Section 3.7—Gas Behavior How does the behavior of gases affect airbags?

What is PRESSURE?

Force of gas particles running into a surface.

Pressure is measured by a Barometer

If pressure is molecular collisions with the container…

As # of moles increase, pressure increases Think about blowing up a balloon!

Pressure and Moles (# of Molecules)

As number of molecules increases, there will be more molecules to collide with the wall

• Collisions between molecules and the wall increase

Pressure increases

# of Gas Particles vs. Pressure

Pressure & Volume

If pressure is molecular collisions with the container…

As volume increases, pressure decreases. Think about how your lungs work! http://www.youtube.com/watch?v=q6-oyxnkZC0

As volume increases, molecules can travel farther before hitting the wall• Collisions between molecules & the wall decrease

Pressure decreases

What is “Temperature”?

Temperature – measure of the average kinetic energy of the molecules

Energy due to motion(Related to how fast the molecules are moving)

As temperature increases, Average Kinetic Energy Increases and Molecular motion increases

Pressure and Temperature

If temperature is related to molecular motion…and pressure is molecular collisions with the container…

As temperature increases, pressure increases

As temperature increases, molecular motion increases

• Collisions between molecules & the wall increase

Pressure increases

Volume and Temperature

If temperature is related to molecular motion…and volume is the amount of space the gas occupies…

As temperature increases, volume increases Think of liquid nitrogen and the balloon. http://www.youtube.com/watch?v=QEpxrGWep4E

As temperature increases, molecular motion increases• molecules will move farther away from each other Volume increases

Pressure In Versus Out

Example: A bag of chips is bagged at sea level. What happens if the bag is then brought up to the top of a mountain.

A container will expand or contract until the pressure inside equals atmospheric pressure outside

The internal pressure is higher than the external pressure.

The bag will expand in order to reduce the internal pressure.

The internal pressure of the bag at low altitude is high

At high altitude there is lower pressureHigher pressure

Lowerpressure

Lower pressure

CanExplodes!

When Expansion Isn’t Possible

Example: An aerosol can is left in a car trunk in the summer. What happens?

Rigid containers cannot expand

The internal pressure is higher than the external pressure.

The can is rigid—it cannot expand, it explodes!

The temperature inside the can begins to rise.

As temperature increases, pressure increases.

Higher pressure

Lowerpressure

Air Pressure Crushing Cans

http://www.csun.edu/scied/4-discrpeant-event/the_can_crush/index.htm

http://www.youtube.com/watch?v=Zz95_VvTxZM

Another cool videohttp://www.youtube.com/watch?v=JsoE4F2Pb20

Air Pressure Crushing “Cans”

Kinetic Molecular Theory(KMT): explains gas behavior based upon

the motion of molecules based on an ideal gas

IDEAL gases are IMAGINARY gases that follow the assumptions of the KMT

1

Assumptions of the KMTAll gases are made of atoms or molecules that are in constant, rapid, random motion

Gas particles are not attracted nor repelled from one another ***All gas particle collisions are perfectly elastic (no kinetic energy is lost to other forms)

The volume of gas particles is so small compared to the space between the particles, that the volume of the particle itself is insignificant***

2

3

4

5

The temperature of a gas is proportional to the average kinetic energy of the particles

So what is a “REAL” gas?Real gases, (like nitrogen), will eventually condense into a liquid when the temperature gets too low or the pressure gets too high BECAUSE:

Assumption #3

Assumption #5

Gas particles do have attractions and repulsions towards one another

Gas particles do take up space

Real Gases Deviate from Ideal Gas Behavior when at high pressure

The gas molecules are compressed making the volume they take up more significant than if they were spread out

Real Gases Deviate from Ideal Gas Behavior when at low temperature.

The lower kinetic energy causes the molecules to move slower and ATTRACTIVE FORCES that really exist start to take effect---------------------------

Polar gases (HCl) deviate more than nonpolar gases (He or H2

At Lower Temperature

Gas Movement: Effusion vs Diffusion

Effusion –gas escapes from a tiny hole in the container

Effusion is why balloons deflate over time!

Diffusion –gas moves across a space from high to low concentration

Diffusion is the reason we can smell perfume across the room

Effusion, Diffusion & Particle Mass

How are particle size (mass) and these concepts related?

As mass of the particles increases, rate of effusion and diffusion is lowered.

As particle size (mass) increases, the particles move slower it takes them more time to find the hole or to go across the room

Rate of Diffusion & Particle Mass

Watch as larger particles take longer to get to your nose

H2

CO2

Section 3.8-3.9 —Gas Laws

How can we calculate Pressure, Volume and Temperature of our airbag?

