Theory of Gases -- Their Properties and Behavior 210 Material/LECTURE_NOTES... · Theory of Gases -- Their Properties and Behavior . The study of gas behavior (~1660’s) lead to

Theory of Gases -- Their Properties and Behavior

The study of gas behavior (~1660’s) lead to many of

today’s modern theories of atomic behavior

The underpinnings of gas behavior provides insight into the physical properties of matter

Where in the world does He come from?

He is an inert, unreactive atom.

It floats to the outer reaches of our atmosphere

The “Three” States of Matter

Ordered Compressible Homogeneous Gases NO YES YES Solids YES NO YES/NO

Liquids NO NO YES

Critical Concepts of Gas Behavior

• About 0.1 % volume of gas is taken up by molecules. There is mostly empty space. In a liquid, there is only 70% empty space.

• Little interaction between neighboring gas molecules/atoms.

• Gases exert a pressure via collisions

Does our atmosphere, which is a ball of gas, exert a pressure?

With the assistance of gravity…YES.

A 1 meter diameter swath of air from ground level to outer space weighs in at 10,300 Kg

or 4,680 lbs.

1 atm = 760 mmHg = 101,325 Pascals (Pa)

The IDEAL GAS LAW

P is pressure;

V is volume;

n = moles of gas

R = constant, either 0.082058 (Lit·atm)/(mole·K) or 8.314472 (J/mole·K)

T is Temperature in Kelvin

We can use the IDEAL GAS equation to derive other gas

laws (this is the reverse of how it was ‘put together’)

Boyle’s law: P1V1 = P2V2 Temp & n held constant

Charles’ Law: 2

2

1

1

TV

TV

= Pressure & n held constant

Avagadro’s Law: 2

2

1

1

nV

nV

= Temp & Pressure constant

Gay-Lussac’s law anybody?

PV = nRT

Let’s consider Boyle’s Law, Temperature and n are held constant

So, we know that PV = nRT

In this case, T, R, and n are constants…

…this requires, nRT = Constant

PV = constant

P1V1 (for state 1) = P2V2 (for state 2) = PxVx =

constant

…as in previous slide,

Boyle’s law: P1V1 = P2V2 Temp & n are held constant

… Derivation of Charles’ Law, P & n are held constant

PV = nRT

Re-expressing (putting all constants on one side)

tConsP

nRTV tan==

tConsTV

TV

TV

x

x tan2

2

1

1 ==

Repeat this process for Avagadro’s Law,

hold P,T constant !

and don’t forget Gay-Lussac’s Law…

Using the Ideal Gas Equation

How would the piston move in each of the scenarios below?

What are STP conditions of a gas …or anything for that matter ?

Temp 0ºC = 273 Kelvin Pressure = 1 atm = 760 mmHg

KNOW THIS !

Standard Ambient Temp and Pressure (SATP)

Temp 25ºC = 298 Kelvin

Pressure = 0.98 atm = 1 bar

STP = Standard Temperature and Pressure

What is the volume of an Ideal Gas (1 mole) at STP ? MOLAR VOLUME

Well, lets do the math! PV = nRT

(1 atm) (V) = (1 mole) (0.08206 lit atm / K mole) (273 K)

…solving for Volume (V)

V = atm1K) (273 mole)K / atmlit (0.08206 mole) (1

What causes “NON-IDEALITY” of gases ?

V = 22.414 liters/mole

Not an error

Let’s integrate the gas laws with Stoichiometry

PV = nRT remembering that n = moles of gas

…and knowing that

moles of gas =

mwmass

PV =

RTmw

mass

Combustion of a Hydrocarbon

C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l)

• 15 liter bottle of propane • Pressure = 4.5 atm • Temperature = 25 C = 298 K

Question: What volume of CO2 would be produced if you burned all the propane? Relevant if you are in a well confined fish

house.

