Chapter 5The Gaseous

State

5 | 2

Gases differ from liquids and solids:They are compressible.

Pressure, volume, temperature, and amount are related.

5 | 3

Pressure, P– The force exerted per unit area.– The SI unit for pressure is the pascal, Pa.

A

FP

5 | 4

Empirical Gas Laws– All gases behave quite simply with respect

to temperature, pressure, volume, and molar amount. By holding two of these physical properties constant, it becomes possible to show a simple relationship between the other two properties.

– The studies leading to the empirical gas laws occurred from the mid-17th century to the mid-19th century.

5 | 5

Boyle’s Law– The volume of a sample of gas at constant

temperature varies inversely with the applied pressure.

– The mathematical relationship:

– In equation form:

PV

1

ffii

constant

VPVP

PV

2211 VPVP

5 | 6

When a 1.00-g sample of O2 gas at 0°C is placed in a container at a pressure of 0.50 atm, it occupies a volume of 1.40 L.

When the pressure on the O2 is doubled to 1.0 atm, the volume is reduced to 0.70 L, half the original volume.

?

5 | 7

A volume of oxygen gas occupies 38.7 mL at 751 mmHg and 21°C. What is the volume if the pressure changes to 359 mmHg while the temperature remains constant?

Vi = 38.7 mLPi = 751 mmHgTi = 21°C

Vf = ?Pf = 359 mmHgTf = 21°C

f

iif P

VPV

2

112 P

VPV

5 | 8

Vi = 38.7 mLPi = 751 mmHgTi = 21°C

Vf = ?Pf = 359 mmHgTf = 21°C

fV

= 81.0 mL

f

iif P

VPV

38.7 mL 751 mm Hg

359 mm Hg

2

112 P

VPV

5 | 9

Charles’s Law– The volume of a sample of gas at constant

pressure is directly proportional to the absolute temperature (K).

– The mathematical relationship:

– In equation form:

TV

f

f

i

i

constant

T

V

T

V

T

V

ViTf = VfTi

V1T2 = V2T1

5 | 10

A balloon was immersed in liquid nitrogen (black container) and is shown immediately after being removed. It shrank because air inside contracts in volume.

As the air inside warms, the balloon expands to its orginial size.

5 | 11

A 1.0-g sample of O2 at a temperature of 100 K and a pressure of 1.0 atm occupies a volume of 0.26 L.

When the absolute temperature of the sample is raised to 200 K, the volume of the O2 is doubled to 0.52 L.

?

5 | 12

You prepared carbon dioxide by adding HCl(aq) to marble chips, CaCO3. According to your calculations, you should obtain 79.4 mL of CO2 at 0°C and 760 mmHg. How many milliliters of gas would you obtain at 27°C?

Vi = 79.4 mLPi = 760 mmHgTi = 0°C = 273 K

Vf = ?Pf = 760 mmHgTf = 27°C = 300. K

i

iff T

VTV

2

122 T

VTV

300 K 79.4 mL

273 K

5 | 13

Vi = 79.4 mLPi = 760 mmHgTi = 0°C = 273 K

Vf = ?Pf = 760 mmHgTf = 27°C = 300. K

fV

= 87.3 mL

i

iff T

VTV

1

122 T

VTV

The Empirical Gas Laws

Gay-Lussac’s Law: – The pressure exerted by a gas at constant

volume is directly proportional to its absolute temperature.

P Tabs (constant moles and V)

i

i

f

f

TP

TP P2T1 = P1T2

A Problem to Consider

Presentation of Lecture Outlines, 5–15

An aerosol can has a pressure of 1.4 atm at 25°C. What pressure would it attain at 1200°C, assuming the volume remained constant?

atm9.6Pf

i

fi

T

TP

f P

1.4 atm 1473 K

298 KfP

1

21

2P TTP

5 | 16

Combined Gas Law– Use this law in the event that all three parameters,

P, V, and T, are changing– The volume of a sample of gas at constant pressure

is inversely proportional to the pressure and directly proportional to the absolute temperature.

– The mathematical relationship:

– In equation form:

P

TV

f

ff

i

ii

constant

T

VP

T

VPT

PV

P1V1T2 = P2 V2T1

?

