GasLaws

DHS ChemistryChapter 14

Do you know…?

What happens when you take a balloon outside on a cold day? _____________________

What happens to the air pressure when you climb a mountain? _____________________

What happens if I stand in the corner of the classroom and spray Febreeze™? _______________

You all have experience with the above situations. In this unit we will explore WHY these things happen.

I. Nature of Gases

A. Kinetic-Molecular Theory

Kinetic Molecular Theory

The kinetic-molecular theory describes the behavior (properties) of gases in terms of particles in motion.

Assumptions of the kinetic-molecular theory:

1. Gases consist of small particles that are separated by empty space. Because gas particles are so far apart, they exhibit no attractive or repulsive forces on one another.

Assumptions of the kinetic-molecular theory:

2. The actual volume of a gas molecule is so small that it is insignificant when compared to the volume of the container it is in.

Assumptions of the kinetic-molecular theory:

3. Gas particles are in constant, random, motion. Collisions between gas particles are elastic (there is no loss of kinetic energy).

Assumptions of the kinetic-molecular theory:

4. The kinetic energy of a gas particle is determined by mass and velocity.

KE = _1 mv2

2 Temperature is a measure of the

average kinetic energy of the particles in a substance.

All gases at a given temperature will have the same average kinetic energy. Temperature is directly proportional to kinetic energy. (In other words, if gas particles have the same temperature, they also have the same energy.)

© prentice hall

C. Deviations from Ideal Gases

•an ideal gas is one whose particles take up no space and have no attractive forces

•no gas is truly ideal•most gases occupy space and

exert attractive forces on one another

C. Deviations from Ideal Gases

•Gases are LEAST ideal at high pressures and low temperatures

•gases get closer to exhibiting ideal behavior when there is little attraction between gas particles.

B. Behavior of Gases

1. Low densitygases have low density because there is a lot of empty space between gas particles.

2. Compression and expansion

gases can be easily compressed (squeezed into a smaller volume) because there is a lot of empty space between gas particles.

3. Diffusion and effusion

Diffusion occurs when two or more gases mix with each other without agitation. Diffusion occurs because of the constant, kinetic motion of gas particles.

3. Diffusion and effusion

The gas particles will move from an area of high concentration to an area of low concentration.

Example:

Spraying air freshener in the classroom

3. Diffusion and effusion

The rate of diffusion depends mainly on mass of particles. (a lighter gas will diffuse faster than a heavier gas)

effusion is the process by which a gas escapes through a tiny opening

T. Graham found that there is an inverse relationship between effusion rates and the square root of the mass.

effusion

(What do you think this equation would look like?)

In other words, the higher the molar mass, the slower the effusion.

ex. Which gas will diffuse or effuse faster? NH3 or Cl2

NH3 has a smaller molar mass, therefore travels

faster

Practice:

Which will diffuse and/or effuse faster?

a. Cl2 or H2

b. He or H2

c. F2 or O2

d. Ne or O2

II. Measuring Gases

A. Temperature

Temperature is a measure of the average kinetic energy of the particles in a material. There will be a range of kinetic energies.

The kinetic energy of a gas is directly proportional to its temperature in Kelvin.

K = °C + 273

The higher the temperature in Kelvin, the higher the energy.

Important Temperatures

°C K

Absolute Zero* -273 0

Freezing Pt of water 0 273

Standard Temp 0 273

Room Temperature (RT)

22 295

Boiling Pt of water 100 373

Absolute Zero is the point at which all motion of particles

completely stops.

Note: Kelvin temperature is directly proportional to the average kinetic energy of the particles of the substance.

Practice:

Convert…1) 33 °C to K

2) 450 K to °C

K = °C + 273

K = 33 °C + 273K = 306K

450K = °C + 273°C = 177 °C

B. Pressure

1. DefinitionPressure is defined as the amount of force applied per unit area.

(why does a person sink into the snow when walking with tennis shoes and not with snow shoes?)

The pressure of a gas is defined as the amount of force exerted by the particles in a gas as the hit the sides of the container.

More collisions = higher pressure

The more often the particles hit the sides of the container, the

higher the pressure.

