BEHAVIOR OF GASES

•Gases have weight

•Gases take up space

•Gases exert pressure

•Gases fill their containers

Gases doing all of these things!

Kinetic Theory of GasesThe basic assumptions of the kinetic molecular theory are:

Gases are mostly empty space

The molecules in a gas are separate, very small and very far apart

Kinetic Theory of GasesThe basic assumptions of the kinetic molecular theory are:

Gas molecules are in constant, chaotic motion

Collisions between gas molecules are elastic (there is no energy gain or loss)

Kinetic Theory of GasesThe basic assumptions of the kinetic molecular theory are:

The average kinetic energy of gas molecules is directly proportional to the absolute temperature

Gas pressure is caused by collisions of molecules with the walls of the container

Measurements of Gases

To describe a gas, its volume, amount, temperature, and pressure are measured.• Volume: measured in L, mL, cm3 (1 mL = 1 cm3)• Amount: measured in moles (mol), grams (g)• Temperature: measured in KELVIN (K)

•K = ºC + 273• Pressure: measured in mm Hg, torr, atm, etc.

•P = F / A (force per unit area)

Moderate Force (about 100 lbs)

Small Area (0.0625 in2)

Enormous Pressure (1600

psi)P = F /A

Bed of Nails

Large Surface Area (lots of nails)

Moderate ForceSmall

Pressure

P = F / A

Units of Pressure

Units of Pressure: 1 atm = 760 mm Hg 1 atm = 760 torr 1 atm = 1.013 x 105 Pa 1 atm = 101.3 kPa 1 atm = 1.013 bar

Boyle’s Law

As P, V (when T and n are constant) and vice versa…. INVERSE RELATIONSHIP

V 1/P

P1V1 = P2V2

For a given number of molecules of gas at a constant temperature, the volume of the gas varies inversely with the pressure.

Example: A sample of gas occupies 12 L under a pressure of 1.2 atm. What would its volume be if the pressure were increased to 3.6 atm? (assume temp is constant)

P1V1 = P2V2

(1.2 atm)(12 L) = (3.6 atm)V2

V2 = 4.0 L

Charles’ LawJacques Charles (1746-1828)

The volume of a given number of molecules

is directly proportional to the Kelvin temperature.

As T, V (when P and n are constant) and vice versa…. DIRECT RELATIONSHIP

V T2

2

1

1

T

V

T

V

Example: A sample of nitrogen gas occupies 117 mL at 100.°C. At what temperature would it occupy 234 mL if the pressure does not change? (express answer in K and °C)

V1 / T1= V2 / T2

(117 mL) / (373 K) = (234 mL) / T2

T2 = 746 K

T2 = 473 ºC

Combined gas law

2

22

1

11

T

VP

T

VP

This is for one gas undergoing changing conditions of temp, pressure, and volume.

Combining Boyle’s law (pressure-volume) with Charles’ Law (volume-temp):

Example 1: A sample of neon gas occupies 105 L at 27°C under a pressure of 985 torr. What volume would it occupy at standard conditions?

P1 = 985 torr

V1 = 105 L

T1 = 27 °C = 300. K

P2 = 1 atm = 760 torr

V2 = ?

T2 = 0 °C = 273 K

P1V1T2 = P2V2T1(985 torr)(105 L)(273K) = (760torr)(V2)

(300K)V2= 124 L

Example 2: A sample of gas occupies 10.0 L at 240°C under a pressure of 80.0 kPa. At what temperature would the gas occupy 20.0 L if we increased the pressure to 107 kPa?

P1 = 80.0 kPa

V1 = 10.0 L

T1 = 240 °C = 513 K

P2 = 107 kPa

V2 = 20.0 L

T2 = ?

P1V1T2 = P2V2T1(80.0kPa)(10.0L)(T2) = (107kPa)(20.0L)

(513K)T2= 1372K≈ 1370K

Example 3: A sample of oxygen gas occupies 23.2 L at 22.2 °C and 1.3 atm. At what pressure (in mm Hg) would the gas occupy 11.6 L if the temperature were lowered to 12.5 °C?

P1 = 1.3 atm = 988 mmHg

V1 = 23.2 L

T1 = 22.2 °C = 295.2 K

P2 = ?

V2 = 11.6 L

T2 = 12.5 °C = 285.5 K

P1V1T2 = P2V2T1(988mm Hg)(23.2L)(285.5K) = (P2)(11.6L)

(295.2K)P2= 1938 mm Hg ≈ 1900 mmHg

Gases: Standard Molar Volume & The Ideal Gas Law

Avogadro’s Law: at the same temperature and pressure, equal volumes of all gases contain the same # of molecules (& moles).

Standard molar volume = 22.4 L @STP

This is true of “ideal” gases at reasonable temperatures and pressures ,the behavior of many “real” gases is nearly ideal.

Example: 1.00 mole of a gas occupies 36.5 L, and its density is 1.36 g/L at a given temperature & pressure.

a) What is its molecular weight (molar mass)?