5 | 17

Divers working from a North Sea drilling platform experience pressure of 5.0 × 101 atm at a depth of 5.0 × 102 m. If a balloon is inflated to a volume of 5.0 L (the volume of the lung) at that depth at a water temperature of 4°C, what would the volume of the balloon be on the surface (1.0 atm pressure) at a temperature of 11°C?

Vi = 5.0 LPi = 5.0 × 101 atmTi = 4°C = 277 K

Vf = ?Pf = 1.0 atmTf = 11°C = 284 K

5 | 18

Vi = 5.0 LPi = 5.0 × 101 atmTi = 4°C = 277 K

Vf = ?Pf = 1.0 atmTf = 11°C = 284 K

fi

iiff PT

VPTV

fV

= 2.6 x 102 L

284 K 5.0 x 101 atm 5.0 L

277 K 1.0 atm

21

1122 PT

VPTV

5 | 19

Avogadro’s Law– Equal volumes of any two gases at the

same temperature and pressure contain the same number of molecules.

– So, the volume of a sample of gas is directly proportional to the number of moles of gas, n.

nV

5 | 20

Standard Temperature and Pressure (STP)– The reference condition for gases, chosen by

convention to be exactly 0°C and 1 atm pressure.

– The molar volume, Vm, of a gas at STP is 22.4 L/mol.

The volume of the yellow box is 22.4 L. To its left is a basketball.

The Ideal Gas Law

From the empirical gas laws, we see that volume varies in proportion to pressure, absolute temperature, and moles.

Law sBoyle' 1/PV

Law sAvogadro' nV Law Charles' TV abs

The Ideal Gas LawThis implies that there must exist a proportionality constant governing these relationships.

– Combining the three proportionalities, we can obtain the following relationship.

)( PnTabs R""V

where “R” is the proportionality constant referred to as the ideal gas constant.

The Ideal Gas LawThe numerical value of R can be derived using Avogadro’s law, which states that one mole of any gas at STP will occupy 22.4 liters.

nTVP R

KmolatmL 0.0821

22.4 L 1.00 atm

1 mol 273 K

5 | 24

Ideal Gas Law

The ideal gas law is given by the equation

PV=nRT

The molar gas constant, R, is the constant of proportionality that relates the molar volume of a gas to T/P.

A Problem to ConsiderAn experiment calls for 3.50 moles of chlorine, Cl2. What volume would this be if the gas volume is measured at 34°C and 2.45 atm?

PnRT V since

atm 2.45K) )(307 1mol)(0.082 (3.50 Kmol

atmL

V then

L 36.0 V then

?

17.1 atm 50.0 L mol K 28.02 g

.08206 L atm 296 K 1 mol

5 | 26

A 50.0-L cylinder of nitrogen, N2, has a pressure of 17.1 atm at 23°C. What is the mass of nitrogen in the cylinder?

V = 50.0 LP = 17.1 atmT = 23°C = 296 K

RT

PVn

n

mass = 986 g

Gas Density and Molar Mass

Using the ideal gas law, it is possible to calculate the moles in 1 L at a given temperature and pressure. The number of moles can then be converted to grams (per liter).

To find molar mass, find the moles of gas, and then find the ratio of mass to moles.

5 | 27

5 | 28

RT

P(MM)d

nRT PV

RTMM

m PV

V

md RT

V

mP(MM)

dRTP(MM)

P

dRT MM

A Problem to ConsiderA 15.5 gram sample of an unknown gas occupied a volume of 5.75 L at 25°C and a pressure of 1.08 atm. Calculate its molecular mass.

PV

mRT MM

g/mol 61.0 MM

15.5 g .08206 L atm 298 K

K mol 1.08 atm 5.75 L

?

5 | 30

What is the density of methane gas (natural gas), CH4, at 125°C and 3.50 atm?

MM = 16.04 g/molP = 3.50 atmT = 125°C = 398 K

RT

MMd

P

d

L

g1.72d

16.05 g 3.50 atm mol K

mol .08206 L atm 398 K

?

5 | 31

A 500.0-mL flask containing a sample of octane (a component of gasoline) is placed in a boiling water bath in Denver, where the atmospheric pressure is 634 mmHg and water boils at 95.0°C.

The mass of the vapor required to fill the flask is 1.57 g.

What is the molar mass of octane?

(Note: The empirical formula of octane is C4H9.)

