure

    • Results from collisions of gas particles with an object.
    • In empty space where there are no particles, there is no pressure and is called a vacuum.
    • Atmospheric pressure (air pressure): due to atoms and molecules in air.
    • Barometer: used to

measure atmospheric

pressure.

slide6

Units for measuring pressure:

      • Pascal (Pa)
      • Standard atmosphere (atm)
      • Millimeters of mercury (mmHg)
    • 1 atm = 760 mmHg = 101.3 kPa
    • 1kpa = 1000 pa
    • Standard pressure: 1 atm
  • Factors affecting gas pressure
    • Amount of gas
    • Volume
    • Temperature
      • Standard temperature : 0C (273K)
slide7

Converting between units of pressure

  • A pressure gauge records a pressure of 450 kPa. What is the measurement expressed in atmospheres and millimeters of mercury?

For converting to atm:

450 kpa x 1 atm = 4.4 atm

1013.kPa

For converting to mmHg:

450kPa x 760 mmHg = 3.4 x 103 mmHg

101.3 kPa

slide8

What pressure in kilopascals and in atmospheres, does a gas exert at 385 mmHg?

  • The pressure on the top of Mount Everest is 33.7 kPa. Is that pressure greater or less than 0.25atm?
slide9

What pressure in kilopascals and in atmospheres, does a gas exert at 385 mmHg?

51.3 kPa, 0.507 atm

  • The pressure on the top of Mount Everest is 33.7 kPa. Is that pressure greater or less than 0.25atm?

33.7 kPa is greater than 0.25 atm

slide10

Kinetic energy and temperature

    • The Kelvin temperature of a substance is directly proportional to the average kinetic energy of the particles of the substance.
    • Directly proportional means that temperatures increases as the average kinetic energy increases or decreases as the average kinetic energy decreases.
    • K = C + 273
slide11

The Atmosphere

an “ocean” of gases

mixed together

Composition

~78%

nitrogen (N2)…………..

~21%

oxygen (O2)……………

~1%

argon (Ar)……………...

Trace amounts of:

~0.04%

carbon dioxide (CO2)…

He, Ne, Rn, SO2,

CH4, NxOx, etc.

water vapor (H2O)…….

~0.1%

slide12

CFCs

refrigerants

aerosol propellants

-- banned in U.S. in 1996

Depletion of the Ozone Layer

O3 depletion is caused by chlorofluorocarbons (CFCs).

Ozone (O3) in upper atmosphere

blocks ultraviolet (UV) light from Sun.

UV causes skin cancer and cataracts.

Uses for CFCs:

O3 is replenished with each strike of lightning.

slide13

Reaction_to_Air_Pressure_Below_Sea_Level.asf

  • Classwork:

Read pages 103-105

Do problems 1-6

gas laws
Gas Laws
  • Objectives
  • Describe the relationships among the temperature, pressure, and volume of a gas
  • Use the gas laws to solve problems
slide15

Boyle’s Law : Pressure and Volume

  • States that for a given mass of gas at constant temperature, the volume of a gas varies inversely with pressure.
  • If pressure increases, volume decreases; if pressure decreases, volume increases.
slide16

P1 x V1 = P2 x V2

P: pressure 1: initial condition

V: volume 2: final condition

slide17

Using Boyle’s Law

  • A balloon with 30.0L of helium at 103kPa rises to an altitude where the pressure is only 25.0kPa. What is the volume of the helium (at constant temperature)?
slide18

Using Boyle’s Law

  • A balloon with 30.0L of helium at 103kPa rises to an altitude where the pressure is only 25.0kPa. What is the volume of the helium (at constant temperature)?

P1=103 kPa V1= 30.0L P2= 25.0kPa V2=?

