Gas Laws

What to do when conditions are ideal

Boyle’s LawWhat was the relationship between pressure and

volume?

When P Then V

When P Then V

Algebraically this is written as P=k/V

When you solved for k; PV=k

Therefore P1V1=P2V2

Gas Laws – Boyle’s LawConstant: Temperature

Relationship: Pressure is inversely proportional to volume

Pressure 1/volume

Written As: P1V1 = P2V2

Pressure is typically in atm or torr

Charles’ LawWhat was the relationship between Temperature and

Volume?

When T Then V

When T Then V

Algebraically this is written as V=kT

When you solved for k; V/T=k

Therefore V1/T1=V2/T2

Gas Laws – Charles’ LawConstant: Pressure

Relationship: Temperature is directly proportional to volume

Temp Volume

Written As: V1/T1 = V2/T2

Charles’ LawWhat unit of measure is needed for these

calculations? C or K?

Temperature is in K (K = 273 + C)

Gas Laws – Gay-Lussac’s LawConstant: Volume

Relationship: Pressure is directly proportional to temperature

Pressure Temperature

Written As: P1/T1 = P2/T2

The IDEAL GAS Law –this is what we will useWhen we put all three laws together:

PV nT (n= number of moles)

PV = nRT (R= ideal gas law constant)

R=62.4 L torr/K mol or .08206 L atm/K mol

Lecture PLUS Timberlake 2000 9

Ideal Gases

Behave as described by the ideal gas equation; no real gas is actually ideal

Within a few %, ideal gas equation describes most real gases at room temperature and pressures of 1 atm or less

In real gases, particles attract each other reducing the pressure

Real gases behave more like ideal gases as pressure approaches zero.

Lecture PLUS Timberlake 2000 10

PV = nRT

R is known as the universal gas constant

Using STP conditions P V

R = PV = (1.00 atm)(22.4 L) nT (1mol) (273K)

n T

= 0.0821 L-atm

mol-K

LecturePLUS Timberlake 11

Combined Gas Law

P1V1 = P2V2

T1 T2

Isolate V2

P1V1T2 = P2V2T1

V2 = P1V1T2

P2T1

Lecture PLUS Timberlake 2000 12

Learning Check G15What is the value of R when the STP value for P is 760 mmHg?

Lecture PLUS Timberlake 2000 13

Solution G15What is the value of R when the STP value for P is 760 mmHg?

R = PV = (760 mm Hg) (22.4 L)

nT (1mol) (273K)

= 62.4 L-mm Hg mol-K

Lecture PLUS Timberlake 2000 14

Learning Check G16

Dinitrogen monoxide (N2O), laughing gas, is used by dentists as an anesthetic. If 2.86 mol of gas occupies a 20.0 L tank at 23°C, what is the pressure (mmHg) in the tank in the dentist office?

Lecture PLUS Timberlake 2000 15

Solution G16

Set up data for 3 of the 4 gas variables

Adjust to match the units of R

V = 20.0 L 20.0 L

T = 23°C + 273 296 K

n = 2.86 mol 2.86 mol

P = ? ?

Lecture PLUS Timberlake 2000 16

Rearrange ideal gas law for unknown P

P = nRT

V

Substitute values of n, R, T and V and solve for P

P = (2.86 mol)(62.4L-mmHg)(296 K)

(20.0 L) (K-mol)

= 2.64 x 103 mm Hg

Lecture PLUS Timberlake 2000 17

Learning Check G17

A 5.0 L cylinder contains oxygen gas at 20.0°C and 735 mm Hg. How many grams of oxygen are in the cylinder?

Lecture PLUS Timberlake 2000 18

Solution G17

Solve ideal gas equation for n (moles)n = PV

RT

= (735 mmHg)(5.0 L)(mol K) (62.4 mmHg L)(293 K)

= 0. 20 mol O2 x 32.0 g O2 = 6.4 g O2

1 mol O2

LecturePLUS Timberlake 19

Learning Check C1

Solve the combined gas laws for T2.