PV = nRT

4.5 atm × 15 lit = n (0.08206 lit atm/K mole) (298K)

n = 2.76 moles of C3H8 consumed (in tank)

CO2 produced = 3 × 2.76 moles C3H8 consumed = 8.28 moles CO2

This is only half the problem, remember, we want to know what volume of CO2 is

produced…

…in other words, what volume does 8.28 moles of CO2 gas occupy at 298 K and 1 atm?

PV = nRT

(1 atm) V = (8.28 mole) (0.082 lit atm/mole K) (298 K)

V = 202.5 liters of CO2

Propane burners in confined spaces consume O2, produce CO2, and in many cases CO (deadly) if

incomplete combustion occurs!

Partial Pressures and Dalton’s Law

PTOT = P1 + P2 + P3 + … + Pn at constant Volume and Temperature

Pair = PN2 + PO2 + Par + PCO2 + … + PHe

Integrating the Ideal Gas equation into the

frey…

P1 = n1

VRT

P2 = n2

VRT

Ptotal = (n1 + n2 + n3 + …) VRT

What is a mole fraction?

Mole Fraction (Xn) = mixtureinmolestotalcomponentofmoles

X 1 = Totalnn

nnn 1

21

1

...=

++

P1 = X1 · Ptotal (Dalton’s Law)

The partial pressure exerted by each

component in a gas mixture is equal to the (mole fraction) · (total pressure)

Partial Pressure Problem…

Question: What is the partial pressure and mole fractions of each of the species in the

given volume?

nG = 4 nY = 2 Ptotal = 600 mmHg nR = 6 nTotal = 12

PG = mmHgmmHg 200)600(124

=

PY = mmHgmmHg 100)600(122

=

PR = mmHgmmHg 300)600(126

=

Kinetic-Molecular Theory of Gases…let’s formalize our discussion and look at gases

from a molecular point of view…rather than an empirical point of view.

1. A gas consists of atoms/molecules moving about in a random fashion

2. The volume that the gas particles

actually occupies is significantly smaller than the volume of the gas

3. There is no attraction/repulsion

between neighboring particles of gas 4. Collisions are elastic

i.e. KE is a constant at constant temp

5. Kinetic Energy (KE) ~ Temperature

…from a “somewhat complex derivation” (wait until Physical Chemistry, ~ 4th year)

2

21

23 mv

NRTE

Ak ==

where NA is Avogadro’s number, m is mass,

v is velocity of particle,

…solving for velocity as a function of the other parameters…

=

mRTv 32

mRTv 3

=

What does this mean in a practical sense?

Life is not fair, not everybody has the same energy….

Mean Free Path for He = 2×10-7 meters or about 1000 He diameters

Problem: What is the average speed of N2 diatomics at two different temperatures (-25

°C and 37 °C).

Remember, mRTv 3

=

…at 37 °C:

( )( )( )( ) sm

moleKgKKmoleJv /525

/028.0310/314.83

==

… at – 25 °C

( )( )( )( ) sm

moleKgKKmoleJv /470

/028.0248/314.83

==

Movement of Gases Through Space

Diffusion Effusion

Gases can permeate (effuse) through porous

materials like membranes made of plastic…this affects bottling of such

things as sodas and beers…

Graham’s Law:

mwmovementofRate 1

∝

mw = molecular weight

Graham’s law applied

2

2

1

2

21

21

gasgas

mvmv

=

( ) ( ) 22

12

gasgas mvmv =

( )( ) 1

22

2

21

mm

vv

=

1

2

1

2

2

1

mm

mm

vv

==

Example: What are the relative rates of N2 (mw

= 28 amu) and C2H2 (26 amu) gases?

04.110.529.5

2628

===

… rate 1 = 1.04 × rate 2

Real (non-ideal) Gases

…attraction starts to occur around 10 atomic/molecular diameters

for an ideal gas… PV = nRT

for the non-ideal gas…

a,b are experimentally determined!