Gas Pressure

• Gases exert pressure on any surface they come in contact with. • Pressure is related to the number of collisions the gas molecules have with wall of a container per unit of area per unit of time. • Pressure = Force / Area 1

Elements that exist as gases at 25 0 C and 1 atmosphere 2 5.1

• • • • Physical Characteristics of Gases Gases assume the volume and shape of their containers.

Gases are the most compressible state of matter.

Gases will mix evenly and completely when confined to the same container.

Gases have much lower densities than liquids and solids.

3 5.1

• The force of impact of a single collision is too small to be sensed. However, taken all together, this large number of impacts of gas molecules exerts a considerable force onto the surface of the enclosure: the

gas pressure

• The larger the number of collisions per area of enclosure, the larger the pressure: 4

Pressure = Force Area ( force = mass x acceleration) Barometer 5 5.2

Units for Pressure

• The SI-unit of pressure is Pascal [Pa] Atmospheres (atm) Millimeters of Mercury (mmHg) Torr (torr) Pressure per square inch (Psi) = lbs/in 2

1atm = 760 mmHg = 760 torr 1 atm = 1.013 x10 5 Pa 1 atm = 14.69 psi

6

Types of Pressure

7

Boyles Law Demo

lets assume that the balloon is tight, so that the amount or mass of air in it stays the same • Density = mass/ volume, • the gas density of the balloon thus varies only with its volume (when mass is held constant). • If we squeeze the balloon, we compress the air and two things will happen: – the air pressure in the balloon will increase. – the density of the air in the balloon will increase. • Since density is mass over volume, and the mass stays constant, the rise in density means that the volume of the balloon decreases:

pressure goes up ( ↑); volume goes down ( ↓)

• Pressure and volume are inversely proportional 8

Boyle’s Law

P ↓ P ↑ V ↑ V ↓

At a constant temperature and a fixed quantity of gas pressure and volume are inversely proportional. P 1 V 1 = P 2 V 2 ( 1 = initial 2 = final) 9

Graphical Explanation

At a constant temperature and a fixed quantity of gas pressure and volume are inversely proportional. P 1 V 1 = P 2 V 2 ( 1 = initial 2 = final) 10

Boyle’s Example

Mini Lab 11

P Depth (ft) 0

Chemistry in Action:

Scuba Diving and the Gas Laws Pressure (atm) 1 33 66 2 3 V 12 5.6

Example

• A 3.0L bulb containing He at 145 mmHg is connected by a value to a 2.0 L bulb containing Ar at 355 mmHg. Calculate the partial pressure of each gas and the total pressure after the valve between the flasks is opened.

3.0 L He 145 mmHg 2.0 L Ar 355 mmHg

13

Answer

First we need to find the total volume of the bulbs: V total = V a +V b = 3+2=5 Next we need to do Boyles’s law twice, once for each bulb to find P of each gas. P 1 V 1 = P 2 V 2 He: 145(3) = P 2 (5) P 2 =87 mmHg Ar = 355 (2) = P 2 (5) P 2 = 142 mmHg 14

Answer Cont.

He: 145(3) P 2 = P 2 (5) =87 mmHg Ar: 355 (2) = P 2 (5) P 2 = 142 mmHg Now we need to find the after the valve between the two flasks is opened.

P total = P 2He = 87 + + P 2Ar 142 = 229 mmHg 15

Bonus

Convert 229 mmHg to atm 229 mmHg x 1 atm 760 mmHg = .303 atm 16

Charlies' Law Demo

• By warming the balloon up, we increase the speed of the moving gas molecules inside it.

• This increases the rate at which the gas molecules hit the wall of the balloon. • Because the balloon’s skin is elastic, it expands upon this increased pushing from inside, and the volume taken up by the same mass of gas increases with temperature. 17

Charlie’s Law

• At constant pressure the volume of gas is direct proportional to its temperature.

V 1 = V 2 T 1 T 2

Note: Temp is ALWAYS in Kelvin!!!!

T ↓ T ↑ V ↓ V ↑

18

Charles's Law Mini Lab

Mini Lab 19

Graphical Explanation

20

Charles's Law Example

A balloon is filled to a volume of 7.00 x 10 2 ml at a temperature of 20.

◦C. The balloon is then cooled at a constant pressure to a temperature of 1.0x10

2 K. What is the final volume of the balloon?

21

Answer

20C + 273 = 293 K 7.00 x 10 2 ml = V 2 293 K 1.0x10

2 K. V 2 = 238.9 ml 22

Avogadro’s Law

• Equal volumes of gas at the same temperature and pressure contain equal numbers of moles. • If temperature and Pressure are constant Volume of a gas is directly proportional to the number of moles. 23

Avogadro’s Law

V ↓ V ↑ n ↑ n ↓

At constant temperature and Pressure the Volume of a gas is directly proportional to the number of moles. V 1 n 1 = V 2 n 2 24

Avogadro’s Law

V

a number of moles (

n

)