LecturePLUS Timberlake 20

Solution C1

Solve the combined gas law for T2.

(Hint: cross-multiply first.)

P1V1 = P2V2

T1 T2

P1V1T2 = P2V2T1

T2 = P2V2T1

P1V1

LecturePLUS Timberlake 21

Combined Gas Law Problem

A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm?

LecturePLUS Timberlake 22

Data Table

Set up Data Table

P1 = 0.800 atm V1 = 0.180 L T1 = 302 K

P2 = 3.20 atm V2= 90.0 mL T2 = ????

LecturePLUS Timberlake 23

Solution

Solve for T2

Enter data

T2 = 302 K x atm x mL = K

atm mL

T2 = K - 273 = °C

LecturePLUS Timberlake 24

Calculation

Solve for T2

T2 = 302 K x 3.20 atm x 90.0 mL = 604 K

0.800 atm 180.0 mL

T2 = 604 K - 273 = 331 °C

LecturePLUS Timberlake 25

Learning Check C2

A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the temperature in °C when the gas has a volume of 0.315 L and a pressure of 802 mm Hg?

LecturePLUS Timberlake 26

Solution G9

T1 = 308 K T2 = ?

V1 = 675 mL V2 = 0.315 L = 315 mL

P1 = 0.850 atm P2 = 802 mm Hg = 646 mm Hg

T2 = 308 K x 802 mm Hg x 315 mL

646 mm Hg 675 mL

P inc, T inc V dec, T dec

= 178 K - 273 = - 95°C

LecturePLUS Timberlake 27

Volume and Moles

How does adding more molecules of a gas change the volume of the air in a tire?

If a tire has a leak, how does the loss of air (gas) molecules change the volume?

LecturePLUS Timberlake 28

Learning Check C3

True (1) or False(2)

1.___The P exerted by a gas at constant V is not affected by the T of the gas.

2.___ At constant P, the V of a gas is directly proportional to the absolute T

3.___ At constant T, doubling the P will cause the V of

the gas sample to decrease to one-half its original V.

LecturePLUS Timberlake 29

Solution C3

True (1) or False(2)

1. (2)The P exerted by a gas at constant V is not affected by the T of the gas.

2. (1) At constant P, the V of a gas is directly proportional to the absolute T

3. (1) At constant T, doubling the P will cause the V of

the gas sample to decrease to one-half its original V.

LecturePLUS Timberlake 30

Avogadro’s Law

When a gas is at constant T and P, the V is directly proportional to the number of moles (n) of gas

V1 = V2

n1 n2

initial final

LecturePLUS Timberlake 31

STP

The volumes of gases can be compared when they have the same temperature and pressure (STP).

Standard temperature 0°C or 273 K

Standard pressure 1 atm (760 mm Hg)

LecturePLUS Timberlake 32

Learning Check C4

A sample of neon gas used in a neon sign has a volume of 15 L at STP. What is the volume (L) of the neon gas at 2.0 atm and –25°C?

P1 = V1 = T1 = K

P2 = V2 = ?? T2 = K

V2 = 15 L x atm x K = 6.8 L

atm K

LecturePLUS Timberlake 33

Solution C4

P1 = 1.0 atm V1 = 15 L T1 = 273 K

P2 = 2.0 atm V2 = ?? T2 = 248 K

V2 = 15 L x 1.0 atm x 248 K = 6.8 L

2.0 atm 273 K

LecturePLUS Timberlake 34

Molar Volume

At STP

4.0 g He 16.0 g CH4 44.0 g CO2

1 mole 1 mole 1mole (STP) (STP) (STP)

V = 22.4 L V = 22.4 L V = 22.4 L

LecturePLUS Timberlake 35

Molar Volume Factor

1 mole of a gas at STP = 22.4 L

22.4 L and 1 mole

1 mole 22.4 L

LecturePLUS Timberlake 36

Learning Check C5A.What is the volume at STP of 4.00 g of CH4?