L

g 36.1

mol 1.00

L 5.36g/mol 6.49

Example: 1.00 mole of a gas occupies 36.5 L, and its density is 1.36 g/L at a given temperature & pressure.

b) What is the density of the gas under standard conditions?

mol

g 6.49

L 22.4

mol 1g/L .212

The IDEAL GAS LAW Shows the relationship among the pressure,

volume, temp. and # moles in a sample of gas.

P = pressure (atm)V = volume (L)n = # moles T = temp (K)R = universal gas constant = 0.0821 Kmol

atmL

The units of R depend on the units

used for P, V & T

Example 1: What volume would 50.0 g of ethane, C2H6, occupy at 140 ºC under a pressure of 1820 torr?

P = (1820 torr)(1 atm/760 torr) = 2.39 atm V = ? n = (50.0 g)(1 mol / 30.0 g) = 1.67 mol T = 140 °C + 273 = 413 K

PV = nRT

(2.39 atm)(V) = (1.67 mol)(0.0821 L·atm/mol·K)(413 K)

V = 23.6 L

Example 2: Calculate (a) the # moles in, and (b) the mass of an 8.96 L sample of methane, CH4, measured at standard conditions.

P = 1.00 atm V = 8.96 L n = ? T = 273 K

PV = nRT

(1 atm)(8.96 L) = (n)(0.0821 L·atm/mol·K)(273 K)

n = 0.400 mol

(a)

Example 2: Calculate (a) the # moles in, and (b) the mass of an 8.96 L sample of methane, CH4, measured at standard conditions.

Or the easier way…

L96.8

L 4.22

mol 1mol 400.0

(a)

Example 2: Calculate (a) the # moles in, and (b) the mass of an 8.96 L sample of methane, CH4, measured at standard conditions.

Convert moles to grams…

0mol40.0

mol 1

g 0.164CH g 40.6

(b)

Example 3: Calculate the pressure exerted by 50.0 g ethane, C2H6, in a 25.0 L container at 25 ºC?

P = ? V = 25.0 L n = (50.0 g)(1 mol / 30.0 g) = 1.67 mol T = 25 °C + 273 = 298 K

PV = nRT

(P)(25.0 L) = (1.67 mol)(0.0821 L·atm/mol·K)(298 K)

P = 1.63 atm

Determining Molecular Weights & Molecular Formulas of Gases:

If the mass of a volume of gas is known, we can use this info. to determine the

molecular formula for a compound.

Example 1: A 0.109 g sample of pure gaseous compound occupies 112 mL at 100. ºC and 750. torr. What is the molecular weight of the compound?

First find # moles Then use mass to determine g/mol…

PV = nRT(0.99 atm)(0.112 L) = (n)(0.0821 L·atm/mol·K)(373 K) n = 0.00362

mol

mol 0.00362g 0.109

MW g/mol 2.30

Example 2: A compound that contains only C and H is 80.0% C and 20.0% H by mass. At STP, 546 mL of the gas has a mass of 0.732 g. What is the molecular (true) formula for the compound?

First find the empirical formula…

g 12.0

mol 1C g 80.0 C mol6.67

g 1.0

mol 1H g 20.0 H mol 20.0

1mol 67.6

3mol 67.6

Empirical formula = CH3

Example 2: A compound that contains only C and H is 80.0% C and 20.0% H by mass. At STP, 546 mL of the gas has a mass of 0.732 g. What is the molecular (true) formula for the compound?

Next, determine the MW of the sample…

0.024 mol

mol 0.024g 0.732

MW g/mol 0.30

L 4.22

mol 1L546.0n

Example 2: A compound that contains only C and H is 80.0% C and 20.0% H by mass. At STP, 546 mL of the gas has a mass of 0.732 g. What is the molecular (true) formula for the compound?

Finally, determine the molecular formula…

Empirical formula = CH3 15.0 g/mol

True MW = 30.0 g/mol

2g/mol 15g/mol 30.0

factor muliplying

CH3 x 2 = C2H6

Partial Pressures and Mole Fractions

In a mixture of gases each gas exerts the pressure it would exert if it occupied the volume alone.

The total pressure exerted by a mixture of gases is the sum of the partial pressures of the individual gases:

Ptotal = P1 + P2 + P3 + …

Example: If 100.0 mL of hydrogen gas, measured at 25C and 3.00 atm, and 100.0 mL of oxygen, measured at 25C and 2.00 atm, what sould be the pressure of the mixture of gases?

Ptotal = P1 + P2 + P3 + …

PT = 3.00 atm + 2.00 atm

PT = 5.00 atm

Notice the two gases are measured at the same temp. and vol.