2. Units of PressureUnits for pressure:

•atmospheres (atm): 1 atm is defined as the average atmospheric pressure at sea level

•millimeters of mercury (mm Hg): older unit

1 atm = 760 mm Hg = 760 torr

•Pascals (Pa): metric unit for pressure 1 atm = 101.3 kPa

•pounds per square inch (psi): tells how much pressure above 1 atm

(1 atm = 14.7 psi)1 atm = 760 mmHg = 101.3 kPa

= 760 torr = 14.7psi

Practice(dimensional analysis helps)

6.2 atm = ______ mmHg

6.2 atm 1 atm

760 mmHg

476.1 kPa 101.3kPa

1 atm

_____ atm = 476.1kPa

= 4712mmHg

= 4.70 atm

3. MeasuringA mercury barometer was an

early device used to measure atmospheric pressure. It consisted of a straight glass tube filled with mercury and closed at one end. It was placed with the open end down in a dish of mercury and the height of mercury that rose in the column was measured.

3. Measuring

At sea level the height of mercury in a mercury barometer is typically 760 mmHg or 1 atm

Have you climbed up a mountain? Did you

notice anything different with regards

to your breathing?

4. Atmospheric Pressure

We do not live in a vacuum. We are surrounded by an atmosphere (air) and like all matter, air has mass . The mass of the air is dependent on the altitude.

4. Atmospheric Pressure

*The higher you are (increase in altitude)

the less air pressing on you (decrease in

atmospheric pressure)*

Altitude & atmospheric pressure are Inversely proportional

Atmospheric pressure is the result of the molecules that make up air bouncing around and applying force against any surface. Atmospheric pressure will change, depending on altitude. As you go up (increase altitude), the atmospheric pressure decreases.

4. Atmospheric Pressure

The effects that we notice are the result of the balances and imbalances between atmospheric pressure in different areas or inside and outside of objects.

© physicalgeography.net

Ex. 1: an inflated balloon

Pinside = Patm

Ex. 2: a popping balloon

Pinside > Patm

Ex. 3 shrinking balloons

Pinside < Patm

5. Factors Affecting Gas Pressure

There are three main factors that affect the

pressure of a gas: amount of gas (mass or

moles), volume of a gas, and temperature of a gas

1). Amount of a Gas

As the amount of gas (moles of gas) in a container increases, pressure increases.

All other variables are constant

Low pressure

Medium pressure

High pressure

Pressure is directly related to

the number of gas particles in a

volume

© prentice hall

2. Volume

As the volume of a container increases, the pressure decreases.

All other variables are constant

high pressure

low pressure

Volume affects Pressure

© prentice hall

3. Temperature

As the temperature of a gas increases, the pressure increases.

All other variables are constant

Low temperature

High temperature

Temperature

C. Standard ValuesSTP stands for standard

temperature and pressure and is defined as 0°C and 1 atm.

SMV stands for standard molar volume and is defined as 22.4L at STP.

STP = 0.00°C & 1.00 atmSMV = 22.4 L at STP

1 mole gas = 22.4 L at STP

D. Proportions

Two quantities are directly proportional if dividing one by the other gives a constant value.

EX: as temperature increases, pressure increases

When a direct proportion is graphed, it would be a straight line with a positive slope.

EX:

Two quantities are inversely proportional to each other if their product is constant.

EX: as volume increases, pressure decreases

A graph of an inverse proportion would produce a curve with a negative slope.

EX:

III. Gas Laws

Gas laws include formulas that help calculate

pressure, volume, and temperature changes.

A. Combined Gas Law

All 3 gas laws can be combined into the combined gas law.

Any of the three laws can be obtained from the combined gas law by holding one quantity constant.

P1V1 = P2V2

T1 T2

By the way, temperature

must be in Kelvin when used in the formulas

Combined Gas Law

P1V1 P2V2

T1

= T1T1

T2

T2 T2

B. Boyle’s Law

Boyle’s Law states that for a given mass of gas at constant temperature, the volume varies inversely with pressure. (As volume increases pressure decreases.)