What is the molecular formula of octane?

1.57 g .08206 L atm 368 K 760 mm Hg

mol K .5000 L 634 mm Hg 1 atm

P = 634 mmHg

P

dRT MM

mol

g114 MM

m = 1.57 g V = 500.0 mL = .5000 L

T = 95.0°C = 368.2 K

MM

V

md

VP

mRT MM

5 | 33

257.13

114n

Molecular formula: C8H18

Molar mass = 114 g/molEmpirical formula: C4H9

Empirical formula molar mass = 57.13 g/mol

5 | 34

Stoichiometry and Gas Volumes– Use the ideal gas law to find moles from a

given volume, pressure, and temperature, and vice versa.

?

5 | 35

When a 2.0-L bottle of concentrated HCl was spilled, 1.2 kg of CaCO3 was required to neutralize the spill. What volume of CO2 was released by the neutralization at 735 mmHg and 20.°C?

5 | 36

First, write the balanced chemical equation:

CaCO3(s) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g)

Moles of CO2 produced = 11.9992007 mol

Second, calculate the moles of CO2 produced:

1.201 kg CaCO3 1000 g CaCO3 1 moles CaCO3 1 mole CO2

1 kg CaCO3 100.09 CaCO3 1 mol CaCO3

11.9992007 mol .08206 L atm 293 K 760 mm Hg

K mol 735 mm Hg 1 atm

5 | 37

n = 11.9992007 molP = 735 mmHgT = 20°C = 293 K

P

nRTV

= 298 L

5 | 38

Gas Mixtures– Dalton found that in a mixture of unreactive

gases each gas acts as if it were the only the only gas in the mixture as far as pressure is concerned.

5 | 39

Originally (left), flask A contains He at 152 mmHg and flask B contains O2 at 608 mmHg. Flask A is then filled with oil forcing the He into flask B (right). The new pressure in flask B is 760 mmHg

5 | 40

Partial Pressure– The pressure exerted by a particular gas in

a mixture.

Dalton’s Law of Partial Pressures– The sum of the partial pressures of all the

different gases in a mixture is equal to the total pressure of the mixture:

– PTotal = PA + PB + PC + . . .

?

5 | 41

A 100.0-mL sample of air exhaled from the lungs is analyzed and found to contain:

0.0830 g N2,

0.0194 g O2,

0.00640 g CO2, and

0.00441 g water vapor at 35°C.

What is the partial pressure of each component and the total pressure of the sample?

5 | 42

nRT PV V

nRTP

(MM)V

mRTP

0.0830 g N2 1 mol N2 .08206 L atm 308 K 1000 mL

28.02 g N2 K mol 100.0 mL 1 L

.74867303 atm N2

0.0194 g O2 1 mol O2 .08206 L atm 308 K 1000 mL

32.00 g O2 K mol 100.0 mL 1 L

.153226535 atm O2

0.00640 g CO2 1 mol CO2 .08206 L atm 308 K 1000 mL

44.01 g CO2 K mol 100.0 mL 1 L

.0367545266 atm CO2

5 | 43

OHCOON 2222PPPPP

P = 1.0005078412 atm

0.00441 g H2O 1 mol H2O .08206 L atm 308 K 1000 mL

18.02 g H2O K mol 100.0 mL 1 L

.0618537496 atm H2O

.74867303 atm N2

.153226535 atm O2

.0367545266 atm CO2

.0618537496 atm H2O

P = 1.001 atm

5 | 44

Collecting Gas Over Water– Gases are often collected over water. The

result is a mixture of the gas and water vapor.

– The total pressure is equal to the sum of the gas pressure and the vapor pressure of water.

– The partial pressure of water depends only on temperature and is known (Table 5.6).

– The pressure of the gas can then be found using Dalton’s law of partial pressures.

5 | 45

The reaction of Zn(s) with HCl(aq) produces hydrogen gas according to the following reaction:

Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)

The next slide illustrates the apparatus used to collect the hydrogen. The result is a mixture of hydrogen and water vapor.

5 | 46

5 | 47

5.6)Table(See

mmHg16.5P

C,19at mmHg769

OH2

P If

mmHg16.5mmHg7692

22

22

H

OHH

OHH

P

PPP

PPP

mmHg752.52H P

mmHg7532H P