1) 5.60 L 2) 11.2 L 3) 44.8 L

B. How many grams of He are present in 8.0 L of

gas at STP?

1) 25.6 g 2) 0.357 g 3) 1.43 g

LecturePLUS Timberlake 37

Solution C5A.What is the volume at STP of 4.00 g of CH4?

4.00 g CH4 x 1 mole CH4 x 22.4 L (STP) = 5.60 L

16.0 g CH4 1 mole CH4

B. How many grams of He are present in 8.0 L of gas at STP?

8.00 L x 1 mole He x 4.00 g He = 1.43 g He

22.4 He 1 mole He

LecturePLUS Timberlake 38

Daltons’ Law of Partial PressuresPartial Pressure

Pressure each gas in a mixture would exert if it were the only gas in the container

Dalton's Law of Partial Pressures

The total pressure exerted by a gas mixture is the sum of the partial pressures of the gases in that mixture.

PT = P1 + P2 + P3 + .....

LecturePLUS Timberlake 39

Gases in the AirThe % of gases in air Partial pressure (STP)

78.08% N2 593.4 mmHg

20.95% O2 159.2 mmHg

0.94% Ar 7.1 mmHg

0.03% CO2 0.2 mmHg

PAIR = PN + PO + PAr + PCO = 760 mmHg 2 2 2

Total Pressure 760 mm Hg

LecturePLUS Timberlake 40

Learning Check C6

A.If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O2 in

the air?

1) 35.6 2) 156 3) 760

B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N2 in the air?

1) 557 2) 9.14 3) 0.109

LecturePLUS Timberlake 41

Solution C6

A.If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O2 in

the air?

2) 156

B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N2 in the air?

1) 557

LecturePLUS Timberlake 42

Partial Pressures

The total pressure of a gas mixture depends

on the total number of gas particles, not on

the types of particles.

P = 1.00 atm P = 1.00 atm

0.5 mole O2

+ 0.3 mole He+ 0.2 mole Ar

1 mole H2

LecturePLUS Timberlake 43

Health NoteWhen a scuba diver is several hundred feet

under water, the high pressures cause N2 from the

tank air to dissolve in the blood. If the diver rises too fast, the dissolved N2 will form bubbles in the

blood, a dangerous and painful condition called "the bends". Helium, which is inert, less dense, and does not dissolve in the blood, is mixed with O2 in

scuba tanks used for deep descents.

LecturePLUS Timberlake 44

Learning Check C7

A 5.00 L scuba tank contains 1.05 mole of O2

and 0.418 mole He at 25°C. What is the partial pressure of each gas, and what is the total pressure in the tank?

LecturePLUS Timberlake 45

Solution C7

P = nRT PT = PO + PHe

V 2

PT = 1.47 mol x 0.0821 L-atm x 298 K

5.00 L (K mol)

= 7.19 atm

Lecture PLUS Timberlake 2000 46

Molar Mass of a gas

What is the molar mass of a gas if 0.250 g of the gas occupy 215 mL at 0.813 atm and 30.0°C?

n = PV = (0.813 atm) (0.215 L) = 0.00703 mol

RT (0.0821 L-atm/molK) (303K)

Molar mass = g = 0.250 g = 35.6 g/mol

mol 0.00703 mol

Lecture PLUS Timberlake 2000 47

Density of a Gas

Calculate the density in g/L of O2 gas at STP. From STP, we know the P and T.

P = 1.00 atm T = 273 K

Rearrange the ideal gas equation for moles/L

PV = nRT PV = nRT P = n

RTV RTV RT V

Lecture PLUS Timberlake 2000 48

Substitute

(1.00 atm ) mol-K = 0.0446 mol O2/L

(0.0821 L-atm) (273 K)

Change moles/L to g/L

0.0446 mol O2 x 32.0 g O2 = 1.43 g/L

1 L 1 mol O2

Therefore the density of O2 gas at STP is

1.43 grams per liter