Pages 422-440

Notes One Unit Five

Characteristics of GasesPressure of fluidsStandard Temperature and PressureConverting PressuresGas Laws

Pressure Versus Molecular Collision

• Pressure is caused by molecular collision

• A molecule colliding creates a force.• Catching a ball creates a force.• P=F/A

• pp 427

Pressure viewed as created in a fluid

• Created by the weight

• The deeper you go, the more weight .

pp 427

Air is a fluid…just like water

pp 427

Torricellian Barometer

760 torr780 torr

Air Pressure

Mercury

740 torr

pp 427

Pop can Demo

Magdeburg plates Demo

Standard Pressure, Temperature and Volume1 atm = 14.7 psi1 atm = 29.92 in Hg1 atm = 760 mm Hg1 atm = 760 torr 1 atm = 101,325 Pa 1 atm = 101.325 kPa 1 atm = 1.01325 bar 273K or 0oC K=oC+273

22.4 Liter/mole for any gas at STP

pp 427

Converting Pressures

• Convert 25 lb/in2 to torr

• Convert 75 Kpa to in Hg

Charles’ and Boyle’s Demo

P1=1.0atmV1=1.0LT1= 273K

P2=2.0L

T2= 546KV2=

1.0atm

Pressure is constant.Moles are constant.

V1

T1

= V2

T2

P1=1.0atmV1=1.0LT1= 273K

P2= 2.0atmV2= 0.50LT2= 273K

P1 x V1 P2 x V2=

Temperature is constant.Moles are constant.

pp 433-4401.0L273K

= 2.0L546K

1.0a x 1.0L 2.0a x0.50L=

Combined Gas Law Equation

• Is……..

pp 433-440

Combined gas Law Problem One• A gas occupies 2.0 m3 at 121.2 K, exerting a pressure

of 100.0 kPa. What volume will the gas occupy at 410.0 K if the pressure is increased to 220.0 kPa?

(2.0m3 )(100.0KPa)(121.2K )

(220.0KPa )(V2 )(410.0K )

=

(2.0m3 )(100.0KPa) (410.0K )(121.2K )(220.0KPa )

= V2

V2= 3.1M3 pp 433-440

Assign variables and calculate V2.

Combined gas Law Problem Two• A 10.0 gram sample of ethane(C2H6) gas is at STP. If

the volume is changed to 26.0 liters, what is the new Kelvin temperature of the gas?

10.0g 30.0g/mol

=22.4L V1 V1= 7.46L

(101.325KPa) (7.46L)(273K)

= (101.325KPa)(26.0L)(T2)

(273K)(101.325KPa)(26.0L)(101.325KPa) (7.46L)

= T2 T2= 951Kpp 433-440

CH

2x6x

12.0 =1.0 =

24.0 6.030.0g/m

E # Mass

10.0 gram

1) Calculate Formula mass.

2) Calculate V1.

3) Assign variables and calculate T2.

Notes Two Unit Five

Grahams’ Law Calculation

Review Mass-Mass Calculation

Mass-Volume Calculation @STP

Volume-Mass Calculation @STP

Pages 441-450

Graham’s Law Demo

17.0g/m 36.5g/m

pp 442

Graham’s Law• Describes how speed compares between gas

molecules with different masses.

• Two different gases:

• 1)Same Temperature

• 2)Different Masses

• Kinetic energy ½ M1V12= ½ M2V2

2

pp 442

V12

½ M2V22= ½ M1V1

2

Graham’s Equation

pp 442

M2V22 M1V1

2

M1 M1

M2V22

M1

V12

= V2

2 V22

= ÷ by M1

÷ by V22

Square root of both sides

Grahams’ Law Problem One• At a high temperature molecules of chlorine gas travel 15.90cm. What is the mass of vaporized

metal (gas) under the same conditions, if the metal travels 8.97cm?

M2= 223g/m

Cl2x35.5 = 71.0g/mE # Mass

pp 442

(15.90cm)

8.97cm= = 14.9 2

Grahams’ Law Problem Two• At a certain temperature molecules of chlorine gas travel at 0.450 km/s. What is the speed of sulfur dioxide gas

under the same conditions?

