lecture 12. gases and kmt
DESCRIPTION
CHEM 16TRANSCRIPT
GENERAL CHEMISTRY
Gases and the
Kinetic Molecular
Theory
LECTURE
12
Gases CONTENTS
14-1 Properties of Gases: Gas Pressure
14-2 The Simple Gas Laws
14-3 Combining the Gas Laws: The Ideal
Gas Equation and The General Gas
Equation
14-4 Applications of the Ideal Gas
Equation
14-5 Gases in Chemical Reactions
14-6 Mixtures of Gases
14-7 Kinetic—Molecular Theory of Gases
14-8 Gas Properties Relating to the
Kinetic—Molecular Theory
14-9 Nonideal (Real) Gases
The Distinction of Gases from
Liquids and Solids
1. Gas expand to fill and assume the shape of their
containers.
The Distinction of Gases from
Liquids and Solids
1. Gas expand to fill and assume the shape of their
containers.
3. Gases are invisible – no visible particles (except Cl2,
Br2, I2)
2. Most gases have relatively low densities under normal
conditions.
The gaseous state of three halogens (group 17)
The Distinction of Gases from
Liquids and Solids
1. Gas expand to fill and assume the shape of their
containers.
5. Gases are miscible.
3. Gases are invincible – no visible particles (except Cl2,
Br2, I2)
2. Most gases have relatively low densities under normal
conditions.
4. Some gases are combustible (H2, CH4) and some are
unreactive (He, Ne)
Gas % by Volume
N2 78.09
O2 20.94
Ar 0.93
CO2 0.03
He, Ne, Kr, Xe
0.002
CH4 0.00015
H2 0.00005
The Concept of Pressure
P (Pa) = Area (m2)
Force (N) Pressure
1.00 atm, 101.325 kPa, 1.01325 bar,
760 torr, 760 mm Hg
Units of Pressure
pascual (Pa) *SI unit
standard atmosphere (atm) 1 atm = 101.325 kPa
millimeter of mercury (mmHg)
Standard Atmospheric Pressure
1 torr = 1 mmHg = atm 1
760
bar (bar) 1 bar = 1000 Pa
Physical behavior of gases
1. Pressure, P
2. Volume, V
3. Temperature, T
4. Amount, (number of moles, n)
Relationship between gas volume and
pressure – Boyle’s Law
PV = constant P a 1 V
14-2 Simple Gas Laws
*(at constant T and n)
A molecular description of Boyle’s law.
Relationship between gas volume and
temperature- Charles’ Law
V = constant x T V a T
*(at constant P and n)
A molecular description of Charles’s law.
V n V = constant x n
Relationship between gas volume and amount of
gas- Avogadro’s Law *(at constant T and P)
A molecular description of Avogadro’s law.
Standard Temperature and Pressure (STP)
•IUPAC defines standard conditions of temperature
and pressure (STP).
•At STP:
P = 1 atm = 760 torr = 760 mmHg
T = 0°C = 273.15 K
Volume of 1 mol of an ideal gas = 22.414 L = 22.4 L
(standard molar volume)
14-3 Combining the Gas Laws:
The Ideal Gas Equation
and the General Gas Equation
• Boyle’s law V 1/P
• Charles’s law V T
• Avogadro’s law V n
V nT
P
PV nT
or
The Ideal Gas Equation
R = PV
nT
PV = nRT
Applying the ideal gas equation
The General Gas Equation
R = = P2V2
n2T2
P1V1
n1T1
= P2
T2
P1
T1
The amount and volume of
the gas are constant:
Using the Gas Laws
Exercise 1
A 50.0 L cylinder contains nitrogen gas at a
pressure of 21.5 atm. The contents of the cylinder
are emptied into an evacuated tank of unknown
volume. If the final pressure in the tank is 1.55 atm,
then what is the volume of the tank?
Exercise 1 (Continued)
•Use Boyle’s Law
P1V1 = P2V2 V2 = P1V1
P2
Vtank = 644 L
V2 = 21.5 atm x 50.0 L
1.55 atm
Exercise 2
A sample of hydrogen, H2, occupies 1.00 x 102 mL
at 25.0oC and 1.00 atm. What volume would it
occupy at 50.0oC under the same pressure?
