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GASES. Paul Gilletti, Ph.D. Mesa Community College. Gases (Vapors). Gases expand to fill any container. Therefore, gases are highly compressible. Kinetic Molecular Theory ( of an Ideal Gas ):.

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slide1

GASES

Paul Gilletti, Ph.D.

Mesa Community College

slide2

Gases (Vapors)

Gases expand to fill any container.

Therefore, gases are highly

compressible.

slide3

Kinetic Molecular Theory (of an Ideal Gas):

1. Gases are composed of molecules or atoms whose size is negligible compared to the average distance between them. (Most of the space in the gas container is empty.)

2. Gas molecules move randomly in straight lines in all directions at various speeds.

3. The forces of attraction or repulsion between gas molecules are very weak or negligible (except during collisions)

4. Collisions between gas molecules are considered elastic.

5. The average kinetic energy of a molecule is proportional to the absolute temperature.

slide4

Pressure and Volume: Boyle’s Law

How is the pressure applied to a gas related to its volume?

Gas molecules

Piston

Let’s apply pressure

slide5

Pressure and Volume: Boyle’s Law

How is the pressure applied to a gas related to its volume?

Gas molecules

Gas molecules

Piston

Piston

Volume is inversely proportional to applied pressure.

Boyle’s Law: P1V1 = P2V2

slide6

The Harder we Push

the smaller the gas

volume gets!

Boyle’s Law: P1V1 = P2V2

slide7

We live in “sea of air”

molecules of air

3

2

Where is the pressure the greatest?

1

Why does a diver get the bends?

slide8

Pressure:

force per unit area of surface

Units

lbs per in2 (psi)

mm of Hg (torr)

atmospheres (atm)

Pascal (Pa)

1 atm = 760 mm of Hg =760 torr = 14.70 psi = 101.325 kPa

Pairs of these can be used as conversion factors.

slide11

Temperature and Volume: Charles’s Law

How is the volume of a gas related to its temperature?

moveable mass

(constant pressure)

gas molecules

What happens if heat is applied to the gas?

slide12

Temperature and Volume: Charles’s Law

How is the volume of a gas related to its temperature?

moveable mass

(constant pressure)

gas molecules

Why did the volume change?

What happens to the average speed of the gas molecules?

.

slide13

Temperature and Volume: Charles’s Law

How is the volume of a gas related to its temperature?

moveable mass

(constant pressure)

gas molecules

The volume of a gas is directly proportional to its Temperature (temperature must be in Kelvin)

Charles’s Law: V1/T1 = V2/T2

slide14

Combined Gas Law (Boyle and Charles):

T must be in Kelvin

Can be rearranged to:

P1V1T2 = P2V2T1

A combined gas law problem can be recognized by having two sets of conditions.

Note: if one set of parameters is unchanged that term

will cancel on each side.

slide15

A balloon contains helium gas with a volume of 2.60 L at 25 oC and 768 mmHg. If the balloon ascends to an altitude where the helium pressure is 590 mmHg and the temperature is 15 oC, what is the volume of the balloon?

What type of

problem

is this?

There are 2 sets of

conditions.

slide16

A balloon contains helium gas with a volume of 2.60 L at 25 oC and 768 mmHg. If the balloon ascends to an altitude where the helium pressure is 590 mmHg and the temperature is 15 oC, what is the volume of the balloon?

P1V1T2 = P2V2T1

P1=

V1=

T1=

768 torr

2.60 L

25 + 273 = 298 K

= (768 torr)(2.60 L)(288 K)

(590 torr)(298 K)

P2=

V2=

T2=

590 torr

?

15 + 273 = 288 K

= 3.27 L

slide17

Ideal Gases and the Ideal Gas Law:

PV = nRT

Temperature in K

*gas constant 0.0821 L•atm = 62.37 L•torr

mol•K mol•K

moles of gas

volume in L

pressure in units to match *R units

Note: there is only one set of conditions.

slide18

Avogadro’s Law:

Equal volumes of any two gases (ideal) at the same temperature and pressure contain the same number of molecules (they also occupy equal volumes).