Pressure Units

Several units are used when describing pressure

Unit Symbol

atmospheres atm

Pascals, kiloPascals

millimeters of mercury

pounds per square inch

Pa, kPa

mm Hg

psi

1 atm = 101300 Pa = 101.3 kPa = 760 mm Hg = 14.7 psi

Conversions Between Different Pressure Units

1 atm = 760 mmHg = 101.3 kPa

Examples

1.Convert 654 mm Hg to atm

2.Convert 879 mm Hg to kPa

3.Convert 15.6 atm to kPa

654 mmHg x 1atm = 760 mmHg

.861 atm

879 mmHg x 101.3 Kpa = 760 mmHg

1.16 Kpa

15.6 atm x 101.3 Kpa = 1atm

1580 Kpa

Temperature Unit used in Gas Laws

Kelvin (K)– temperature scale with an absolute zero

Temperatures cannot fall below an absolute zero

KC 273

Examples1.Convert 15.6 °C into K

2. Convert 234 K into °C

15.6 + 273 = K 288.6 289 K

°C + 273 = 234 -39 °C

Standard Temperature & Pressure (STP)

the conditions of:1 atm (or the equivalent in another unit) 0°C (273 K)

Problems often use “STP” to indicate quantities…don’t forget this “hidden” information when making your list!

GAS LAWS: “Before” and “After” This section has 5 gas laws which have “before” and “after” conditions.

For example:2

2

1

1

n

P

n

P

1= initial amount2= final amount

P= PressureV= VolumeT=Temperaturen= moles(molecules)

Boyle’s Law

Pressure Increases as Volume Decreases

Boyles’ LawVolume & Presssure are INVERSELY proportional when temperature and moles are held constant

P = pressureV = volume2211 VPVP

The two pressure units must match and the two volume units must match!

Example: A gas sample is 1.05 atm when at 2.5 L. What volume is it if the pressure is changed to 0.980 atm?

Boyles’ Law 2211 VPVP ***The two pressure units must match & the two volume units must match!

Example: A gas sample is 1.05 atm when 2.5 L. What volume is it if the pressure is changed to 0.980 atm?

P1 = 1.05 atm

V1 = 2.5 L

P2 = 0.980 atm

V2 = ? L

V2 = 2.7 L

2980.05.205.1 VatmLatm

2980.0

5.205.1V

atm

Latm

Boyles Law: Graph

Charles’ Law

Volume Increases as Temperature Increases

Charles’ LawVolume & Temperature are DIRECTLY proportional when pressure and moles are held constant.

V = VolumeT = Temperature

2

2

1

1

T

V

T

V

The two volume units must match & temperature must be in Kelvin!

Example: What is the final volume if a 10.5 L sample of gas is changed from 25C to 50C?

V1 = 10.5 L

T1 = 25C

V2 = ? L

T2 = 50C

Temperature needs to be in Kelvin!

25C + 273 = 298 K

50C + 273 = 323 K

Charles’ Law2

2

1

1

T

V

T

V

***The two volume units must match & temperature must be in Kelvin!

Example: What is the final volume if a 10.5 L sample of gas is changed from 25C to 50C?

V1 = 10.5 L

T1 = 25C

V2 = ? L

T2 = 50C

V2 = 11.4 L

= 298 K

= 323 K

K

V

K

L

323298

5.10 2

2298

5.10323V

K

LK

Charles Law: Graph

Gay-Lussac’s Law

Temperature decreases as Pressure decreases

Gay-Lussac’s LawPressure & temperature are DIRECTLY proportional when moles and volume are held constant

P = PressureT = Temperature

The two pressure units must match and temperature must be in Kelvin!

Example: A sample of hydrogen gas at 47C exerts a pressure of .329 atm. The gas is heated to 77C at constant volume and moles. What will the new pressure be?

P1 = .329 atm

T1 = 47C

P2 = ? atm

T2 = 77C

Temperature needs to be in Kelvin!

47C + 273 = 320 K

77C + 273 = 350 K

Gay-Lussac’ Law

Example: A sample of hydrogen gas at 47C exerts a pressure of .329 atm. The gas is heated to 77C at constant volume and moles. What will the new pressure be?

P1 = .329 atm

T1 = 47C

P2 = ? atm

T2 = 77C

P2 = .360 atm

= 320 K

= 350 K

Gay Lussac Law: Graph

Avogadro’s LawMoles and Volume are directly proportional when temp. & pressure are held constant

V = Volumen = # of moles of gas

2

2

1

1

n

V

n

V

Example: A sample with 0.15 moles of gas has a volume of 2.5 L. What is the volume if the sample is increased to 0.55 moles?