P1V1 = P2V2

(constant temperature)

Boyle’s Law

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Boyle’s Law

C. Charles’ Law

Charles’ Law states that the volume of a fixed mass of gas is directly proportional to its temperature in Kelvin (at constant pressure)

Charles’ Law

V1 = V2

T1 T2

(constant pressure)

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D. Gay-Lussac’s Law

Gay-Lussac’s Law states that the pressure of a fixed mass of gas is directly proportional to its temperature in Kelvin. (at constant volume)

Gay-Lussac’s Law

P1 = P2

T1 T2

(constant volume)

Low temperature

High temperature

Combined Gas Law

P1V1 P2V2

T1

=T2

Boyle’s

Gay-Lussac

Charles

Make a Mnemonic Device or Acronym

Boyles – T constantCharles – P constant

Gay-Lussac – V constant

“Boy That Charles Picks Goofy Videos”

EX 1 : Brad Pitt is in a hot air balloon that contains 44.0 L of helium at 114 kPa. What is the volume when the balloon rises to an altitude where the

pressure is only 37.0 kPa? (Assume constant n and T.)

44.0L114kPa

? L37.0 kPa

1 2PVT

44.0L

114kPa 37.0kPa

V2

EX 1 : Brad Pitt is in a hot air balloon that contains 44.0 L of helium at 114kPa. What is the volume when the balloon rises to an altitude where the pressure is

only 37.0 kPa? (Assume constant n and T.)

P1V1 = P2V2

(114kPA)(44.0L)= (37.0kPA)

V2

(37.0kPA)

(37.0kPA) V2

=136L

T1 T2

1 2PV44.0L

114kPa V2

37.0kPa

Double check your answers by:

-plugging it back into the equation

-ask yourself, “does it make sense?”

-check if the units cancelled out

EX 2: The gas left in an aerosol can is at a pressure of 0.75 atm at 25°C. If this container is thrown onto a fire, what is the pressure of the

gas when its temperature reaches the fire temperature of 780°C?

1 2PVT25 °C

0.75 atm P2

780 °C

Temp in K always!!

°C + 273

298 K 1053 K

Doesn’t mention volume… but think about it. When does a can change volume?

P1 = P2

T1 T2

0.75atm = P2_____ (25+273) (780 + 273)

0.75atm = P2 (298K) (1053K)(you can use cross multiplication)

2.65atm = P2

EX 3: The volume of a gas-filled balloon is 30.0 L at 40˚C and 153 kPa. What volume will the

balloon have at STP?

P1V1 = P2V2

T1 T2 *STP = 0ºC (273K), 1atm

1 2

153 kPa P 1 atm

30.0 L V V2

40 °C T 273 K Remember: 1 atm = 101 kPa

313 K

101.3 kPa

EX 3: The volume of a gas-filled balloon is 30.0 L at 40˚C and 153 kPa. What volume will the

balloon have at STP?

P1V1 = P2V2

T1 T2

(153kPa)(30.0L) = (101.3kPa) V2

(313K) (273K)

V2 = 39.5 L

EX 3: The volume of a gas-filled balloon is 30.0 L at 40˚C and 153 kPa. What volume will the

balloon have at STP?

P1V1 = P2V2

T1 T2 *STP = 0ºC, 1atm

P1V1T2 = P2V2T1

Tip: Change temp. to Kelvin

V2 =(153kPa)(30.0L)(273K) =39.6 L (101kPa)(313K)

EX 1 : A high altitude balloon contains 30.0 L of helium at 103 kPa. What is the volume when the

balloon rises to an altitude where the pressure is only 25.0 kPa?

(Assume constant m and T)

P1 = 103kPa

V1 = 30.0L

P2 = 25.0kPa

P1V1 =P2V2

(103kPA)(30.0L)= (25.0kPA)

V2

(25.0kPA)

(25.0kPA) V2

=123.6L

EX 2: The gas left in an aerosol can is at a pressure of 1 atm at 25°C. If this can is thrown

onto a fire, what is the pressure of the gas when its

temperature reaches the fire temperature of 928°C? T2

T1

P1

P1 = P2

T1 T2

1atm = P2_____ (25+273) (928 + 273)

1atm = P2 (298K) (1201K)(you can use cross multiplication)

4.03atm = P2

EX 3: The volume of a gas-filled balloon is 30.0 L at 40ºC and 153 kPa. What volume will the

balloon have at STP?