STP

Pressure 1 atm (760 mm Hg)

Temperature 0oC (273 K)

Standard

At STP one mole of ideal gas occupies 22.4 L

slide19

A 12.25 L cylinder contains 75.5 g of neon at 24.5 oC. Determine the pressure in the cylinder.

What type of

problem

is this?

Only one

set of

conditions

slide20

A 12.25 L cylinder contains 75.5 g of neon at 24.5 oC. Determine the pressure in the cylinder.

P = nRT

V

PV = nRT

= (3.74 mol)(62.4L•torr)(297.5K)

(12.25 L) mol•K

?

P =

V =

n =

R =

T =

12.25 L

mol

= 5667.7 torr

75.5 g = mol

3.74

20.18 g

= 5670 torr

62.4 L•torr

mol•K

How many atmospheres is this?

24.5 + 273 = 297.5 K

slide21

What is the density of carbon dioxide gas at 25 oC and 725 mmHg pressure?

Density = g/L = g  L

so if we can find g and L, division will work!

P =

V =

n =

R =

T =

725mmHg

What do we

do now?

62.4 L• torr

mol•K

25 + 273 = 298 K

slide22

What is the density of carbon dioxide gas at 25 oC and 725 mmHg pressure?

Density = g/L = g  L

so if we can find g and L division will work!

P =

V =

n =

R =

T =

725mmHg

Two variables! Let’s pick an amount for one and calculate the other!

Let’s choose 1 mol of CO2 and find the number of Liters.

62.4 L•torr

mol•K

25 + 273 = 298 K

slide23

What is the density of carbon dioxide gas at 25 oC and 725 mmHg pressure?

Density = g/L = g  L

so if we can find g and L division will work!

V = nRT

P

P =

V =

n =

R =

T =

725mmHg

= (1 mol) (62.4 L•torr) (298 K)

( mol•K ) (725 torr)

1.0 mol (44.0 g)

62.4 L•torr

mol•K

= 25.6 L

44.0 g

___________ = g

L

NOW:

1.72

25 + 273 = 298 K

25.6 L

slide24

A 2.50 gram sample of a solid was vaporized in a 505 mL vessel. If the vapor pressure of the solid was 755 mmHg at 155 oC, what is the molecular weight of the solid?

molecular weight ~ molar mass = g/mol = g  mol

..so if we can find grams and moles and divide....

...we already have grams!! We’re halfway there!

P =

V =

n =

R =

T =

755 torr

n = PV

RT

0.505 L

= 755 torr | 0.505 L | mol•K_____|______

| 62.4 L•torr | 428 K

62.4 L•torr

mol•K

= 0.01428 mol

155 + 273 = 428 K

NOW: 2.50 g = g

0.01428 mol mol

175.1

slide25

So Density is g/L (g ÷ L)

and

molar mass

is g/mol (g ÷ mol).

slide26

Dalton’s Law of Partial Pressures:

He

H2N2

Ptotal = P1 + P2 + P3 +...

slide27

Dalton’s Law of Partial Pressures:

Ptotal = P1 + P2 + P3 +...

Since they are considered to be ideal gases, the attractions and repulsions between molecules are ignored. ... and...

PV=nRT

so: PV = (n1 + n2 + n3)RT

or:

We also refer to mole fractions:

slide28

To find the gas pressure, the pressure of the water

vapor must be subtracted from the total pressure.

slide29

A 250.0 mL flask contains 1.00 mg of He and 2.00 mg of H2 at 25.0oC. Calculate the total gas pressure in the flask in atmospheres.