T1 = 25oC + 273 = 298 K
T2 = 50oC + 273 = 323 K
Using Charles’ Law:
= V2
T2
V1
T1
V2 = V1T2
T1 =
100 mL x 323 K
298
V2 = 108 mL
Exercise 3
A steel tank used for fuel delivery is fitted with a safety valve
that opens if the internal pressure exceeds 1.00 x 103 torr. It
is filled with methane at 23oC and 0.991 atm and placed in
boiling water at exactly 100oC. Will the safety valve open?
P1 = 0.991 atm (convert to torr) P2 = unknown
T1 = 23oC (convert to K) T2 = 100oC (convert to K)
P1V1
n1T1
P2V2
n2T2 =
V and n remain constant
P1
T1
P2
T2 =
0.991 atm x 1 atm
760 torr = 753 torr
P2 = P1 x T2
T1 = 753 torr x
373 K
296 K
P2 = 949 torr
Exercise 4
• A steel tank has a volume of 438 L and is filled with
0.885 kg of O2. Calculate the pressure of O2 at 21oC.
V = 438 L T = 21oC (convert to K)
n = 0.885 kg O2 (convert to mol) P = unknown
21oC + 273.15 = 294 K
0.885 kg x 103 g
kg
mol O2
32.00 g O2 = 27.7 mol O2 x
P = nRT
V =
27.7 mol 294 K L-atm mol-K
0.0821 x x
438 L
P = 1.53 atm
14-4 Applications of the Ideal Gas
Equation
PV = nRT and n = m
M
PV = m
M
RT
M = m
PV
RT
Molar Mass Determination
Gas Density
= m
V
PV = m
M
RT
MP
RT V
m = =
KEEP IN MIND
that gas densities are
typically much smaller than
those of liquids and solids.
Gas densities are usually
expressed in grams per liter
rather than grams per
milliliter.
Sample Problem 5 Finding the Molar Mass of a Volatile Liquid
An organic chemist isolates a colorless liquid from a petroleum sample. She places the liquid in a flask and puts the flask in a boiling water bath, which vaporizes the liquid and fills the flask with gas. She closes the flask, reweighs it, and obtains the following data:
SOLUTION:
Volume (V) of flask = 213 mL
Mass of flask + gas = 78.416 g
T = 100.0oC
Mass of empty flask = 77.834 g
P = 754 torr
m = (78.416 - 77.834) g = 0.582 g
M = m RT
VP =
0.582 g atm•L
mol•K 0.0821 373.2 K x x
0.213 L x 0.992 atm = 84.4 g/mol
PV = nRT PV = m
M
RT M = m
PV
RT
Sample Problem 6 Calculating Gas Density
A chemical engineer uses waste CO2 from a manufacturing process, instead of chlorofluorocarbons, as a “blowing agent” in the production of polystyrene containers. Find the density (in g/L) of CO2 (a) at STP (0oC and 1 atm) and (b) at room conditions (20.oC and 1.00 atm).
SOLUTION: PV = nRT d =
RT
M x P
d = 44.01 g/mol x 1.00 atm
atm•L
mol•K 0.0821 x 273 K
= 1.96 g/L (a)
PV = m
M
RT
Sample Problem 6 (continued) Calculating Gas Density
(b) = 1.83 g/L d = 44.01 g/mol x 1.00 atm
x 293 K atm•L
mol•K 0.0821
A chemical engineer uses waste CO2 from a manufacturing process, instead of chlorofluorocarbons, as a “blowing agent” in the production of polystyrene containers. Find the density (in g/L) of CO2 (a) at STP (0oC and 1 atm) and (b) at room conditions (20.oC and 1.00 atm).
d = RT
M x P PV = nRT PV =
m
M
RT
12-5 Gases in Chemical Reactions
• Ideal gas equation relates the amount of a gas to
volume, temperature and pressure
• Stoichiometric factors relate gas quantities to
quantities of other reactants or products.
Sample Problem 7 Using Gas Variables to Find Amounts of Reactants or Products
Copper reacts with oxygen impurities in the ethylene used to produce polyethylene. The copper is regenerated when hot H2 reduces the copper(II) oxide, forming the pure metal and H2O. What volume of H2 at 765 torr and 225oC is needed to reduce 35.5 g of copper(II) oxide?
