Warm UpWarm Up 2-2-7-147-14

1.What is STP?

2. How much space does 1 mole of hydrogen gas occupy at STP?

3. How many torr is in 5 atm?4. Convert 30 Celsius to Kelvin.

Agenda-Pick up binders (40pts)-Notes Chp 12-3 (gas laws)-WS The gas law

HomeworkQuiz next class

12-312-3The Gas LawsThe Gas Laws

Mathematical relationships between volume, temperature,

pressure and amount of gas

Combined Gas LawCombined Gas Law:: Pressure, Volume Pressure, Volume

&Temperature&Temperature• Combines all three gas laws into one

equation!!• Memorize this equation!!!

P1V1=

P2V2

T1 T2Remember: all

temperatures must be in Kelvin!!!

Combined LawCombined Law:: Example 1Example 1

• A helium filled balloon has a volume of 50.0 L at 25 ºC and 1.08 atm. What volume will it have at 0.855 atm and 10 ºC?

Given: VV11 = 50.0 L = 50.0 L

PP11 = 1.08 atm = 1.08 atm

TT11 = 25 ºC + 273 = 298 K = 25 ºC + 273 = 298 K

P P22 = 0.855 atm = 0.855 atm

TT22 = 10 ºC + 273 = 283 K = 10 ºC + 273 = 283 K

VV2=2=

Find: VV22

Combined LawCombined Law:: Example Example 22

• A sample of air has a volume of 140.0 mL at 67 ºC under 2 atm. At what temperature will its volume be 50.0 mL under 2 atm?

Given: VV11 = 140.0 L = 140.0 L

TT11 = 67 ºC + 273 = 340 K = 67 ºC + 273 = 340 K

V V22 = 50.0 mL = 50.0 mL

PP11 = P = P22

Find: TT22

BoyleBoyle’’s Laws Law:: Pressure & VolumePressure & Volume

• As pressure increases, volume decreases

↑P ↓V

↓P ↑V

P1V1 = P2V2

BoyleBoyle’’s Laws Law

• Think about why this is true:– Pressure is caused by gas molecules hitting the

container– If the volume of the container is decreased, the same

number of gas molecules are moving in a much smaller area & will hit the container more often

BoyleBoyle’’s Laws Law:: Example 1 Example 1• A 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?

Given: VV11 = 150 mL = 150 mL

PP11 = 0.947 atm = 0.947 atm

PP22 = 0.987 atm = 0.987 atm

Find: VV22

BoyleBoyle’’s Laws Law:: Example 1 Example 1Given: VV11 = 150 mL = 150 mL

PP11 = 0.947 atm = 0.947 atm

PP22 = 0.987 atm = 0.987 atm

Find: VV22

Plan: PP11VV11 == PP22VV22

PP11VV11== VV22

PP22Solve:

VV22 ==(0.947atm)(150 mL)(0.947atm)(150 mL)

0.987 atm0.987 atm

VV22 == 144 mL144 mL

Check: ↑P ↓V ? yes↑P ↓V ? yes

BoyleBoyle’’s Laws Law:: Example 2 Example 2• A gas has a pressure of 1.26 atm and

occupies a volume of 7.40 L. If the gas is compressed to a volume of 2.93 L, what will its pressure be?

Given: PP11 = 1.26 atm = 1.26 atm

VV11 = 7.40 L = 7.40 L

VV22 = 2.93 L = 2.93 L

Find: PP22

BoyleBoyle’’s Laws Law:: Example 2 Example 2Given: PP11 = 1.26 atm = 1.26 atm

VV11 = 7.40 L = 7.40 L

VV22 = 2.93 L = 2.93 L

Find: PP22

Plan: PP11VV11 == PP22VV22

PP11VV11== PP22

VV22Solve:

PP22 ==(1.26atm)(7.40 L)(1.26atm)(7.40 L)

2.93 L2.93 L

PP22 == 3.18 atm3.18 atm

Check: ↓V ↑P ? yes↓V ↑P ? yes

CharlesCharles’’s Laws Law:: Volume & Volume & TemperatureTemperature

• When temperature increases, volume increases

↑T ↑V

↓T ↓ V

V1=

V2

T1 T2

All temperatures must be in Kelvin!!!