Vapor Pressure of a Liquid

The pressure exerted by its gaseous molecules in equilibrium with the liquid; increases with temperature

Vapor Pressure of a Liquid

Patm = Pgas + PH2O

or

Pgas = Patm - PH2O

Vapor Pressure of a Liquid

Temp. (C)

v.p. of water(mm Hg)

Temp. (C)

v.p. of water(mm Hg)

18 15.48 21 18.65

19 16.48 22 19.83

20 17.54 23 21.07

See A-2 for a complete table

Example 1: A sample of hydrogen gas was collected by displacement of water at 25 C. The atmospheric pressure was 748 mm Hg. What pressure would the dry hydrogen exert in the same conditions?

PH2 = Patm - PH2O

PH2 =748 mm Hg – 23.76 mm Hg

PH2 = 724.24 mm Hg

PH2 724 mm Hg

Example 2: A sample of oxygen gas was collected by displacement of water. The oxygen occupied 742 mL at 27 C. The barometric pressure was 753 mm Hg. What volume would the dry oxygen occupy at STP?

PO2 = Patm - PH2O

PO2 =753 mm Hg – 26.74 mm Hg

PO2 = 726 mm Hg

P1V1T2 = P2V2T1

(726 mm Hg)(0.742 L)(273K) = (760 mm Hg)(V2)(300K)

V2 = 0.645 L

Example 3: A student prepares a sample of hydrogen gas by electrolyzing water at 25 C. She collects 152 mL of H2 at a total pressure of 758 mm Hg. Calculate:

(a) the partial pressure of hydrogen, and (b) the number of moles of hydrogen collected.

PH2 = Patm - PH2O

PH2 =758 mm Hg – 23.76 mm Hg

PH2 = 734 mm Hg

Example 3: A student prepares a sample of hydrogen gas by electrolyzing water at 25 C. She collects 152 mL of H2 at a total pressure of 758 mm Hg. Calculate:

(a) the partial pressure of hydrogen, and (b) the number of moles of hydrogen collected.

PV = nRT

(0.966 atm)(0.152 L) = (n)(0.0821 L·atm/mol·K)(298 K)

n = 0.00600 mol H2

Graham’s Law of Diffusion & Effusion

Where, Rate = rate of diffusion or effusion MM=molar mass

1

2

2

1

MMMM

raterate

Stoichiometry of Gaseous Reactions

A balanced equation can be used to relate moles or grams of substances taking part in a reaction. (AND VOLUME!)

Example: Hydrogen peroxide is the active ingredient in commercial preparations for bleaching hair. What mass of hydrogen peroxide must be used to produce 1.00 L of oxygen gas at 25 C and 1.00 atm?

2H2O2 O2 + 2H2O

PV = nRT(1.00 atm)(1.00 L) = (n)(0.0821 L·atm/mol·K)(298 K) n = 0.0409 mol

O2

2O mol 0409.0 2

22

O mol 1

OH mol 2

22OH mol 1

g 0.34

22OH g 78.2

Real Gases

&

Ideal Gases

The “Ideal” in the ideal gas law refers to the gas in question.

The assumption is that all gasses behave in ideal ways. In real life, such is not the case.

For the most precise work, we need to correct for this

deviation from ideal behavior.

The attractive forces among molecules operate at relatively short distances. For a gas at atmospheric pressure, the molecules are far apart and these attractive forces are negligible.

At high pressures, the density of the gas increases, and the molecules are much closer to one another.

Then intermolecular forces can become significant enough to affect the motion of the molecules, and the gas will no longer behave ideally.

Deviation from Ideal Behavior

Another way to observe the nonideality of gases is to lower the temperature. Cooling a gas decreases the molecules’ average kinetic energy, which in a sense deprives molecules of the drive they need to break away from their mutual attractive influences.

Hot is more ideal!

The nonideality of gases requires some modifications to the ideal gas law. Such

modifications were first made by me. J.D. van der Walls in

1873.

Dutch Physicist

The speed of a molecule that is moving toward the container wall is reduced by the attractive forces exerted by its neighbors. Consequently, the impact this molecule will make on the wall is not as great as it would be if no intermolecular forces were present.

Net Force

The measured gas pressure is always lower than the pressure the gas would have if it behaved ideally.

2

2

V

anPP realideal

The interaction between molecules that gives rise to non-ideal behavior depends on how frequently any two molecules approach each other closely.

The number of such “encounters” increases as the square of the number of molecules per unit volume, (n/V)2, because the presence of each of the two molecules in a particular region is proportional to n/V.

Pressure Correction

2

2

V

anPP realideal

The quantity Pideal is the pressure we would measure if there were no intermolecular attraction. The quantity a then is just a proportionally constant in the correction term for pressure.

Pressure Correction

Volume Correction

Each real molecule occupies a small intrinsic volume, so the effective volume of a gas becomes

(V-nb) where…

n is the number of moles

b is a constant (the volume of a single molecule)

The term nb represents the volume occupied by n moles of of the gas.

Having taken into account the corrections for pressure and volume, we can write the ideal gas law as …..

nRTnbVV

anP

2

2

This equation is known as van der Waals equation. The constants a and b are selected for each gas to give the best possible agreement between the equation and actual observed behavior.

Van der Waals Equation

Get Real!