chapter 12 the behavior of gases 12.3 the gas laws

Chapter 12The Behavior of gases

12.3The Gas Laws

All gases

• Are less dense when warmer• Are less dense when under less pressure• Are less dense when there are less particles in

a volume• Obey laws, within limits, which allow us to

predict their behavior under most conditions

Things you will learn

• Understand Boyle’s Law• Understand Charles’s Law• Understand Gay-Lussac’s Law• Understand and be able to solve problems

using the combined gas laws

Boyle’s LawThe pressure-volume relationship of gases

Pressure and volume

Pressure and volume have an inverse relationship

gas-properties_en.jar

Boyles Law essentials

• The product of a pressure and volume of any two sets of conditions at a given temperature is constant:

• P1 x V1 = P2 x V2

• 100 kPa @ 1 L = 50 kPa @ 2 L = 200 kPa @ .5L

• The two conditions are inversely proportional• P~1/V

A high altitude balloon contains 30 L of He gas at 103 kPa. What is the volume when it gets to an altitude where the pressure is 25

kPa?


kPa?

• Knowns:– V1 = 30 L– P1 = 103 kPa– P2 = 25 kPa

• Unknown:– V2


kPa?

• Our equation is V1 x P1 = V2 x P2• 30L x 103 kPa = v2 x 25 kPa• 30 L x 103 kPa 25 kPa • V2 = 124 L

V2

Charles’s LawThe temperature-volume relationship of gases

Charles’s Law esentials

• An increase in the temperature of a gas yields an increase in the volume of a gas

• V1/T1 = V2/T2

• 1 L @ 300K = 2 L @ 600k

• The two conditions are directly proportional• V~T remember all temperature measurements are in kelvins

A balloon inflated in a room at temperature of 24°C has a volume of 4 L. The balloon is then heated to a temperature of 58°C. What is

the new volume if the temperature remains constant?

A baloon inflated in a room at temperature of 24°C has a volume of 4 L. The balloon is then heated to a temperature of 58°C. What is


• Knowns:– V1 = 4 L– T1 = 24°C– T2 = 58°C

• Unknown:– V2

A baloon inflated in a room at temperature of 24°C has a volume of 4 L. The balloon is then heated to a temperature of 58°C. What is


• Our equation is V1/T1= V2 /T2• 4L /24°C = xL/58°C• 58C x 4L / 24°C = xLwait, we haven’t converted to Kelvins!

331K x 4L = 4.46 L 297K

Gay-Lussac The temperature-volume relationship of gases

Gay-Lussac’s Law

• The pressure of a gas is directly proportional to the temperature of a gas (in Kelvins) if the volume remains constant

• Because these relations are directly related, they obey the formula:

• P1/T1 = P2/T2

Sample problemThe gas in an aerosol can may be 103 kPa, meaning it won’t squirt, at 25°C, but if thrown in a fire, the

pressure could be quite dangerous. How high is the pressure if the fire

is 928°C ?

Sample problemThe pressure in a tire is 198 kPa at the start of a trip at 27°C. At the

end of the trip, it is 225 kPa. What is the internal temperature of the

tire

Combining the gas laws

• Boyles’ Law is• Charles’s Law is• Gay-Lussac’s Law is

Combining the gas laws

• Boyles’ Law is P1 x V1 = P2 x V2

• Charles’s Law is V1/T1 = V2/T2

• Gay-Lussac’s Law is P1/T1 = P2/T2

The combined gas law

P1 x V1

T1

P2 x V2

T2

If you hold the temperature constant, you have Boyle’s LawIf you hold volume constant, you have Gay-Lussac’s LawIf you hold the pressure constant, you have Charles’s Law

Sometimes you are not able to hold any of the variables constant

• The volume of a gas filled balloon is 30L and 153 kPa. What volume will the balloon be at STP?


• Knowns:V1=30 LT1=40CT2= 273K (standard temp)P1= 153 kPaP2=101.3 kPa (standard pressure


• Knowns:V1=30 LT1=40CT2= 273K (standard temp)P1= 153 kPaP2=101.3 kPa (standard pressure

Change all temps to KelvinsIsolate V2 and solve

Practice problems

• A gas at 155 kPa and 25C occupies a container with an initial volume of 1 L. By changing the volume, the pressure of the gas increases to 605 kPa as the temperature is raised to 125C. What is the new volume?

Practice problems

• A 5 L air sample at a temperature of -50C has a pressure of 107 kPa. What will be the new pressure if the temperature is raised to 102C and the volume expands to 7 L?

Ideal gas law

• Up to this point, we have left out another thing that can change pressure, volume and temperature

• We have used the combined gas law to find changes in a system when we change conditions such as temperature, pressure or volume.

