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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 ----------
mol
28 g
1
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 = gfm
S 1 x 32 = 32O 3 x 16 = 48
80
volume = 22.4 L