V

= constant x

n V

1 /

n

1 =

V

2 /

n

2 Constant temperature Constant pressure 25 5.3

Example

• How many liters of O 2 prepare 100 L of CO 2 gas are required to gas by the following reaction. 2 CO (g) + O 2 (g) 2 CO 2 (g) 26

100 = V 2 2 1 V 2 = 50L

Answer

V1 = V2 n1 n2 27

Homework

Chang: pg 210-211 #”S : 13,15, 18,19,21,23 BL 3,17, 19, 20, 21 28

Combined Gas Law

• This is used when nothing is constant in an experiment. P 1 V 1 T 1 = P 2 V 2 T 2 29

Example

A gas is contained in a cylinder with a temperature of 281 K and a volume of 2.1 ml at a pressure of 6.4 atm. The gas is heated to a new temperature of 298 K and the pressure decreases to 1 atm.

What is the new volume of the gas. 30

Answer

P 1 V 1 T 1 P 1 = 6.4atm V 1 = 2.1 ml T 1 = 281 k P 2 = 1 atm V 2 = ?

T 2 = 298 K = P 2 V 2 T 2 6.4 (2.1) = 1 (V 2 ) 281 298 V 2 = 14ml 31

Ideal Gas Law 10.4

Ideal gas: a hypothetical gas whose pressure (atm), volume (L), and temperature (K) behave as predicted every time. (Perfect like each and everyone of you!) One can visualize it as a collection of perfectly hard spheres which collide but which otherwise do not interact with each other. Ideal Gas Law: PV = nRT gas constant: R= 0.0821 L x Atm/mol x K 32

Temperature and KE

• Temperature is a measure of the Average kinetic energy of the molecules of a substance.

• Higher temperature faster molecules.

• At absolute zero (0 K) all molecular motion would stop.

e c u le % o f M ol s Kinetic Energy High temp .

Low temp.

34

Kinetic Molecular Theory

•

The Kinetic Molecular Theory explains the forces between molecules and the energy that they possess. This theory has 3 basic assumptions.

35

KMT

1.

Matter is composed of small particles (molecules). The measure of space that the molecules occupy (volume) is derived from the space in between the molecules and not the space the molecules contain themselves.

36

•

The molecules are in constant motion. This motion is different for the 3 states of matter.

Solid - Molecules are held close to each other by their attractions of charge. They will bend and/or vibrate, but will stay in close proximity.

Liquid - Molecules will flow or glide over one another, but stay toward the bottom of the container. Motion is a bit more random than that of a solid.

Gas - Molecules are in continual straightline motion. The kinetic energy of the molecule is greater than the attractive force between them, thus they are much farther apart and move freely of each other.

37

KMT 3

•

When the molecules collide with each other, or with the walls of a container, there is no loss of energy.

• Random Fact: The Average speed of an oxygen molecule is 1656 km/hr at 20ºC 38

Example

• A 50.0L cylinder of acetylene C 2 H 2 has a pressure of 17.1 atm at 21C . What is the mass of acetylene in the cylinder. 39

Answer

PV = nRT 21 + 273 = 294 K 17.1 (50.0) = n (0.0821) (294) n = 35.4 mol Need answer in grams • 35.4 C 2 H 2 x 26g C 2 H 2 1 mol C 2 H 2 = 920 g C 2 H 2 40

A short Way to do that

• mw = mRT = dRT PV P An unknown gas weighs 34g and occupies 6.7L at a pressure of 2 atm at temperature of 245K.What is its average molecular weight. 41

• 51 g/mol

Answer

42

• Density = mass = P mw vol RT 43

STP

• Standard Temperature and Pressure • P = 1 atm = 760 torr • T = 273 K • Molar volume of ideal gas at STP = 22.42L 44

What is the volume (in liters) occupied by 49.8 g of HCl at STP?

Given:

T

= 0 0 C = 273.15 K

P = 1 atm PV = nRT V = nRT P n

= 49.8 g x 1 mol HCl 36.45 g HCl = 1.37 mol

V

= 1.37 mol x 0.0821 x 273.15 K mol •K 1 atm

V

= 30.6 L 45 5.4

Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0 C is heated to 85 0 C at constant volume. What is the final pressure of argon in the lightbulb (in atm)?

PV

=

nRT n, V

and R are constant

nR V

=

P T

= constant

P 1

= 1.20 atm

T 1 =

291 K

P 2

= ?

T 2 =

358 K

P T 1 1 P 2

= =

P 2 T 2 P 1 x T 2 T 1

= 1.20 atm x 358 K 291 K = 1.48 atm 46 5.4

Dalton’s Law of Partial Pressure

• The sum of the partial pressures of the system is equal to the total pressure of the system. • P total = P 1 + P 2 + P 3 47

P

1 Dalton’s Law of Partial Pressures

V

and

T

are

constant

P

2

P

total

= P

1 +

P

2 48 5.6

Homework

Chang pg 211 DAY 1 #’s 27,29,31,33 Day 2 #’s 38,41,49,51-53,62,66,71 Write out the combined gas law and ideal gas law formula 5 times each and then do problems 23,27,28, 38 49