Day 1

• I CAN…– Understand and apply Boyle’s Law– Understand and apply Charles’ Law– Observe and explain demos using gas laws

1. Convert 687 torrs into atmospheres.

2. If temperature is constant, what happens to volume (V) as pressure (P) increases?

1. What if P decreases?

1 torr = 1mm Hg

1 atm = 760 mm Hg

ANSWERS1. Pressure (P) is COLLISIONS and is force

per unit area

2. 687 torr x 1 atm = 0.904 atm

760 torr

3. If T is constant…

as P increases, V decreases!

4. If T is constant…

as P decreases, V increases!

F

A

V = volume of the gas (liters, L)

P = pressure (atmospheres, atm)

T = temperature (Kelvin, K)

n = amount (moles, mol)

Gases can be described using the following four variables:

Inverse and DirectProportions

Inverse / Indirect Relationship

• What variables did you observe to have an indirect relationship yesterday with the simulation?

Direct Relationship

• What variables did you observe to have a direct relationship yesterday with the simulation?

INVERSE PROPORTIONS• As one variable goes up, the other

goes down!

• Produces a

curved graph…

Tconstant

P

V

Demo Time

• Vacuum pump

Inversely ProportionalEx: Boyle’s Law

(V as a function of P)

0

500

1000

1500

2000

2500

3000

3500

4000

4500

5000

0 200 400 600 800 1000 1200 1400 1600

Pressure (torr)

Vo

lum

e (

mL

)P (torr) V (mL)

100 4560

200 2280

300 1520

400 1140

500 912

600 760

700 651

800 570

900 507

1000 456

1100 414

1200 380

1300 351

1400 326

1500 304

P1 x V1 = P2 x V2

P may change and V may change, but their product

stays the same!

Multiplying the two variables equals a constant.

Boyle’s Law

A gas filled syringe has a volume of 150mL and a starting pressure of 0.947atm. If the pressure is increased to 0.987 atm what is the new volume?

Given:

P1= 0.947 atm

V1= 150mL= 0.15L

P2= 0.987 atm

V2 = ????

P1V1=P2V2

V2 = 0.947 atm (.15L) = 0.987 atm

Demo Time

• Liquid nitrogen

Temperature and Volume

Directly ProportionalEx: Charles’ Law

(V as a function of T)

0

200

400

600

800

1000

1200

1400

1600

0 200 400 600 800

Temperature (K)

Vo

lum

e (m

L)

T (K) V (mL)

0 0

50 100

100 200

150 300

200 400

250 500

300 600

350 700

400 800

450 900

500 1000

550 1100

600 1200

650 1300

700 1400

DIRECT PROPORTIONS• As one variable goes

up, so does the other!

• Produces a straightline graph…

• Dividing the one variableby the otherequals a constant.

T

V

V1 V2

T1 T2 =

YES!!!

Charles’ LawThe volume of a balloon is 2.45 L when at a temperature of 273 K. What happens to the volume when the temperature is raised to 325K?

Given:

V1= 2.45L

T1= 273 K

V2= ?

T2 = 325 K

V1 V2

T1 T2 =

V2 = 2.45 L (325K) = 273 K

Charles’ Law

• Temperature MUST be in Kelvins!• Kelvins is the unit of temperature that gives

the direct relationship to energy and the other units. Celsius does not!

• 0°C = 273 K• So… what is 23°C in K?• 23 + 273 = 296 K

Stop here for today!

• Let’s practice

Demo Time

• Candle

Practice #1

• A sample of helium gas has a pressure of 3.54 atm in a container with a volume of 23.1 L. This sample is transferred to a new container and the pressure is measured to be 1.87 atm. What is the volume of the new container? Assume constant temperature.

43.7 L

Practice #2

• A 2.45 L sample of nitrogen gas is collected at 273 K and heated to 325 K. Calculate the volume of the nitrogen gas at 325 K. Assume constant pressure.

2.92 L

Day 2

• I CAN…– Understand and apply Gay-Loussac’s Law– Understand and apply Avogadro’s Law– Understand and apply the Combined Gas Law

Pressure depends on TempPressure depends on Temp

Pressure Gauge

Pressure Gauge

Today’s temp: 35°FToday’s temp: 35°F

Pressure Gauge

Pressure Gauge

Today’s temp: 85°FToday’s temp: 85°F

Gay-Loussac’s Law

• P1

Gay Loussac’s law

A bike tire has a pressure of 0.987 atm at a temperature of 25°C. What temperature would bring the pressure down to 0.795 atm?

Given:P1= 0.987 atmT1= 298 KP2= 0.795 atm

T2 = ??? T2 = 0.795atm (298K) = 0.98.7atm

Avogadro’s Law:

V1 V2

n1 n2 =

Avogadro’s Law

A 59.5 L cylinder has 2.55 moles of hydrogen gas. More hydrogen is added so that there are now 7.83 moles, what is the new volume?

Given:V1= 59.5 Ln1= 2.55 molV2= ???n2 = 7.83 mol

V1 V2

n1 n2 =

V2 = 59.5 L (7.83mol) = 2.55 mol

P1V1 = P2V2

n1T1 n2T2

P1V1 = P2V2

n1T1 n2T2

Combined Gas LawWhich is a direct relationship?Which is inverse?

Combined Gas LawWhich is a direct relationship?Which is inverse?

A little reviewA little review

ExamplesExamples

A 15 L cylinder of gas at 4.8 atm A 15 L cylinder of gas at 4.8 atm pressure at 25ºC is heated to 75ºC pressure at 25ºC is heated to 75ºC and compressed to 17 atm. What is and compressed to 17 atm. What is the new volume?the new volume?

