Chapter 12.2

The Gas Laws

1. State Boyle’s law, and use it to solve problems involving pressure and volume.

2. State Charles’s law, and use it to solve problems involving volume and temperature.

3. State Gay-Lussac’s law, and use it to solve problems involving pressure and temperature.

4. State Avogadro’s law, and explain its importance in determining the formulas of chemical compounds.

Objectives

The Gas LawsSimple mathematical relationships

betweenVolumeTemperaturePressureAmount of gasGas Law

Program: http://intro.chem.okstate.edu/1314F00/Laboratory/GLP.htm

Shows the relationship of all four of the above on gases

Constant :Volume and amount of gas

Shows:Change in pressure and temperature

Boyle’s LawStates – volume of a fixed mass of gas

varies inversely with the pressure at constant temperature.Constant temperature and amount of gasIf you double volume, pressure is cut in halfIf you cut volume in half, pressure doubles

Boyle’s LawPressure caused by

Moving molecules hitting container wallsSpeed of particles (force) and number of

collisionsBoth increase pressure

Mathematically:

V1

k=P

PVor =k

k is constant

P is pressure

V is volume

Interactive graph

Volume-Pressure Data for Gas SampleVolume Pressure

P x V (mL) (atm) 1200 0.5 600 600 1.0 600 300 2.0 600 200 3.0 600 150 4.0 600 120 5.0 600 100 6.0 600

Boyle’s Law

P1V

1

=k

Boyle’s Law Equation:

P2V

2

=k

So:P1V

1

= P2V

2Sample Problem 1A sample of oxygen gas has a volume of 150. mL when its pressure is 0.947 atm. What will the volume of the gas be at a pressure of 0.987 atm if the temperature remains constant.P1

V1

P2

V2

=

===

0.947 atm150. mL0.987 atm?

P1V

1

= P2V

2 P2P2

P1V

1

= V2P2

V2 =(0.947 atm)

(150. mL)0.987

atm=144

mL

Charles’s LawVolume-temperature Relationship

When using temperature, Must use absolute zero, Kelvin Temperature scaleTemperature -273.15oC is absolute zero

All molecular movement would stop.

Fahrenheit

Celsius Kelvin

Temperature Scales

212

32

100

0

373

273

K = oC + 273

So : How many Kelvin is 10 oC?K = oC + 273

= 10oC + 273 = 283 K

Charles’s LawVolume-Temperature Data for Gas Sample

Temperature Kelvin Volume V/T

(oC) (K) (mL) (mL/K) 273 546 1092 2 100 373 746 2

10 283 566 2

1 274 548 2

0 273 546 2

-1 272 544 2

-73 200 400 2

-173 100 200 2

-223 50 100 2

Charles’s LawStates that the volume of a fixed mass of gas at constant

pressure varies directly with the Kelvin temperature.Constant Pressure and amount of gas If you double temperature, the volume will double If you cut the temperature in half, the volume will be half as

much.V k=

VorkT

T=

k is constant

T is temperature V is volume

Charles’s Law Equation:

V1

T1

=kV2

T2

=k

So: V2

T2

V1

T1

=

Charles’s LawSample Problem 2

A sample of neon gas occupies a volume of 752 mL at 25oC. What volume will the gas occupy at 50oC if the pressure remains constant?

V1

T1

V2

T2

=

===

752 mL

25oC ?50oC

**Always convert to Kelvin!!

+ 273 =+ 273 =

298 K

323 K

V2

T2

V1

T1

= x T2xT2

V2 =V1T2

T1

=(752 mL)

(323 K)298 K

=815 mL

Gay-Lussac’s LawPressure-temperature Relationship

Increasing temperature, increases the speed of the gas particlesThus, more collisions with the container walls

Causing an increase in pressure

Gay-Lussac’s LawStates that the pressure of a fixed mass of gas at

constant volume varies directly with the Kelvin temperature. Constant volume and amount of gas

If you double temperature, pressure doublesIf you cut temperature in half, pressure is also cut in

halfP k= orkT

T=

k is constant

T is temperature P is pressure

P1

T1

=kP2

T2

=k

So: P2P1 =

P

T1 T2

Gay-Lussac’s LawSample Problem 3

The gas in an aerosol can is at a pressure of 3.00 atm at 25oC. Directions on the can warn the user not to keep the can in a place where the temperature exceed 52oC. What would the gas pressure in the can be at 52oC?

P1

T1

P2

T2

=

===

3.00 atm25oC ?52oC

**Always convert to Kelvin!!

+ 273 =+ 273 =

298 K

325 K

T2T1

= x T2xT2

P2 =P1T2

T1

=(3.00 atm)

(325 K)298 K

=3.27 atm

P1 P2

Avogadro’s LawEqual Volumes of gases at the same temperature and pressure contain equal numbers of molecules.

How does it work!

Hydrogen Gas + Chlorine gas Hydrochloric Acid

H2 Cl2 2 HCl

1 Volume 1 Volume 2 Volumes

1 Molecule 1 Molecule 2 Molecules 1 Mol 1 Mol 2 Mol

V = kn Direct Relationship

n = amount of gas

Molar Volume of Gas

1 mol = 6.022 x 10 23 molecules1 mol O2 = 6.022 x 10 23 molecules O2 = 32.00 g

O2

1 mol H2 = 6.022 x 10 23 molecules H2 = 2.02 g H2Standard Molar Volume of gasesVolume occupied by 1 mol of a gas at

STPAt STP, 1 mol of any gas = 22. 4 L

Recall