Ch 5 student notes.pptx

Gases

Introduction



We live in a solution of gases;

nitrogen, oxygen and other

gases surround and support

us. In this chapter we will

focus on the behavior of

5.1 Pressure

Properties of Gases



Gases are compressible



Gases uniformly fill any

container



Gases mix easily with other

Barometer

-

In 1643, Evangelista

Torricelli invented the

barometer, an instrument

for measuring

atmospheric pressure.

Atmospheric pressure

results from gravity’s pull

on air masses.

 dish filled with mercury with a

closed tube

 outside pressure causes



Manometer- instrument used for

measuring pressure



Gas pressure less than

atmospheric pressure



gas pressure = atm pressure –

h



Gas pressure greater than

atmospheric pressure



gas pressure= atm pressure +

h

The Manometer



mm Hg is also called torr



760mm Hg = 760 torr

= 1 atm



SI unit of pressure is

N/m2 or Pascal (Pa)



1 atm = 101,325 Pa or

101.325 kPa

Boyle’s Law

1

quantitative study of gases, 1600s.

Pressure and volume are inversely related.

P

= V

P

V

= P

V

P

V

Ideal gas- gas that obeys Boyle’s law

We “Boyle” peas and vegetables

An

ideal

gas

is

expected to

have a

constant

value of PV,

as

CO

shows the

largest

change in

PV, but this

Charles’ Law 1700’s

-volume of a gas is directly proportional to

Kelvin temperature

V

= V

**

Temp must be in

Kelvin!!!!!!

T

T

The volume of a gas at absolute zero is zero.

Notice how the balloon shrinks in the cold water and expands in the warm water.

A Charlie Brown

Avogadro’s

Law

-equal volumes of gases at the same

temperature and pressure contain

the same # of particles

-for a gas at constant temp. and

pressure, the volume is directly

Gay-Lussac’s Law

Pressure of a gas is directly proportional to

Kelvin temperature

P

= P

Temp must be in Kelvin!!!!!!

T

T

Gayle

drive s

a PT c ruise

Combined Gas Law

P

1

V

1

= P

2

V

2

T

1

T

2

Ideal Gas

Law

Combining Boyle’s, Charles’,

and Avogadro’s laws we get

PV= nRT.

R = 0.08206 L

atm/K

mol

(proportionality constant)

Most gases behave

We can use the ideal gas law for all gas law

problems by putting changing variables

on one side and the constant on the other.

Ex. If P&V change w/ others constant:

P

V

= nRT and P

V

= nRT

so P

V

= P

V

Ex. If V&T change with others constant:

V

= nR and V

= nR so V

= V

T

P T

P T

T

If the volume is held constant, then the

formula P

V

/T

= P

V

/T

shortens down to

P

/T

= P

/T

.

1.5atm = x atm

298K 725K

We use P

V

/T

= P

V

/T

and

solve for V

(45.6atm)(16.5L) = (1 atm)(x L)

(351K) (273K)

X = 585 L

Ex. A quantity of helium gas occupies a

volume of 16.5 L at 78

C and 45.6 atm.

_{We will use PV = nRT and solve for n or moles so that}

we can convert to mass.

(2.039 atm)(50.0L) = (n)(0.08206L.atm/mol.K)(296K)

n = 4.20 mol O₂

4.20 mol O₂ 32.0g O₂ = 134 g O₂

Molar Volume = 22.42 L

of an ideal gas at STP

(Some gases behave

more ideally than

others.)

In this problem, we must first find the mol of hydrogen gas produced (this gas causes the pressure in the

balloon) and then convert to grams of CaH₂. PV=nRT

(1.12 atm)(5.50L) = (n)(0.08206L._atm/mol._K)(288K)

N = 0.261 mol H2

0.261 mol H2 1 mol CaH2 42.1 g CaH2 = 5.49g CaH2

2 mol H2 1 mol CaH2

Ex. CaH

reacts with H

O to produce

H

gas.

CaH

(s)

+ 2H

O

(l)



2H

(g)

+ Ca

(aq)

+ 2OH

(aq)

Assuming complete rxn with water,

how many grams of CaH

are

required to fill a balloon to a total

pressure of 1.12 atm at 15

C if its

First, we must write the balanced equation.

N2 + 3H2



2NH3

Now, we convert, using 1 mol=22.4L as a

conversion factor because this reaction is

occurring at STP.