P1V1 = P2V2

T1 T2 *STP = 0ºC, 1atm

P1V1T2 = P2T1V2

Tip: Change temp. to Kelvin

V2 =(153kPa)(30.0L)(273K) =39.6 L (101kPa)(313K)

Practice 1. The pressure on a 2.50 L cylinder of

anesthetic gas changes from 105 kPa to 40.5 kPa. What will the new volume be if the temperature of the gas remains constant?

6.48 L2. A gas at 155 kPa and 25oC occupies a

container with an initial volume of 1.00 L. By changing the volume, the pressure of the gas increases to 605 kPa as the temperature is also raised to 125oC. What is the new volume?

0.342 L3. A gas with a volume of 3.00 x 102 mL at

150.0oC is heated until its volume is 6.00 x 102 mL (at constant pressure). To what temperature was this gas heated?

846 K

The pressure on a 2.50 L cylinder of anesthetic gas changes from 105 kPa to 40.5 kPa. What will the new volume be if the temperature of the gas remains constant?

start with combined: = P1V1 = P2V2

T1 T2

hint: temperature remains constant

P1V1 = P2V2

T1 T2

final formula: P1V1 = P2V2 (boyle’s law)

V1

P2

P1

V2

P2 V1 P2The pressure on a 2.50 L cylinder of

anesthetic gas changes from 105 kPa to 40.5 kPa. What will the new volume be if the temperature of the gas remains constant?

P1V1 = P2V2

(105kPa)(2.50L) = (40.5kPa) V2

(105kPa)(2.50L) = (40.5kPa) V2

(40.5kPa) (40.5kPa)

6.48L = V2

2. A gas at 155 kPa and 25ºC occupies a container with an initial volume of 1.00 L. By changing the volume, the pressure of the gas increases to 605 kPa as the temperature is also raised to 125ºC. What is the new volume?

start with combined: = P1V1 = P2V2

T1 T2

change all Celsius to Kelvin T1 = 25 + 273 = 293KT2 = 125 + 273 = 398K

(155kPa)(1.00L) = (605kPa)V2

293K 398K

P1 T1 V1

P2

T2V2

3. A gas with a volume of 3.00 x 102 mL at 150.0ºC is heated until its volume is 6.00 x 102 mL (at constant pressure). To what temperature was this gas heated?

start with combined: P1V1 = P2V2

T1 T2

V1 = V2 3.00 x 102 mL = 6.00 x 102 mL

T1 T2 423K T2

846 K or 573 °C

E. Ideal Gas Law

E. Ideal Gas Law

The ideal gas law is a formula used when gases are under certain conditions. The particles in an ideal gas are far enough apart that they don’t have a lot of opportunity for attractive or repulsive forces between each other to form.

PV = nRTP = pressure (depends on R)V = volume (L)T = temperature (K)n = moles (can be converted from

grams)R = Ideal Gas Constant0.0821 L atm OR 8.31 L kPa OR 62.4 L

mmHg mol K mol K mol K

Example 1: Calculate the amount of gas, in moles, contained in a 3.00 L container

at 300 K with a pressure of 1.50 atm.

PV = nRT

(1.50 atm) (3.00 L) = (n)0.0821 L atm mol K (300 K)

0.183 moles = n

Example 2: What is the pressure, in atmospheres, of a 0.108 mole sample of helium gas at a temperature of 20C if its

volume is 0.505 L?

PV = nRT

(P atm) (0.505 L) = (0.108 mol)0.0821 L atm mol K (293 K)

5.14 atm=P

F. Dalton’s Law of Partial Pressures

Dalton’s Law of Partial Pressures says that the total pressure of a mixture of gases is equal to the sum of the partial pressures of all the gases in the mixture.

The partial pressure of the gas is determined by the number of moles of gas, the size of the container, and the temperature of the mixture. It is not determined by the identity of the gas.

Partial pressures can only be added if they have the same unit of pressure PTOT = P1 + P2 + P3 + …….

