• April 24, 2013 Perseverance

• Aficionado: a devotee; a fan; an enthusiastic person about a sport or hobby

• Do Now: Quad Card• Topic: Air pressure

Basic Gas Basic Gas LawsLaws

(Boyle’s, Charles’s & Gay-Lussac’s)(Boyle’s, Charles’s & Gay-Lussac’s)

What makes a hot air balloon inflate?

HINT: look at the name!

Part 1: What Is a Gas Law?• the gas laws are simple, mathematical relationships

between the pressure (P), volume (V), temperature (T), and moles (n), of a gas. Basic gas laws involve P, V, and T only.

• the 5 basic gas laws:

• gas laws use the Kelvin temperature scale (K). Why? The Celsius scale (C) has negative values and a zero value (the Kelvin scale does not). If we used the Celsius scale, we might calculate a zero or negative volume/pressure from it, which can’t exist!• to convert a Celsius temp into Kelvin, just add 273• notice that Kelvin temps do not have a degree sign,

just a “K”

C + 273 = __K

C + 273 = __K

Boyle’s LawP1V1 = P2V2

Boyle’s LawP1V1 = P2V2

Charles’s Law

V1 = V2

T1 T2

Charles’s Law

V1 = V2

T1 T2

Gay-Lussac Law

P1 = P2

T1 T2

Gay-Lussac Law

P1 = P2

T1 T2

Combined Gas Law

P1V1 = P2V2

T1 T2

Combined Gas Law

P1V1 = P2V2

T1 T2Partial Pressures

LawPT = P1 + P2 +

P3

Partial Pressures Law

PT = P1 + P2 + P3

• to convert a Celsius temp into Kelvin, just add 273• notice that Kelvin temps do not have a degree sign,

just a “K”• there is a certain temperature that is considered

“standard,” as well as a standard pressure. The values for standard temp and pressure (STP) are 273 K and 1 atm.

• all the gas laws (except Charles’s law) involve pressure. Most people are not familiar with the many units pressure can be measured in (except maybe psi). So here they are:

• remember: units of volume = milliliters (mL), liters (L), and cubic centimeters (cm3).

C + 273 = __K

C + 273 = __K

Unit Abbr. STP valueatmospheres atm 1 atmmillimeters of

mercurymmH

g760

mmHgpounds per square

inchpsi 14.7 psi

kilopascals kPa101.325

kPa

NOTE: the STP values shown here can be used to convert one pressure unit to another, which will need to be done quite a bit in your calculations!

Part 2: Boyle’s Law (1662)• Boyle’s Law states that the pressure of a fixed

mass of gas varies inversely with the volume at a constant temperature.

• this means if you compare the initial volume and pressure of a gas with the new conditions of the gas, you will get an inverse relationship every time• P1 and V1 indicate initial (or starting) conditions• P2 and V2 indicate new (or final) conditions

Steps for Solving ANY Gas Law Problem:1.Write out a column of information down the left-hand

side. Make sure all of your variable’s units match (i.e. if P1 is in kPa, then P2 must be in kPa as well). If one doesn’t match, convert it to match the other, using a conversion table. Put a question mark in the space for the variable you are trying to solve for (what you DON’T have).

2.Write the original equation for the gas law you will be using.

3.Rearrange the equation to solve for the variable you need.

Boyle’s LawP1V1 = P2V2

Boyle’s LawP1V1 = P2V2

P

V

Steps for Solving ANY Gas Law Problem:1.Write out a column of information down the left-hand

side. Make sure all of your variable’s units match (i.e. if P1 is in kPa, then P2 must be in kPa as well). If one doesn’t match, convert it to match the other, using a conversion table. Put a question mark in the space for the variable you are trying to solve for (what you DON’T have).

2.Write the original equation for the gas law you will be using.

3.Rearrange the equation to solve for the variable you need.

4.Plug in the values and units you have in to the rearranged equation, and make sure all your units will cancel except for one. This will be the unit for your answer.

5.Calculate, then box your answer!

Ex1: Using 14.3 L of N2 as the initial volume, calculate the volume that would result if the pressure was raised from 150 kPa to 250 kPa.

Ex1: Using 14.3 L of N2 as the initial volume, calculate the volume that would result if the pressure was raised from 150 kPa to 250 kPa.P1 = __________V1 = __________P2 = __________V2 = __________

Part 3: Charles’s Law (1787)• Charles’s Law states that the volume of a fixed mass

of gas varies directly with the temperature at a constant pressure.

• this means that as the volume of gas increases, so does the temperatureEx2: A sample of gas occupied a volume of 5.0L at a temp of 37.0C. If the temp were to increase by 6C, what would be the volume of the gas under this new condition?

150 kPa14.3 L

250 kPa?

P1V1 = P2V2

____ ____ P2 P2

V2 = P1V1

P2

V2 = (150 kPa)(14.3 L) = 250 kPa

V2 = 8.58 L

V

T

Charles’s Law

V1 = V2

T1 T2

Charles’s Law

V1 = V2

T1 T2

V2 = 150 14.3 ÷ 250 =

Ex2: A sample of gas occupied a volume of 5.0L at a temp of 37.0C. If the temp were to increase by 6C, what would be the volume of the gas under this new condition?V1 = __________T1 = ____C____KV2 = __________T2 = ____C____K

Part 4: Gay-Lussac’s Law (1802)•Gay-Lussac’s Law states that the pressure of a fixed mass of gas varies directly with the temperature at a constant volume.•this means that as the pressure of gas increases, so does the temperatureEx3: The gas left in a used aerosol can is at a pressure of 125.3 kPa at 17C. If the can is thrown into a fire, what will the pressure be inside the can at 1045C?

5.0 L 37 ? 43

V1T2 = V2T1

____ ____ T1 T1

V2 = V1T2

T1

V2 = (5.0 L)(316 K) = 310 K

V2 = 5.10 L

Charles’s Law

V1 = V2

T1 T2

Charles’s Law

V1 = V2

T1 T2

V2 = 5.0 316 ÷ 310 =

310 316

Gay-Lussac Law

P1 = P2

T1 T2

Gay-Lussac Law

P1 = P2

T1 T2

P

T

Ex3: The gas left in a used aerosol can is at a pressure of 125.3 kPa at 17C. If the can is thrown into a fire, what will the pressure be inside the can at 1045C?P1 = __________T1 = ____C____KP2 = __________T2 = ____C____K

Gay-Lussac Law

P1 = P2

T1 T2

Gay-Lussac Law

P1 = P2

T1 T2 125.3 kpa 17 ?1045

____ ____ T1 T1

290 1318

P1T2 = P2T1

P2 = P1T2

T1

P2 = (125.3 kPa)(1318 K) = 290 K

P2 = 569.47 kPa

P2 = 125.3 1318 ÷ 290 =