4.9 L4.9 L

ExamplesExamples If 6.2 L of gas at 723 mm Hg at If 6.2 L of gas at 723 mm Hg at

21ºC is compressed to 2.2 L at 21ºC is compressed to 2.2 L at 4117 mm Hg, what is the 4117 mm Hg, what is the temperature of the gas?temperature of the gas?

590K590K

When measured at STP, a quantity of gas has a volume of 500 L. What volume will it occupy at 0 oC and 93.3 kPa?

P1 = 101.3 kPaT1 = 273 KV1 = 500 LP2 = 93.3 kPaT2 = 0 oC + 273 = 273 KV2 = ?

V2 = 542.9 L

1 1 2 2

1 2

PV PV

T T

If I initially have a gas at a pressure of If I initially have a gas at a pressure of 12 atm, a volume of 23 liters, and a 12 atm, a volume of 23 liters, and a temperature of 200 K, and then I raise temperature of 200 K, and then I raise the pressure to 14 atm and increase the pressure to 14 atm and increase the temperature to 300 K, what is the the temperature to 300 K, what is the new volume of the gas?new volume of the gas?

V2=30 L

Combined

1        

A gas has a temperature of 14 A gas has a temperature of 14 00C, and C, and a volume of 4.5 liters. If the a volume of 4.5 liters. If the temperature is raised to 29 temperature is raised to 29 00C and the C and the pressure is not changed, what is the pressure is not changed, what is the new volume of the gas?new volume of the gas?

VV22 = = 4.7 L

Charles’

A sample of gas under a pressure of 720 mm Hg has a volume of 300. mL. The pressure is changed to 800. mmHg. What volume will the gas then occupy?

VV22 = = 0.270 L0.270 L

Boyle’sBoyle’s

If I have 2.9 L of gas at a pressure of 5.0 atm and a temperature of 50 0C, what will be the temperature of the gas if I decrease the volume of the gas to 2.4 L and decrease the pressure to 3.0 atm?

T2= 160 K   

CombinedCombined

Tonights HW problem + 21, 29, 43Tonights HW problem + 21, 29, 43

A gas that has a volume of 28 liters, a A gas that has a volume of 28 liters, a temperature of 45 temperature of 45 00C, and an unknown C, and an unknown pressure has its volume increased to pressure has its volume increased to 34 liters and its temperature 34 liters and its temperature decreased to 35 decreased to 35 00C. If I measure the C. If I measure the pressure after the change to be 2.0 pressure after the change to be 2.0 atm, what was the original pressure of atm, what was the original pressure of the gas?the gas?

PP11= = 2.5 atm

Day 3

• I CAN…– Understand and apply the Ideal Gas Law– Explain how ideal gases and real gases differ in

their behavior

IF WE COMBINE ALL OF THE LAWS TOGETHER INCLUDING AVOGADRO’S LAW MENTIONED EARLIER WE GET:

IF WE COMBINE ALL OF THE LAWS TOGETHER INCLUDING AVOGADRO’S LAW MENTIONED EARLIER WE GET:

PVPVTTnn

= R= R

WHERE R IS THE UNIVERSAL GAS

CONSTANT

WHERE R IS THE UNIVERSAL GAS

CONSTANT

NORMALLYWRITTEN ASNORMALLY

WRITTEN AS PVPV=nRT=nRT

Ideal gas lawIdeal gas law

Ideal Gas LawIdeal Gas Law

PV = nRTP = pressure (in atm!!)P = pressure (in atm!!)

V = volume (in L!!)V = volume (in L!!)

n = number of molesn = number of moles

R = universal gas constant = R = universal gas constant =

T = temperature (in K!!)T = temperature (in K!!)

0.08206 0.08206 L atmL atm

K molK mol

Argon is an inert gas used in lightbulbs to slow the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm, temperature of 18.0 0C, and a volume of 0.500L. How many moles of argon are present in the lamp? Grams?

PV = nRT

Need to rearrange to solve for n:PV = n

RT

Given:P = 1.20 atmT = 18.0°CR = 0.08206 L atm

K molV = 0.500L

(1.20 atm)(0.500L) = n(0.08206 L*atm/mol*K)(291K) = 0.0251 mol Ar

What volume does 9.45g of C2H2 occupy at STP?

What volume does 9.45g of C2H2 occupy at STP?

P P

V V T T

1atm1atm

?? 273K273K

R R

n n = .3635 mol C2H2

= .3635 mol C2H2

0.08206 0.08206L•atmL•atm

mol•Kmol•K

9.45g9.45g

26g/mol26g/mol

V = nRT P

V = nRT P

(1 atm)(1 atm)VV

(.3635mol

)(.3635mol

)(273K)(273K)

V = 8.14LV = 8.14L

==(0.08206 )(0.08206 )L•atm

mol•KL•atmmol•K

Volume of gas under STPVolume of gas under STP

L L Mol under STP conditions: Mol under STP conditions:

22.4 L/mol of any gas22.4 L/mol of any gas

0.03635 mol C0.03635 mol C22HH22 x x 1 mol 1 mol = 8.14 L C= 8.14 L C22HH22

22.4 L22.4 L

Using PV=nRTUsing PV=nRT

A camping stove propane tank holds 3000 g of C3H8. How large would a container have to be to hold the same amount of propane gas at 25 ºC and a pressure of 303 kPa?