P1 x V1

T1

P2 x V2

T2

• But the amount of gas- the number of moles- can change the pressure and volume also

• It makes sense that the volume and pressure in a container must be proportional to the number of moles

• We can add the term for moles (n) to the combined gas laws that we just looked at :so P1V1/T1 becomes P1V1 / T1n1

• This shows that the term P x V/ T x n is a constant

• This will allow us to find unknowns in a system where we are not changing conditions

P1 x V1

T1 x n1

P2 x V2

T2 x n2

P1V1 = T1n1

• This is called the Ideal Gas Law• We need to evaluate this in order to come up

with a constant which will make the equation work

• We do this at STP• What is STP

We use STP as a starting point

• P= 101.3 kPa (one ATM)• V= 22.4 Liters (volume of 1 mole at STP)• N= 1 mole • T= 273K (melting point of ice in K)

• The constant, R, = (P x V) / (T x n) or 8.31 (L x kPa) / (K x mol)

The ideal gas law becomes:

• PV=nRT• Pressure x volume = # moles x temp x

constant• This will allow us to figure any one of the

variables in any system if we know the other three

Sample problem

• You fill a rigid steel cylinder that has a volume of 20.0 L with nitrogen gas to a final pressure of 2 x 104 kPa at 28C. How many moles of gas are in the cylinder?


• Knowns:P = 2x104 kPaV = 20 LT = 28C

• Unknown:N = ?


• Convert 28C to K 301K• Isolate n (number of moles)

n = P x V R x T


n = 2x104kPa x 20L 8.31 L x kPa x 301K

K x mol


n = 1.60 x 102 mol N2

When the pressure of a rigid hollow sphere containing 685 L of helium gas is held at 621K, the pressure of the gas is 1.89 x 103 kPa. How many moles of

helium does the sphere contain?

What pressure will be exerted by .450 mol of a gas at 25C if it is

contained in a .650 L vessel?

A deep underground cavern contains 2.24 x 106 L of methane (CH4) at a pressure of 1.50 x 103 kPa and a temperature of 42C.

How many kilograms of CH4 does this deposit contain?

What volume will 12 g of oxygen gas (O2) occupy at 25C and a

pressure of 52.7 kPa?

Ideal gas??

• The ideal gas law assumes that gas particles occupy no volume and that there are no attractions between particles. This is never true.

• Gases under high pressure and low temperatures turn to liquids. Gases do occupy volume (obviously), and there are attractions (intermolecular forces) as we learned in chapter 10.

• Boyle’s law implies that gases are infinitely compressible, but this is not the case

Partial pressures (Dalton’s Law)

• The pressure exerted by a mixture of gases (our atmosphere, for example) equals the sum of the pressures of each individual gas in the mixture

• Ptotal = Pa + Pb + Pc

Partial pressures

• Each gas in a mixture acts independently of each other

• The total pressure exerted by the gases in a mixture is directly related to the number of moles in the mixture Ptotal ~ ntotal

• Ptotal =ntotalRT / V

The total pressure of a mixture of hydrogen, argon and nitrogen is 120 kPa. The partial

pressure of hydrogen is 32 kPa and the partial pressure of argon is 58 kPa. What is the partial

pressure of the nitrogen?

The total pressure of a mixture of hydrogen, argon and nitrogen is 120 kPa. The partial

pressure of hydrogen is 32 kPa and the partial pressure of argon is 58 kPa. What is the partial

pressure of the nitrogen?

• Using Ptotal = Pa + Pb + Pc

• 120 kPa = 32 kPa + 58 kPa + Pnitrogen

• Pnitrogen = 30 kPa

A mixture of .25 mole CO2 and .30 mole of O2 is in a 10 L container at 298K. What is the total pressure of the mixture?What is the partial pressure of the CO2?What is the partial pressure of the O2?


• The total pressure depends on the total number of moles (.55)



• The total pressure depends on the total number of moles (.55)


• Ptotal = .55 mole x 8.3 x 298K / 10L• = 136kPa


• The partial pressure of each gas depends on the number of moles of each gas (.25 mole CO2)

• PCO2 =nCO2RT / V


• The partial pressure of each gas depends on the number of moles of each gas (.25 mole CO2)

• PCO2 =nCO2RT / V

• PCO2 =.25 mol x 8.3 x 298 / 10• = 62 kPa



• 136 kPa = 62 kPa + PO2

• PO2 = 74 kPa

The gas laws

• Boyle’s Law• Charle’s Law• Gay-Lussac’s Law• Combined gas law• Ideal gas law• Dalton’s Law

Boyle’s Law

• P1V1 = P2V2• Pressure and volume change• Temperature is held constant

Charle’s Law

• V1/T1 = V2/T2• Volume and temperature change• Pressure remains constant

Gay-Lussac’s Law

• P1/T1 = P2/T2• Pressure and temperature change• Volume remains constant

Combined gas law

• P1V1/T1 = P2V2/T2• Pressure, volume and temperature change• # moles remains constant

Ideal gas law• PV = nRT• Used for finding conditions where nothing is

changing. • One variable can be missing• Units must be:– P (kPa)– V (L)– N (moles)– T (Kelvins)– R (8.3)

Dalton’s Law


• The total pressure of a mixture of gases is the sum of the pressures of each different gas (partial pressures)

chapter 12 the behavior of gases 12.3 the gas laws

Documents

volume slide

v2 slide

tire slide

p2t2 slide

kpa v2

new volume

kpa p2

kpa unknown