The total pressure is due to the partial pressures of each of these gases.

so:

For He:

1.00 x 10-3 g He

mol

_____________________ = mol He

2.50 x 10-4

4.00 g

For H2:

2.00 x 10-3 g H2

mol

______________________ = mol H2

9.92 x 10-4

2.016 g

slide30

A 250.0 mL flask contains 1.00 mg of He and and 2.00 mg of H2 at 25.0oC. Calculate the total gas pressure in the flask in atmospheres.

so:

1.00 x 10-3 g He

mol

_____________________ = mol He

For He:

2.50 x 10-4

4.00 g

For H2:

2.00 x 10-3 g H2

mol

______________________ = mol H2

9.92 x 10-4

2.016 g

And: Ptotal = (2.50 x 10-4 + 9.92 x 10-4)(RT/V)

= (0.001242 mol)(0.0821 L•atm)(25 + 273)K

mol•K (0.2500 L)

Ptotal= 0.1216 atm

slide31

A 250.0 mL flask contains 1.00 mg of He and and 2.00 mg of H2 at 25.0oC. Calculate the total gas pressure in the flask in atmospheres.

so:

1.00 x 10-3 g He

mol

_____________________ = mol He

For He:

2.50 x 10-4

4.00 g

For H2:

2.00 x 10-3 g H2

mol

______________________ = mol H2

9.92 x 10-4

2.016 g

Calculate the pressure due just to He (you have 37 seconds):

= 0.0245 atm

and Phydrogen= ?

0.1216 - 0.0245 = 0.0971 atm

slide32

Magnesium is an active metal that replaces hydrogen from an acid by the following reaction:

Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)

How many g of Mg are needed to produce 5.0 L of H2 at a temperature of 25 oC and a pressure of 745 mmHg?

Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)

? g

5.0 L

Hint: find moles of H2 using PV = nRT then work as a stoichiometry problem.

n = PV

RT

745 mmHg

5.0 L

mol•K

=____________________________________

62.4 L•mmHg

298 K

n = 0.20 mol

slide33

Magnesium is an active metal that replaces hydrogen from an acid by the following reaction:

Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)

How many g of Mg are needed to produce 5.0 L of H2 at a temperature of 25 oC and a pressure of 745 mmHg?

Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)

? g

5.0 L

0.20 mol

0.20 mol H2

1 mol Mg

24.3 g Mg

____________________________________ = g Mg

4.87

1 mol H2

mol Mg

slide34

Molecular Speeds:

K.E. = ½ mv2

Average kinetic energy of a gas molecule:

 = ½ m2

Where = the rms (root-mean-square) speed of the molecules at each temperature.

From kinetic-molecular theory: At any given temperature the molecules of all gases have the same average kinetic energy.

Which molecules travel

faster, big or little?

slide36

Molecular diffusion and effusion:

Diffusion:

“gas molecules spreading out to fill a room are diffusing.”

Its not easy since an average gas

molecule at room temperature and

pressure will experience about

10 billion collisions per second!

It only travels about 60 nm

between collisions!

slide37

Effusion:

“A Helium filled balloon loses He by effusion.”

escaping molecule

Small hole or pore

slide38

Which molecules will effuse faster from this semiporous

container?

Graham’s Law of effusion: effusion rate is inversely proportional to the square root of its molar mass.

For 2 gases:

slide39

r = rate of effusion

u = root mean speed (~average speed) of molecules

M = molar mass

Compare the rates of effusion of He and N2.

He effuses 2.65 times as fast

as N2.

slide40

He

N2

Which balloon will lose pressure sooner?

slide41

N2

He

Little molecules

(escape more

easily)

Big molecules

Which balloon will lose pressure sooner?

slide42

Real Gases: When do gases become non-ideal?

As they approach the

liquid state, attractions

between molecules

increase and they

become less ideal.

Temperature: low

Pressure: high

van der Waal’s equation is one equation used to treat

non-ideal gases.

a and b are constants found in tables for each gas.

slide43

Which gas would deviate the most from the ideal gas law at room temperature (25oC)?

C3H8 boiling Pt. 231K

PH3 boiling Pt. 188K

SiH4 boiling Pt. 161K

CO boiling Pt. 81K

slide44

Which gas would deviate the most from the ideal gas law at room temperature (25oC)?

300K

298K

C3H8(l) boiling Pt. 231

200K

PH3(l) boiling Pt. 188K

SiH4(l) boiling Pt. 161K

100K

CO(l) boiling Pt. 81K