Gas Laws

Chapter 14

Kinetic Molecular Theory of Gases

No attraction/repulsion between particlesVolume of particles is 0Constant random motionElastic collisionsEqual kinetic energy at same temperature

Things to Remember…

Units of Pressure1 atm = 101.3 KPa = 760 mm Hg = 14.7 psi= 760 Torr

Temperature ConversionK= oC + 273

Boyle’s Law

At constant temperature, pressure and volume are indirectly related.

P1V1 = P2V2

Practice

A gas at 0.8atm is compressed from 3L to 1.2L. What is the resulting pressure in atm? In KPa?

Charles’s Law

At constant pressure, volume and temperature are directly related.

V1/T1 = V2/T2

Practice

5.2 L of Helium is heated from 3oC to 76oC. What is the volume of the gas after heating?

What About Pressure and Temperature?

At constant volume, pressure and temperature are directly related

P1/T1 = P2/T2

Practice

A gas is pressurized in a 2L bottle to 57 KPa. If the bottle is heated from room temperature (298K) to 312K, what will be the resulting pressure in KPA? In mm Hg?

Combined Gas Law

A closed gas system initially has pressure and temperature of 1.56atm and 629K with the volume unknown. If the same closed system has values of 195torr, 9940mL and 523K, what was the initial volume in L?

Ideal v. Real Gases

Particles of real gases have volume and experience attractive and repulsive forces

They’re the same except at:EXTREMELY LOW T

Kinetic energy decreases. Particles are close enough to allow for intermolecular bonding.

EXTREMELY HIGH P Compressed gases (gas particles are so close that there

is little space between them so their own volume becomes significant)

IDEAL GAS LAW

PV=nRT n=moles

n=m/MM m=(n)(MM) MM=m/n

Ideal Gas ConstantsR= 0.0821 L*atm/mol*KR= 8.314 L*KPa/mol*KR= 62.4 L*mm Hg/mol*K

Avogadro’s Principal

Equal volumes of gases at the same temperature and pressure contain an equal number of particles.I mole of any gas at STP = 22.4 LSTP (standard temperature and pressure)=0oC and 1atm

What volume does 24g of oxygen occupy at STP?

22 8.161

4.22

32

124 OLmole

L

g

molegO

Gas Stoichiometry – STP Conditions

OHOH 222 22

222

222 0.10

1

4.22

1

2

4.22

10.5 LH

molH

L

molO

molH

L

moleOLO

Volume-Volume problems (two substances have to be gases)

Determine the volume of hydrogen gas needed to completely react with 5.0L of oxygen gas to form water at STP.

Volume-Mass problems (at least one substance has to be a gas)

Calculate the volume of oxygen gas at STP that is required to complete react with 52.0g or iron.

322 234 OFeOFe

22

2 7.151

4.22

4

3

8.55

10.52 LO

molO

L

molFe

molO

gFe

moleFegFe

STP vs. Non-STP Gas Stoichiometry Problems

Calculate the volume of oxygen gas at STP that is required to completely react with 52.0g of iron.

322 234 OFeOFe

22

2 7.151

4.22

4

3

8.55

10.52 LO

molO

L

molFe

molO

gFe

moleFegFe

VS.

Calculate the volume of oxygen gas that is required to completely react with 52.0g of iron at 85kPa and 273K.

22 6.18

85

1

1

27331.8

4

3

8.55

10.52 LO

kPa

K

molK

kPaL

molFe

molO

gFe

moleFegFe