Ideal Gas LawIdeal Gas Law

PV=nRTPV=nRT

Remember…Remember…

Boyle’s Law Boyle’s Law

Charles’ Law: Charles’ Law:

Combined Gas Law:Combined Gas Law:(Units MUST Match(Units MUST Match Temp in Kelvin!!!) Temp in Kelvin!!!)

2

2

1

1

T

V

T

V

2

22

1

11

T

VP

T

VP

2211 VPVP

A gas with a volume of 350 ml is collected A gas with a volume of 350 ml is collected at 15at 15o o C and 120 kPa. If the temperature C and 120 kPa. If the temperature changes to 30changes to 30o o C, what pressure would be C, what pressure would be required to put this gas in a 300 ml required to put this gas in a 300 ml container?container?

K

mlP

K

mlkPa

303

300

288

350120 2

kPaP 3.1472

A balloon has a volume of 500 ml at a A balloon has a volume of 500 ml at a temperature of 22temperature of 22ooC and a pressure of 755 C and a pressure of 755 mmHg. If the balloon is cooled to 0mmHg. If the balloon is cooled to 0o o C and C and a pressure of 145 mmHg, what is its new a pressure of 145 mmHg, what is its new volume?volume?

mlV 3.24092

K

VmmHg

K

mlmmHg

273

145

295

500755 2

Avogadro’s PrincipleAvogadro’s Principle

Under similar conditions (same Temp and Under similar conditions (same Temp and Pressure) equal volumes of gases contain Pressure) equal volumes of gases contain equal numbers of particles.equal numbers of particles.

10 L of H10 L of H22 (g) and 10 L of O (g) and 10 L of O22 (g) (g)

Both at Standard Temperature and Both at Standard Temperature and Pressure (STP) containPressure (STP) contain……

The same number of particles!The same number of particles!

Molar VolumeMolar Volume

The volume of 1 mole of gas particles The volume of 1 mole of gas particles

at STP at STP

is is 22.4 L22.4 L

Ideal Gas Law Animation:Ideal Gas Law Animation:

Try this:Try this:

1 mole of gas occupies 22.4 L at STP1 mole of gas occupies 22.4 L at STP

= __________ ml= __________ ml

(22400)(22400)

= ___________ moles of gas= ___________ moles of gas

(1 mole)(1 mole)

= ___________ particles= ___________ particles

(6.02 x 10(6.02 x 102323))

How many particles in 11.2 dmHow many particles in 11.2 dm33 of gas at STP? of gas at STP? 0.5 moles = 3.01 x 100.5 moles = 3.01 x 102323 particles particles22,400 cm22,400 cm3 3 of NHof NH33 gas at STP weighs? gas at STP weighs? = 22.4 L = 1 mole = 17 grams (add up MW)= 22.4 L = 1 mole = 17 grams (add up MW)44.8 L of NH44.8 L of NH33 at STP weighs? at STP weighs? = 2 moles = 34 grams= 2 moles = 34 grams

_____ grams = 1 mole of nitrogen gas _____ grams = 1 mole of nitrogen gas

= _____ L at STP? = _____ L at STP?

mole

LL

4.22

8.44

28.00

22.4

How many NHow many N22 molecules are molecules are

in 22.4 dmin 22.4 dm3 3 at STP?at STP?

= 1 mole = 6.02 x 10= 1 mole = 6.02 x 102323

What volume will 1.2 x 10What volume will 1.2 x 102424 H H2 2 molecules occupy molecules occupy

at STP?at STP?

= 2 moles = 44.8 L at STP= 2 moles = 44.8 L at STP

Ideal Gas EquationIdeal Gas EquationUse when Use when NOTNOT at STP!!! at STP!!!

PV= nRTPV= nRTP = Pressure (in kPa)P = Pressure (in kPa)

V = Volume (V = Volume (in Liters or dmin Liters or dm33))

n = number of molesn = number of moles

T = Temperature (T = Temperature (in Kelvinin Kelvin))

R =R = 8.31 8.31 L• kPaL• kPa

mole • Kmole • K

Development of R inDevelopment of R in

Kmole

LkPa

Kmole

LkPa

nT

PVR

31.82731

4.223.101

Kmole

LkPa

1. What volume will 2 moles of NO1. What volume will 2 moles of NO22 occupy occupy at 300 Kelvin and 90 kPa?at 300 Kelvin and 90 kPa?

LV

KmolesVkPa

nRTPV

4.5590

30031.82

30031.8290

What will be the temp of What will be the temp of 2 grams2 grams of of HH22 if 5000 cm if 5000 cm33 is at 5 atm? is at 5 atm?