P1V1= P2V2

V2= P1V1 = (103 kPa)(30.0L) = 124 L

P2 (25.0 kPa)

Since pressure decreases, you expect volume to increase.

slide19

2. A gas with a volume of 4.00L at a pressure of 205 kPa is allowed to expand to a volume of 12.0L. What is the pressure of the container now (at constant temperature)?

slide20

2. A gas with a volume of 4.00L at a pressure of 205 kPa is allowed to expand to a volume of 12.0L. What is the pressure of the container now (at constant temperature)?

P1= 205 kPa V1= 4.00L P2=? V2=12.0L

P1V1=P2V2

P2= P1V1 = (205 kPa)( 4.00 L) = 68.3 kPa

V2 (12.0L)

Classwork: p 121 # 2-6

slide21

Charles’s Law: Temperature and Volume

  • States that the temperature of an enclosed gas varies directly with the volume at constant pressure.
  • As temperature increases, volume increases.

V1 = V2

T1 T2

V1: initial volume V2: final volume

T1: initial temperature T2: final temperature

Temperature has to be in Kelvin scale.

volume and temperature

As a gas is heated, it expands.

This causes the density of the

gas to decrease.

Volume and Temperature
slide23

Using Charles’s Law

  • A balloon inflated in a room at 24C has a volume of 4.00L . The balloon is then heated to a temperature of 58C. What is the new volume ?
slide24

Using Charles’s Law

  • A balloon inflated in a room at 24C has a volume of 4.00L . The balloon is then heated to a temperature of 58C. What is the new volume ?

V1 = V2 V1= 4.00L V2= ?

T1 T2 T1= 24C +273= 297 K

T2= 58C + 273 = 331 K

V2= V1T2 = (4.00L)(331K) = 4.46 L

T1 (297K)

Since temperature increases, you expect the volume to increase.

Classwork: p124 # 11, 12 (a-c), 13

slide25

Combined Gas Law

  • Describes the relationship among the pressure, temperature and volume, when the amount of gas is constant.
  • P1V1 = P2V2

T1 T2

  • Standard temperature and pressure (STP): 0C, 1 atm
  • Useful conversions:

1L =1000 mL ; 1mL =1cm3 ; 1dm3 = 1 L

slide26

Using the combined gas law:

  • The volume of a gas filled balloon is 30.0L at 313K and 153 kPa. What would the volume be at standard temperature and pressure (STP)?
slide27

Using the combined gas law:

  • The volume of a gas filled balloon is 30.0L at 313K and 153 kPa. What would the volume be at standard temperature and pressure (STP)?

P1= 153 kPa V1= 30.0 L T1= 313K

P2= 1 atm=101.3kPa V2= ? T2= 0C= 273K

P1V1 = P2V2

T1 T2

V2= P1V1T2 = (153 kPa)(30.0L)(273K)= 39.5 L

P2T1 (101.3kPa)(313K)

Classwork: p 126 #14(a,b), 15 (c,d), 16

ideal gases
Ideal Gases
  • Objectives
  • Compute the value of an unknown using the ideal gas law.
  • Compare and contrast real and ideal gases.
slide29

Avogadro’s Principle and Molar Volume

  • Avogadro’s principle states that equal volume of all gases, measured under the same conditions of pressure and temperature, contain the same number of particles.
  • At STP, the volume of one mole of gas is 22.4 (this is called the molar volume)
  • From last year, to convert between grams and moles of a substance we used its molar mass.
slide31

Converting between moles and grams:

How many moles are 98.32g CO2?

Calculate molar mass CO2 (use periodic table)

Molar mass= 12 + (2 x 16) =44.0 g/1 mol

To convert grams to moles:

98.32g x 1 mol = 2.23 mol CO2

44g

slide32

Using Avogadro’s Principle

  • How many grams of carbon dioxide, CO2, will occupy a volume of 500.0 mL at STP?

