Chapter 11

The Behavior of Gases

Kinetic Theory

• Kinetic Theory – all molecules are in constant motion.– Collisions between gas molecules are perfectly

elastic.• Diffusion – movement of molecules from

areas of high concentration to low concentration.

• Rate of diffusion – the size and mass of the molecule.– Smaller, lighter molecules move faster.

Pressure

• Gas pressure – due to collisions of gas molecules on an object.

• Atmospheric pressure – due to collisions of air molecules on an object.– 1 atm = 760 mm Hg = 30 in Hg = 14.7 psi

• Partial pressure – the portion of pressure that one gas contributes to the total pressure in a mixture of gases.

Dalton’s Law of Partial Pressure

• The total pressure of a mixture of gases is equal to the sum of the partial pressures.

• PT = P1 + P2 + P3

• Pair = PN2 + PO2 + PCO2

Pressure vs. Moles (at constant volume)

• Same volume containers at constant temperature:

• If 1 mole of gas exerts 1 atm of pressure and we add another mol of gas twice as many particle will have twice as many collisions exert twice the pressure (2atm)

1mol: 2 mol. Directly proportional # moles, P

Pressure vs. Volume (at constant Temperature)

Start with 1 L of gas at 1 atm.

P V1 1.5 22 .5

V P½ volume 2x P

V P2x volume ½ P

Pressure vs. Volume (at constant Temperature)

- As volume decreases, the pressure increases proportionally.

- As volume increases, the pressure decreases proportionally.

- As one goes up, the other goes down: P and V are Inversely Proportional.

- P1V1 = P2V2

Boyle’s Law

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

P1V1 = P2V2

Heat the gas themolecules speed upand hit the top, pushing it tomaintain constant pressure.

Volume vs. Temperature (at constant Pressure)

Start with 1 L of gas at 100 K and 1 atm.

K = oC + 273

200 K = 2 L2x T 2x V

T = V

Cool the gas themolecules slowdown, fewer collisions w/the top so it falls.

Volume vs. Temperature (at constant Pressure)

Start with 1 L of gas at 100 K and 1 atm.

K = oC + 273

50 K = ½ L½x T ½x V

T = V

Charles’ Law

• For a given mass of gas, at constant pressure, the volume of the gas varies directly with its Kelvin temperature.

V1T2 = V2T1

Pressure vs. Temperature(at constant volume)

Start w/ 1 L at 100 K and 1 atm.

Heat the gas themoles speed up andincrease the # of collisions,which increases the pressure.

2x T = 2x P T = P

Pressure vs. Temperature(at constant volume)

Start w/ 1 L at 100 K and 1 atm.

Cool the gas themoles slow down anddecrease the # of collisions,which decreases the pressure.

½x T = ½x P T = P

Gay-Lusaac’s Law

• For a given mass of gas, at constant volume, the pressure of the gas varies directly with its Kelvin temperature.

P1T2 = P2T1

Combined Gas Law

• Combines Boyle’s, Charles’, and Gay-Lusaac’s Laws into one equation.

•P1V1T2 = P2V2T1

• When using the combined gas law, UNIT MUST AGREE and all temperatures must be in Kelvin.

Moles Meets Gas Laws

• We know that the volume of a gas is proportional to its number of particles and the pressure of a gas is proportional to its number of particles, which means:

• V~ # mol and P ~ # mol or

V ~ n and P ~ n

Moles Meets Gas Laws

• We also know that if the temperature of a gas increases, its pressure increases and if the temperature of a gas increases, its volume increases. This means:

• T ~ P and T ~ V • so we can write PV ~ nT

Moles Meet Gas Laws

• In order to make this proportion useful as a mathematical expression we can derive a constant by solving PV/nT using the values for 1 mole of a gas at STP. This constant will be called “R”.

• Substituting into the equation we get:

( atm) ( L)( mol) ( K)

1 22.41 273

= .0821 atm Lmol K = R

Ideal Gas Law

• PV = nRT• When using this equation, units MUST be

the same as those of the R value therefore:– Pressure must be in ________– Volume must be in _____– n must be in ________– Temperature must be in _____

atmL

molK

The Ideal Gas Law applies to real andIdeal gases under ALL conditions.