The two volume units must match!

Avogadro’s Law2

2

1

1

n

V

n

V

Example: A sample with 0.15 moles of gas has a volume of 2.5 L. What is the volume if the sample is increased to 0.55 moles?

The two volume units must match!

n1 = 0.15 moles

V1 = 2.5 L

n2 = 0.55 moles

V2 = ? L

mole

V

mole

L

55.015.0

5.2 2

215.0

5.255.0V

mole

Lmole

V2 = 9.2 L

Combined Gas Law

P = PressureV = Volumen = # of molesT = Temperature22

22

11

11

Tn

VP

Tn

VP

Each “pair” of units must match and temperature must be in Kelvin!

Example: What is the final volume if a 15.5 L sample of gas at 755 mmHg and 298 K is changed to STP?

Combined Gas Law

P = PressureV = VolumeT = Temperature

Moles is not mentioned so remove it from equation!

Example: What is the final volume if a 15.5 L sample of gas at 755 mmHg and 298K is changed to STP?

P1 = 755 mmHg

V1 = 15.5 L

T1 = 298 K

P2 = 760mmHg

V2 = ? L

T2 = 273 K V2 = 14.1 L

STP is standard temperature (273 K) and pressure (1 atm)

22

22

11

11

Tn

VP

Tn

VP

22

12

11

11

Tn

VP

Tn

VP

The combined gas law can be used for all “before” and “after” gas law problems!

For example, if volume is held constant, then

and the combined gas law becomes:

21 VV

22

2

11

1

Tn

P

Tn

P

Why you really only need 1 of these

Watch as variables are held constant and the combined gas law “becomes” the other 3 laws

22

22

11

11

Tn

VP

Tn

VPHold pressure and

temperature constantAvogadro’s Law

22

22

11

11

Tn

VP

Tn

VPHold moles and

temperature constantBoyles’ Law

22

22

11

11

Tn

VP

Tn

VPHold pressure and moles

constantCharles’ Law

Transforming the Combined Law

Dalton’s Law

Dalton’s Law

• Each gas in a mixture exerts its own pressure called a partial pressure = P1, P2….

• Total Pressure = PT

Example: A gas mixture is made up of oxygen(2.3 atm) and nitrogen(1.7 atm) gases. What is the total pressure?

....321 PPPPT

PT = 2.3 atm + 1.7 atm

4.0 atm

2KClO3 (s) 2KCl (s) + 3O2 (g)

PT = PO + PH O2 2

Modified Dalton’s Law

When a gas is Collected over water, the total pressure of the mixture collectedis a combination of water vapor and the gas you are collecting!

Modified Dalton’s Law

Example: What is the pressure of the water vapor if the total pressure of the flask is 17.5 atm and the pressure of the oxygen gas is 16.1 atm?

watergas PPPT

17.5 = 16.1 atm + PH2O

1.4 atm

The Ideal Gas Law (an“AT NOW”equation)The volume of a gas varies directly with the number of moles and its Kelvin temperature

P = PressureV = Volume n = molesR = Gas Law ConstantT = Temperature

nRTPV

There are three possibilities for “R”!

Kmole

atmL

*

*0821.0

Choose the one with units that match your pressure units!

Volume must be in Liters when using “R” to allow the unit to cancel!

The Ideal Gas Law nRTPV Example: A sample with 0.55 moles of gas is at 105.7 kPa

and 27°C. What volume does it occupy?

The Ideal Gas Law

Example: A sample with 0.55 moles of gas is at 105.7 kPa and 27°C. What volume does it occupy?

n = 0.55 moles

P = 105.7 kPa

T = 27°C + 273 = 300 K

V = ?

R = 8.31 L kPa / mole K

)300(**31.8)55.0()7.105( KkmolekPaLmoleVkPa

V2 = 13 L

The Ideal Gas Law does not compare situations—it describes a gas in one situation.

)7.105(

)300(**31.8)55.0(

kPa

KkmolekPaLmole

V

nRTPV

Chosen to match the kPa in the “P” above

The Ideal Gas LawExample 2:

What mass of hydrogen gas in grams is contained in a 10.0 L tank at 27°C and 3.50 atm of pressure?

n = ?

P = 3.50 atm

T = 27°C + 273 = 300 K

V = 10.0 L

R = .0821 L atm /mole K

)300(**0821.)0.10()50.3( KkmoleatmLnLatm

)*

*0821(.)300(

0.10)50.3(

kmoleatmLxK

Latmn

nRTPV

Chosen to match the atm in the “P” above

n = 1.42 mol 1.42 mol x 2.02 g = 2.87 g 1 mol