V= 500.0mL =0.5 L

Molar mass CO2= 44g/mol

At STP, molar volume is : 22.4L/1 mol

0.5 L x 1 mol x 44 g = 0.982g CO2

22.4L 1 mol

Classwork: p 132 # 1 (a-c), 3 (a-d)

slide33

The ideal gas law

  • Considers that amount of gas varies.
  • New variable
    • n: number of moles of gas (mol)
  • Ideal gas constant (R)
    • R= 8.314 L kPa (when pressure is measured in kPa)

mol K

    • R= 0.0821 L atm (when pressure is measured in atm)

mol K

  • PV= nRT(T must be in Kelvin)
slide34

Using the ideal gas law

1. A deep underground cavern contains 2.24x106 L of methane gas (CH4) at a pressure of 1500 kPa and a temperature of 315K. How many kilograms of CH4 does the cavern contain?

slide35

Using the ideal gas law

  • A deep underground cavern contains 2.24x106 L of methane gas (CH4) at a pressure of 1500 kPa and a temperature of 315K. How many moles of CH4 does the cavern contain?

P= 1500kPa V= 2.24x106 L R= 8.314 L kPa

n= ? T= 315 K mol K

PV=nRT

n= PV = (1500 kPa) (2.24x106 L ) (mol K)

RT (8.314 L kPa)(315K)

Classwork: p141 # 1-4

slide36

Ideal gas variations

  • For calculating molar mass
    • M= mRT M: molar mass (g/mol)

PV m: mass (g)

  • For calculating density
    • D= MP D: density (g/L)

RT

slide37

Sample problem

  • What is the molecular mass of sulfur dioxide, SO2, if 300 mL of the gas has a mass of 0.855 g at STP?
slide38

Sample problem

  • What is the molecular mass of sulfur dioxide, SO2, if 300 mL of the gas has a mass of 0.855 g at STP?

M= ? V= 300mL=0.3 L T=273K, P= 101.3kPa m=0.855g

M= mRT = (0.855g )x(8.314 L kPa)x(273K)

PV (mol K ) (101.3kPa)(0.3L)

= 63.8g/mol

Classwork: p 133 #5,6, 9

gases mixtures and movements
Gases: Mixtures and Movements

Objectives:

  • Relate the total pressure of a mixture of gases to the partial pressures of the component gases.
  • Explain how the molar mass of a gas affects the rate at which the gas diffuses and effuses
slide40

Dalton’s Law

  • Partial pressure: the contribution each gas in a mixture makes to the total pressure.
  • Dalton’s law of partial pressure: at constant volume and temperature, the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the component of the gases.
  • P total= P1 + P2 + P3 + …
slide41

Graham’s Law

  • Diffusion: the tendency of molecules to move toward areas of lower concentration until the concentration is uniform throughout.
  • Effusion: a gas escapes through a tiny hole inits container.
  • Particles of lower molar mass diffuse and effuse faster than gases of higher molar mass.
slide42

Using Dalton’s and Graham’s Laws:

  • Air contains oxygen, nitrogen, carbon dioxide, and trace amounts of other gases. What is the partial pressure of oxygen (PO2) at 101.3kPa of total pressure if the partial pressures of nitrogen, carbon dioxide, and other gases are 79.10kPa, 0.040kPa, and 0.94kPa respectively?

Ptotal = 101.3 kPa PN2= 79.10kPa PCO2= 0.040kPa

Pothers= 0.94kPa PO2= ?

Ptotal = PN2 + PCO2 + PO2 + Pothers

PO2 = Ptotal- (PN2+ PCO2+ PO2) =101.3kPa-(79.1kpa+0.040kPa+0.94kPa)

= 21.22 kPa

slide43

2. A nitrogen,N2, molecule travels at about 505 m/s at room temperature. Find the velocity of a helium, He, atom at the same temperature.

slide44

2. A nitrogen,N2, molecule travels at about 505 m/s at room temperature. Find the velocity of a helium, He, atom at the same temperature.

rateHe /rateN2=  (molar mass N2/molar massHe)

rateHe = rateN2  (molar mass N2/molar massHe)

= 505m/s (28/4)

= 1336 m/s