Pressure Conversions

1 atm = 760 mm Hg = 30 in Hg = 14.7 psi = 101.3 kPa

Problem 1

• .05 moles of a gas at a temperature of 20oC is contained in a 150 mL vessel. What is the pressure of this gas inside the vessel?

P =V = n =R = T =

150 mL = .150 L.05 mol.0821 atmL/mol K20oC + 273 = 293 K

PV = nRT

Problem 1: Answer

P =V = n =R = T =

.150 L

.05 mol

.0821293 K

PV = nRT

P(.150) = (.05)(.0821)(293)

P = atm8.02

Problem 2

• How many grams of bromine gas at – 10oC and 1277 mm Hg would be contained in a 3000 mL vessel?

P =V = n =R = T =

1277 mm x (1 atm/760 mm) =3000 mL = 3 L

.0821 atmL/mol K-10oC + 273 = 263 K

1.68 atm

Problem 2: Answer

P =V = n =R = T =

3 L

.0821263 K

1.68 atm PV = nRT(1.68)(3) = n(.0821)(264)

5.04 = 21.59n n = .23 mol Br2

Br2 = 2(80) = 160 g

.23 mol x = 160 g1 mol 36.8 g

Problem 3

• 110 g of carbon monoxide at a pressure of 35.4 in Hg and a volume of 782 mL would be at what temperature? Express your answer in degrees Celsius.

P =V = n =R = T =

35.4 in x (1 atm/30 in) =782 mL = .782 L

.0821 atmL/mol K3.93 mol

1.18 atm

C 1 x 12 = 12O 1 x 16 = 16

= 28110 g COx ----------

mol28 g1

Problem 3: Answer

P =V = n =R = T =

1.18 atm.782 L3.93 mol.0821

PV = nRT

(1.18)(.782) = (3.93)(.0821)T

T = K2.86.92 = .32T

oC = 2.86 K – 273 = -270.14 oC

Real vs. Ideal Gases

• Ideal Gas• Follows the gas laws

at all conditions of temp. and pressure.

• Particles are infinitely small (have no vol.)

• Particles are not attracted to one another.

• DO NOT EXIST!

• Real Gas• Do not follow gas

laws at all conditions of temp. and pressure.

• Particles have volume.

• Particles may attract one another when very close.

Real Gases

• Conditions at which real gases do NOT behave as ideal gases and therefore do not obey the gas laws:

1. At extremely high pressures do not obey Boyle’s Law.

2. At extremely low temperatures do not obey Charles’ Law.

Reasons:

• This occurs because under these two conditions the gas molecules are close enough together that they begin to exert forces on one another and behave similarly to a liquid.

• Gas Law equations are still extremely useful because under common conditions the behavior of a real gas is the same as the behavior of an ideal gas.

Density and Molecular Weight of Gases

• Density (D) = mass/volume = m/V = g/L

• Molecular Weigh (MW) = gram/mol

• For gases we know that at STP :– 1 mol = gfm = 22.4 L = 6.02x1023 molecules

– STP is defined as _________ and _________.1 atm 0 oC

Problem 1

• What is the density of a gas with a mass of 28 g and a volume 31 L? What is its MW?

D = M =V

28 g31 L

= .9 g/L

MW = g = mol

28 g31 L

x 22.4 L

1 mol

= 20.23 g/mol

Problem 2

• Calculate the molecular weight of a gas with a mass of 45 g and a volume of 6.8 L.

MW = g = mol

45 g6.8 L

x 22.4 L

1 mol

= 148.24 g/mol

Problem 3

• What is the density of oxygen at STP?

D = MV

= 1.43 g/L

32 g22.4 L

D =

O2: mass = gfm mass = 2(16)

mass = 32 g

volume = 22.4 L

Problem 4

• What is the density of sulfur trioxide at STP?

D = MV

= 3.57 g/L

80 g22.4 L

D =

SO3: mass = gfmS 1 x 32 = 32O 3 x 16 = 48

80

volume = 22.4 L