CharlesCharles’’s Laws Law

• Think about why this is true:– Increase in temp. causes molecules to move faster

– Faster molecules more collisions

– More collisions more pressure inside container

– More pressure bigger volume

CharlesCharles’’s Laws Law:: Example 1Example 1

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

Given: VV11 = 752 mL = 752 mL

TT11 = 25 ºC + 273 = 298 K = 25 ºC + 273 = 298 K

TT22 = 50 ºC + 273 = 323 K = 50 ºC + 273 = 323 K

Find: VV22

CharlesCharles’’s Laws Law:: Example Example 11

Given: VV11 = 752 mL = 752 mL

TT11 = 25 ºC + 273 = 298 K = 25 ºC + 273 = 298 K

TT22 = 50 ºC + 273 = 323 K = 50 ºC + 273 = 323 K

Find: VV22

Plan: VV11 == VV22

TT11 TT22

VV1 1 TT22== VV22

TT11Solve:VV22 ==

(752 mL)(323 K)(752 mL)(323 K)

298 K298 K

VV22 == 815 mL815 mL

Check: ↑T ↑ V ? yes↑T ↑ V ? yes

CharlesCharles’’s Laws Law:: Example 2Example 2

• A helium balloon has a volume of 2.75 L at 20 ºC. If you bring it outside on a cold day, the volume decreases to 2.46 L. What is the outside temperature in ºC?

Given: VV11 = 2.75 L = 2.75 L

TT11 = 20 ºC + 273 = 293 K = 20 ºC + 273 = 293 K

VV22 = 2.46 L = 2.46 L

Find: TT22

CharlesCharles’’s Laws Law:: Example 2 Example 2Given: VV11 = 2.75 L = 2.75 L

TT11 = 20 ºC + 273 = 293 K = 20 ºC + 273 = 293 K

VV22 = 2.46 L = 2.46 L

Find: TT22

Plan: VV11 == VV22

TT11 TT22

VV2 2 TT11== TT22

VV11Solve:TT22 ==

(2.46 L)(293 K)(2.46 L)(293 K)

2.75 L2.75 L

TT22 == 262 K – 273 = -11 ºC 262 K – 273 = -11 ºC

Check: ↓↓ V V ↓↓ T ? yesT ? yes

Gay-LussacGay-Lussac’’s Laws Law:: Pressure & TemperaturePressure & Temperature

• When temperature increases, pressure increases

↑T ↑P

↓T ↓P

P1=

P2

T1 T2

Remember: all temperatures must

be in Kelvin!!!

Gay-LussacGay-Lussac’’s Laws Law

• Think about why this is true:– Increase in temp. causes molecules to move faster

– Faster molecules more collisions

– More collisions more pressure inside container

Gay-LussacGay-Lussac’’s Laws Law:: Example 1Example 1

• Gas in an aerosol can is at a pressure of 3.00 atm 25 ºC. What would the pressure be in the can at 52 ºC?

Given: PP11 = 3.00 atm = 3.00 atm

TT11 = 25 ºC + 273 = 298 K = 25 ºC + 273 = 298 K

TT22 = 52 ºC + 273 = 325 K = 52 ºC + 273 = 325 K

Find: PP22

Gay-LussacGay-Lussac’’s Law s Law :: Example 1Example 1

Given: PP11 = 3.00 atm = 3.00 atm

TT11 = 25 ºC + 273 = 298 K = 25 ºC + 273 = 298 K

TT22 = 52 ºC + 273 = 325 K = 52 ºC + 273 = 325 K

Find: PP22

Plan: PP11 == PP22

TT11 TT22

PP1 1 TT22== PP22

TT11Solve:PP22 ==

(3.00 atm)(325 K)(3.00 atm)(325 K)

298 K298 K

PP22 == 3.27 atm 3.27 atm

Check: ↑T ↑P ? yes↑T ↑P ? yes

Gay-LussacGay-Lussac’’s Laws Law:: Example 2Example 2

• Before you leave on a road trip, the pressure in your car tires is 1.8 atm at 20 ºC. After driving all day, the pressure gauge read 1.9 atm. What temperature (in ºC) are your tires?

Given: PP11 = 1.8 atm = 1.8 atm

TT11 = 20 ºC + 273 = 293 K = 20 ºC + 273 = 293 K

PP22 = 1.9 atm = 1.9 atm

Find: TT22

Gay-LussacGay-Lussac’’s Law s Law :: Example 2Example 2

Given: PP11 = 1.8 atm = 1.8 atm

TT11 = 20 ºC + 273 = 293 K = 20 ºC + 273 = 293 K

PP22 = 1.9 atm = 1.9 atm

Find: TT22

Plan: PP11 == PP22

TT11 TT22

PP2 2 TT11== TT22

PP11Solve:TT22 ==

(1.9 atm)(293 K)(1.9 atm)(293 K)

1.8 atm1.8 atm

TT22 == 310 K – 273 = 37 ºC 310 K – 273 = 37 ºC

Check: ↑P ↑P ↑↑ T ? yesT ? yes

Quiz Topics Next ClassQuiz Topics Next Class

1. KMT

2. Properties of Gases

3. Combined gas law calculation

4. P, V, T relationship

5. Converting Celsius to Kelvin