115 g NH3 1 mol NH3 1 mol N2 22.4L

N2 = 75.8L N2

17.0g NH3 2 mol NH3 1 mol N2

n = mass so P = m R T

MW (MW)V

Since m/V = density (g/L),

P = dRT

MW

MW = dRT

P

“Molecular Weight Kitty Cat”

MW = dRT/P

d= mass/volume = 0.608g/0.750L = 0.811 g/L T = 35°C + 273.15 = 308 K

P = 385mmHg 1 atm = 0.507 atm 760 mm Hg

MW = [(0.811g/L)(0.08206L.atm/mol.K)(308

K)]/0.507 atm

MW = 40.4 g/mol

Dalton’s Law of Partial Pressures

For a mixture of gases in a container, the

total pressure exerted is the sum of the

pressures that each gas would exert if it

were alone.

P

= P

+ P

+ P

+ …

P

= n

RT + n

RT + n

RT +…

V V V

P

= n

(RT)

-the ratio of the number of moles

of a given component in a mixture

to the total number of moles in the

mixture.



is used to symbolize mole fraction



= n

n

+ n

+ n

+ …

The partial pressure of a particular

component of a gaseous mixture is

the mole fraction of that component

times the total pressure.

P

=



(P

)

When gases are collected over

water, we must adjust for the

pressure of the water vapor.

In this problem, the total volume should be the sum of the partial pressures of each gas.

However, we must take into account the pressure of each gas as it fills the new container.

Therefore, we will use P1V1=P2V2 to find each sample’s partial pressure.

Oxygen (1.0atm)(0.20L)=(P2)(0.40L) P2 = 0.50 atm

Nitrogen (2.0atm)(0.10L)=(P2)(0.40L) P2 = 0.50 atm

P_total = P_O2 + P_N2 = 0.50 + 0.50 = 1.0 atm

Ex. If a 0.20 L sample of O

at 0

C and 1.0 atm

pressure and a 0.10 L sample of N

at 0

C and

2.0 atm pressure are both placed in a 0.40 L

container at 0

C, what is the total pressure in

Kelvin temp. is an index of the

random motions of the particles

of a gas, with higher temp.

meaning greater motion.

KE (avg) = 3/2 RT

R = 8.3145 J/K mol

Units are J/mol

u

= avg of the squares of the

particle velocities

= root mean square velocity =

u

u

= where M = mass of

a mole of

gas in

kg

The units of u

are m/s.

Root mean square

velocity

2

u

M

RT

)

/

3

At any given temp., lighter

molecules have higher root mean

square velocities.

Ex. Calculate the root mean square

velocity in m/s of O

molecules at 27

C.

s

m

kg

M

RT

u

483

.

6

/

KE = 3/2RT

KE = 3/2 (8.3145)300

=

3740 J

Calculate the avg. KE of

the same molecules.

Ex. Calculate the root mean square

Real gases have many collisions

between particles. The average

distance a particle travels

between collisions in a particular

gas sample is called the mean

free path. These collisions

produce a huge variation in

velocities. As temperature

A Plot of the Relative Number of O₂

A Plot of the

Relative

Number of N

Molecules That

Have a Given

Velocity at

Three

Effusion and

Diffusion

Diffusion- mixing of gases

Graham’s Law of Effusion

-The rate of effusion of a gas is

inversely proportional to the square

root of the mass of its particles

.

____

Rate of effusion for gas 1 =



MW

Rate of effusion for gas 2



MW

-diffusion rates can be

calculated the same way

Real gases

*No gas exactly follows the

ideal gas law.

*A real gas exhibits behavior

closest to ideal behavior at low

pressures and high

At high temperatures, there is less

interaction between particles because

they are moving too fast.

At high concentrations, gases have

much greater attractive forces between

particles. This causes particles to hit

At high pressure (small volume), the

volume of the particles becomes significant,

so that the volume available to the gas is

significantly less than the container

volume.

We can use the Van der Waals

equation to adjust for departures

from ideal conditions.

PV = nRT becomes:

[P

+ a(n/V)

]V-nb = nRT

corrected corrected

pressure volume

a

= correction for pressure that takes into

b

= correction for the finite volume

of the gas molecules. It is a measure

of the actual volume occupied by the

gas molecules. It increases with an

increase in mass of the molecule or

in the complexity of the structure.

-major gases are N

and O

-see others in Table 5.4, pg 225

-heaviest gases are concentrated

closest to the surface

Chemistry

in the

Read about ozone layer protection.

Most air pollution centers around

NO, Nitric oxide, which is emitted

into the air from exhaust of engines.

NO + O



NO

NO

(g)

NO + O

O + O





O

S (from coal) + O



SO

2SO

+ O



2SO

SO

+ H

O



H

SO

Read about scrubbers,

etc.

Ozone reacts with

unburned hydrocarbons in

polluted air to produce

irritating chemicals. This

process creates