Gas Laws

Day 3

Gas Law Foldable

Fold the left and right to the middle.

Dalton’s Law of Partial Pressure

The pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases.

PTotal = P1 + P2 + P3….

Example

A balloon is filled with air (O2,

CO2, & N2) at a pressure of

1.3 atm.

If PO2 = 0.4 atm and PCO2 =

0.3 atm, what is the partial pressure of the nitrogen gas?

PTotal = P1 + P2 + P3….

Ptotal = PO2 + PCO2 + PN2

1.3 atm = 0.4 atm + 0.3 atm + PN2

PN2 = 0.6 atm

Boyle’s Law: Pressure vs. Volume

At a constant temperature, the volume of a fixed mass of gas varies inversely with the pressure

P1V1 = P2V2

*use if temperature is constant

Boyle's Law

0

500

1000

1500

1 2 3 4 5 6 7

Pressure (atm)

Vo

lum

e (

mL

)

InverseIndirect

Pressure and Volume with constant Temperature

Changing the volume of the container changes the amount of space between the particles. The less space, the more the particles collide with each other and the walls.

Uses of Boyle’s LawTesting materials stability and ability to

maintain their shape under force.Compressing gases for use in cooking

cylinders, SCUBA tanks, and shaving cream.Used to describe density relationships

between gases.

ExampleIf a gas expands from a volume of 5 L

to 25 L at an initial pressure of 3.5 atm, what will be its new pressure?

P1 = 3.5 atm

V1= 5 L

P2 = ?

V2 = 25 L

Charles’ Law: Volume vs Temperature

At constant pressure, the volume of a fixed mass of gas varies directly with the Kelvin temperature

V1 = V2

T1 T2

*use if pressure is constant

Charles' Law ( Vol v. Temp)

0

200

400

600

800

1000

1200

0 200 400 600

Volume (mL)

Te

mp

era

ture

(K

) Direct

Charles’ Law Graph

Temp decreases

Vol decreases.

Temperature (K)

Volume (mL)

546 1092

373 746

283 566

274 548

273 546

272 544

200 400

50 100

0 0

Temperature and Volume with constant Pressure

Changing the temperature but requiring the pressure to stay the same causes the volume to increase.

Uses of Charles’ Law

Describing the properties of gases, liquids, and solids at extremely low temperatures.

Hot air ballooningUsed to describe density

relationships of gases.

Example

If 22.4 L of oxygen is heated from 23˚C to 50 ˚C, what is its new volume?

V1 = 22.4 L

T1 = 23 + 273 = 296 K

V2 = ?

T2 = 50 + 273 = 323 K

Absolute Zero

Temperature at which all molecular motion stops.

It is defined by 0 K or -273C. Scientists used Charles Law to

extrapolate the temperature of absolute zero.

Avogadro’s Law: Volume vs. Moles

At a constant temperature & pressure, the volume of a gas varies directly with the moles

V1 = V2

n1 n2

*use if temperature and pressure are constant

Avogadro’s Law

As the number of moles increases, the volume expands to make room for the additional gas

Example

If 4.65 L of CO2 increases from 0.8 moles to 3.75 moles, what is the new volume of the gas?

V1 = 4.65 L

n1 = 0.8 moles

V2 = ? L

n2 = 3.75 moles

V1 = V2

n1 n2

V2 = V1n2

n1

V2 = (4.65 L)(3.75 mol) = 21.79 L

(0.8 mol) (20 L)

V1 = V2

n1 n2

The Combined Gas Law

P1 V1 = P2 V2

n1T1 n2T2

The Combined Gas LawExpresses a relationship between

pressure, volume, and temperature (and moles) of a fixed amount of gas.

It takes all three gas laws: (Boyle’s,

Charles’s, & Avogadro’s) and combines them form one usable equation.

Example 1A gas is cooled from 45˚C to 20˚C. The

pressure changes from 103 kPa to 101.3 kPa as the volumes settles to 16.0 L. What was the initial volume?

P1 = 103 kPa

V1 = ?

T1 = 45oC + 273 = 318 KP2 = 101.3 kPa

V2 = 16.0 L

T2 = 20oC + 273 = 293 K

(103 kPa)(V1) = (101.3 kPa)(16.0 L)

(n1) (318 K) (n2) (293 K)

V1 = 17.1 L

P1 V1 = P2 V2

n1T1 n2T2

Example 2 (on back of foldable)A 1.5 mole sample of methane was originally

0.5 L at 25˚C at 1.1 atm. If we decreased the volume of the container to 0.25 L, increased the pressure to 2.0 atm and added 2.5 moles, what would be the new temperature in ˚C?

P1 = 1.1 atm

V1 = 0.5 L

T1 = 25 + 273= 298 K

n1 = 1.5 moles

P2 = 2.0 atm

V2 = 0.25 L

T2 = ?

n2 = 1.5+2.5 = 4 moles

T2 = 101.59 K – 273

= -171.41 ˚C

P1 V1 = P2 V2

n1T1 n2T2*** easier if you solve for T2 first THEN plug in the #’s ***

P1 V1n2T2 = P2 V2n1T1

P1 V1n2 P1 V1n2

= (2.0 atm)(0.25 L)(1.5 mol)(298K)

(1.1 atm)(0.5 L)(4.0 mol)