EX: A gas mixture containing oxygen, nitrogen, and carbon dioxide has a total

pressure of 257 mm Hg. If PO2 = 52 mm Hg

and PN2 = 171 mm Hg, what is PCO2

?PTOT = PCO2

+ PO2 + PN2

257 mmHg = PCO2 + 52 mmHg + 171 mmHg

PCO2 = 34 mm Hg

EX 2: A mixture of oxygen, carbon dioxide, and nitrogen has a total

pressure of 0.97 atm. What is the partial pressure of oxygen, if the

partial pressure of carbon dioxide is 0.70 atm and the partial pressure of

nitrogen is 0.12 atm?PTOT = PO2

+ PCO2 + PN2

0.97 atm = PO2 + 0.70 atm + 0.12

atm

PO2 = 0.15 atm

EX 3: In the lab Jill mixed 3 noble gases together. The resulting pressure was 930 torr. Helium had a partial pressure of 0.42 atm and

Neon had a partial pressure of 0.33 atm. What is the partial pressure of Argon in atm?

PTOT = PHe + PNe + PAr

1.22 atm = 0.42 atm + 0.33 atm + PAr

PAr = 0.470 atm

930 torr1

x760 torr1 atm = 1.22 atm

Toxic levels of oxygen: PO2 > 1.4atm

The fractional contribution to pressure exerted by each gas in a mixture does not change as temperature, pressure, or volume changes.

EX: At sea level

PTOT = 101.32 kPa

PO2 = 21.22 kPa

PO2 21.22 kPa = part

PTOT 101.32 kPa whole

or PO2 20.9% of total

On top of Mt. Everest

PTOT = 33.73 kPa

PO2 = ?????

PO2 = 21.22 kPa = ??? PTOT 101.32 kPa 33.73 kPa

or PO2 20.9%

of the PTOT is 7.06 kPa is

7.06 kPa (PO2)

On top of Mt. Everest:

PTOT = 33.73 kPa

PO2 = 20.9% of total

pressure PO2

= ?

PO2 = 7.06 kPa Most humans need PO2 > 10.67 kPa for respiration.

P = 4 atm

1. Oxygen and Hydrogen are mixed in a balloon. Hydrogen has a partial pressure of 0.2 atm and oxygen has a partial pressure of 680 torr. What is the total pressure in atm?

PTOT = PO2 + PH2

PO2 = 680 torr = 0.8947 atm PTOT = 0.8947 atm + 0.2 atm

PTOT =1.09 atm

Extra Practice

Practice

The total pressure of a mixture of gases is 257.4 kPa. The partial pressure of gas A in that mixture is 46.5 kPa. If the pressure of the mixture is reduced to 100.2 kPa, what is the new partial pressure of gas A?

PracticeThree gases, A, B, and C, are contained in a tank. What is the total gas pressure in atmospheres if PA = 3.5 atm, PB = 550 mmHg, and PC = 45.6 kPa?

PTOT = PA + PB + PC

Ptot = 3.5 atm + 0.724 atm + 0.450 atm

PTOT = 28.8 atm

PracticeThe total pressure of a mixture of 4 gases is 150.6 kPa. What is the pressure of the 4th gas if P1 = 2.5 kPa, P2 = 57.0 kPa, and P3 = 62.3 kPa?PTOT = P1 + P2 + P3 + P4

150.6 kPa = 2.5 kPa + 57 kPa + 62.3 kPa + P4

P4 = 28.8 kPa

Practice I

A sample of gas occupies 12.0 L under a pressure of 1.20 atm. What would the volume be if the pressure were increased to 2.2 atm?

Practice II

A sample of nitrogen occupies 117 mL at 100 ˚C. At what temperature (in ˚C) would the sample occupy 234 mL if the pressure did not change?

Practice III

A gas in an expandable box that is 2 cm X 2 cm X 2 cm has a pressure of 6.58 kPa at 539 K. What will the volume be at STP?

Practice IV

1. Which law is used to determine the total pressure in a mixture of gases?

2. Pressure and temperature are directly proportional when volume is held constant.

3. Pressure and volume are inversely proportional when temperature is held constant.

Practice V

4. Temperature and volume are directly proportional when pressure is held constant.

5. This gas law uses temperature, pressure, and volume to determine the behavior of gases.

True of False?

1. Reducing the quantity of a gas in a container increases the pressure.

2. Pressure increases if a fixed amount of gas is cooled while the volume is held constant.

3. Halving the number of particles in a given volume of gas decreases the pressure by one-half if the temperature is kept constant.