V2= 0.474Km/s pp 442

Cl2x35.5 = 71.0g/mE # Mass

SO

1x2x

32.1 =16.0 =

32.1 32.064.1g/m

E # Mass

V2=

Review Mass-Mass Calculation

1) grams H2 to moles H2

5.00g÷

2) moles H2 to moles O2

2.5m H2x

3) moles O2 to grams O2

1.3mO2x

2.0gH2 /m= 2.5m H2

32.0g/m = 42gO2

(1mO2)(2mH2)

=1.3mO2

2 1 2

H 2 x 1.0 = 2.0g/mE # Mass

O 2x16.0 = 32.0g/mE # Mass

How many grams of oxygen will react with 5.00 grams of hydrogen to make water?

H2(g) + O2(g) H2O(l)

425.0g

X32.0g/m___m

___g

2.5 ___m1.3

÷2.0g/m

Mass-Volume Calculation @STP

1) grams H2 to moles H2

5.00g÷

2) moles H2 to moles O2

2.5m H2x

3) moles O2 to liters O2

1.3mO2x

2.0gH2 /m= 2.5m H2

22.4L/m = 29LO2

(1mO2)(2mH2)

=1.3mO2

2 1 2

H 2 x 1.0 = 2.0g/mE # Mass

O 2x16.0 = 32.0g/mE # Mass

How many liters of oxygen will react with 5.00 grams of hydrogen to make water?

H2(g) + O2(g) H2O(l)

295.0g

X22.4g/m___m

___L

2.5 ___m1.3

÷2.0g/mpp 449

Volume-Mass Calculation @STP

1) Liters H2 to moles H2

56.0L÷

2) moles H2 to moles O2

2.50m H2x

3) moles O2 to grams O2

1.25mO2x

22.4LH2 /m= 2.50m H2

32.0g/m = 40.0gO2

(1mO2)(2mH2)

=1.25mO2

2 1 2

H 2 x 1.0 = 2.0g/mE # Mass

O 2x16.0 = 32.0g/mE # Mass

How many grams of oxygen will react with 56.0 liters of hydrogen to make water?

H2(g) + O2(g) H2O(l)

40.056.0L

X32.0g/m____m

____g

2.50 ____m1.25

÷22.4L/mpp 449

Notes Three Unit Five

• Kinetic theory of gases

• Molar volume @ Non-STP Conditions

• R is Universal Gas Constant

Pages 452-459

THE KINETIC THEORY OF GASES• Large number of particles

6.022x1023atoms/mole

pp 426

THE KINETIC THEORY OF GASES• Large number of particles • Elastic collisions For a collision KEBefore=KEAfter

pp 426

THE KINETIC THEORY OF GASES• Large number of particles • Elastic collisions • No external forces

pp 433-440

THE KINETIC THEORY OF GASES• Large number of particles • Elastic collisions • No external forces • Separated by large distances

1.6x1011 times diameter

pp 426

THE KINETIC THEORY OF GASES• Large number of particles • Elastic collisions • No external forces • Separated by large distances• No forces between particles

pp 426

Finding volumes @ Non-STP Conditions

pp 446

PV=nRTWhat is…P?V?n?

T?

moles

R? Universal Gas Constant

P Vn T

R=(101.325kpa)(22.4L)

(1 m)(273.15K)R=

Ideal Gas Equation

R =(8.314 )Kpa- Lm- K

Finding Volume at Non-STPOxygen is made reacting 15.00g water at 209.0Kpa and 20.0oC.

18.0g/m= 0.833m

22.4L/m = 9.33LO2

( 1mO2 )(2m H2O) =0.417mO2

a) What volume of oxygen would be made at STP?1) grams H2O to moles H2O

2) moles H2O to moles O2

3) moles O2 to liters O2

15.00g÷

0.833m x

0.417mO2x

2H2O(l) 2H2(l) + 1O2(l)

15.00g

X22.4L/m______m

____L

0.833 0.417

÷18.0g/m

_____m

9.33

@STP

E # Mass

O 16.0 = 16.01 x18.0g/m

H 2 x 1.0 = 2.0

Finding Volume at Non-STP

PV=nRTV= (0.417m)(8.314 L•Kpa•m-1•K-1)(293.2K)

(209.0KPa)

V= 4.86L

n= 0.417mO2

n x g/m= g0.417m x 32.0g/m=13.3gO2

pp 450

c) How many moles O2 gas are produced?

d) How many grams O2 gas are produced?

b) What is the volume of gas at reaction conditions?

O 2x16.0 = 32.0g/mE # Mass

P Vn TR=