KT

TmoleLkPa

nRTPV

8.30431.81

55.506

31.8155.506

Finding Molecular Weight of a GasFinding Molecular Weight of a Gas

Remember: MW = grams / molesRemember: MW = grams / moles

Converting grams to molesConverting grams to moles– Divide grams by the molecular weightDivide grams by the molecular weight

1) 5.0 L of a gas weighs 30.00 g at 201) 5.0 L of a gas weighs 30.00 g at 20o o C & C & 92 kPa. What is the mole weight of the gas?92 kPa. What is the mole weight of the gas?

g/mol 581mol 0.19

g 30.00MW

PV=nRT92kPa • 5.0 L= n • 8.31 • 293 K

n = 0.19 mol

mol ?

g 30.00

moles

gramsMW

2)If the mole weight of a gas is 26 g/mol and 2)If the mole weight of a gas is 26 g/mol and 18.00 g of the gas is 30 L at 2118.00 g of the gas is 30 L at 21oo C, what is the C, what is the pressure of the gas?pressure of the gas?

PV= nRT

P x 30 L=0.69 mol x 8.31 x 294 K

P = 56.2kPa

mol 0.69 g/mol 26

g 18.00

MW

g n

StoichiometryStoichiometry

Solving Steps.Solving Steps.– Balance the EquationBalance the Equation– Change grams to molesChange grams to moles– Use mole ratio to solveUse mole ratio to solve– Change moles to volumesChange moles to volumes

Use mole ratio (coefficients)

Use 22.4 L/mol @ STPOr PV = nRT

Use MWon P.T.

Use MWon P.T.

Use 22.4 L/mol@ STPOr PV = nRT

Mg (c) + HCl (ag) Mg (c) + HCl (ag) MgCl MgCl22 + H + H22 (g) (g)

If 2.43 g Mg react what volume of HIf 2.43 g Mg react what volume of H22 is is

produced? (at STP)produced? (at STP)

Mg (c) + HCl (ag) Mg (c) + HCl (ag) MgCl MgCl22 + H + H22 (g) (g)

2.43 g Mg2.43 g Mg

0.1 mole0.1 mole

molgg

3.24

43.2

1:1

molesmol

L1.04.22

0.1 moles0.1 moles

2

2.24 L H2.24 L H22? L

)313)(31.8)(1.0())(85( KmolVkPa

nRTPV

Mg (c) +Mg (c) +22 HCl (ag) HCl (ag) MgCl MgCl22 + H + H22 (g) (g)

If 2.43 g Mg react what volume of HIf 2.43 g Mg react what volume of H22 is produced at is produced at

4040ooC and 85 kPa?C and 85 kPa?

Mg (c) + HCl (ag) Mg (c) + HCl (ag) MgCl MgCl22 + H + H22 (g) (g)

2.43 g Mg2.43 g Mg

0.1 mole0.1 mole

molgg

3.24

43.2

1:10.1 moles0.1 moles

2

3.06 L H3.06 L H22? L

If 250 ml of HIf 250 ml of H22 is produced at 20 is produced at 20oo C & C &

100 kPa, what mass of Mg reacted?100 kPa, what mass of Mg reacted?

KnLkPa 29331.825.0100

2H 01.0 molesn

Mg (c) + HCl (ag) Mg (c) + HCl (ag) MgCl MgCl22 + H + H22 (g) (g)2

? g 250 mlnRTPV

0.01 moles0.01 moles0.01 mole0.01 mole 1:1

0.243 g Mg0.243 g Mg

molesmol

g01.03.24

Volume to VolumeVolume to Volume

THE MOLE RATIO IS THE SAME AS THE THE MOLE RATIO IS THE SAME AS THE VOLUME RATIO.VOLUME RATIO.

Liters BUse mole ratio (coefficients)

Liters A

Burning of methane:Burning of methane:

What vol. of oxygen is needed to completely What vol. of oxygen is needed to completely burn 1 L of methane?burn 1 L of methane?

CHCH44 + O + O22 CO CO22 (g) + H (g) + H22O (l)O (l)

2.0 L2.0 L 1:2

1.0 L1.0 L

2 2

Burning of methane:Burning of methane:

To produce 11.2 L of COTo produce 11.2 L of CO22 requires how requires how

many moles of Omany moles of O22 at STP? at STP?

CHCH44 + O + O22 CO CO22 (g) + H (g) + H22O (l)O (l)

22.4 L22.4 L 2:111.2 L11.2 L1 mole = 1 mole =

2 2