SOLUTION:
CuO(s) + H2(g) Cu(s) + H2O(g)
35.5 g CuO x mol CuO
79.55 g CuO
1 mol H2
1 mol Cu = 0.446 mol H2
0.446 mol H2 x
498 K atm•L
mol•K 0.0821 x
1.01 atm = 18.1 L
x V = nRT/P
PV = nRT
Sample Problem 8 Using the Ideal Gas Law in a Limiting-Reactant Problem
The alkali metals [Group 1A(1)] react with the halogens [Group 7A(17)] to form ionic metal halides. What mass of potassium chloride forms when 5.25 L of chlorine gas at 0.950 atm and 293 K reacts with 17.0 g of potassium?
SOLUTION: 2K(s) + Cl2(g) 2KCl(s)
n = PV
RT Cl2
5.25 L =
0.950 atm x
atm•L mol•K
0.0821 x 293 K = 0.207 mol Cl2
17.0 g K x 39.10 g K
1 mol K = 0.435 mol K
V = 5.25 L
T = 293K n = unknown
P = 0.950 atm
Cl2
Sample Problem 8 Using the Ideal Gas Law in a Limiting-Reactant Problem
SOLUTION:
0.207 mol Cl2 x 2 mol KCl 1 mol Cl2
= 0.414 mol KCl formed
0.435 mol K x 2 mol KCl
2 mol K = 0.435 mol KCl formed
0.414 mol KCl x 74.55 g KCl
mol KCl = 30.9 g KCl
continued
Cl2 is the limiting reactant.
12-6 Mixtures of Gases
• Gas laws apply to mixtures of gases.
• Partial pressure
–Each component of a gas mixture exerts a
pressure that it would exert if it were in the
container alone.
Dalton’s law of partial pressures
Ptotal = P1 + P2 + P3 + ...
Dalton’s Law of Partial Pressure
Partial Pressure
The mole fraction of a component in a mixture is described by the equation:
c1 =
n1 n1 + n2 + n3 +... =
n1 ntotal
P1= c1 x Ptotal where c1 is the mole fraction
Σcn = c1 + c2 + c3 +... = 1
Sample Problem 9 Applying Dalton’s Law of Partial Pressures
In a study of O2 uptake by muscle at high altitude, a physiologist prepares an atmosphere consisting of 79 mol N2, 17 mol 16O2, and 4.0 mol 18O2. (The isotope 18O will be measured to determine the O2 uptake.) The pressure of the mixture is 0.75 atm to simulate high altitude. Calculate the mole fraction and partial pressure of 18O2 in the mixture.
SOLUTION:
c 18O2 = 4.0 mol 18O2
100 = 0.040
= 0.030 atm P = c x Ptotal = 0.040 x 0.75 atm 18O2
18O2
P = c x Ptotal 18O2 18O2
c =
n
ntotal
18O2 18O2
A gaseous mixture contains 3.23 g of chloroform, CHCl3, and 1.22 g
of methane, CH4. Assuming that both compounds remains as
gases, what pressure is exerted by the mixture inside a 50.0-mL
metal container at 275 oC? What pressure is contributed by CHCl3?
Sample Problem 10 Applying Dalton’s Law of Partial Pressures
Ptotal= ntotalRT
V
3.23 g CHCl3 mol CHCl3
119.35 g x = 0.027 mol CHCl3
SOLUTION:
1.22 g CH4 mol CH4
16.05 g x = 0.076 mol CH4
ntotal = 0.103 mol
= 0.103 mol
atm•L
mol•K 0.0821 548 K x x
0.050 L
Ptotal = 92.68 atm
A gaseous mixture contains 3.23 g of chloroform, CHCl3, and 1.22 g
of methane, CH4. Assuming that both compounds remains as
gases, what pressure is exerted by the mixture inside a 50.0-mL
metal container at 275 oC? What pressure is contributed by CHCl3?
Sample Problem 10 (continued) Applying Dalton’s Law of Partial Pressures
n CHCl3 = 0.027 mol
n CH4 = 0.076 mol ntotal = 0.103 mol
Ptotal = 92.68 atm
PCHCl3= cCHCl3 x Ptotal SOLUTION:
cCHCl3 = 0.027 mol
0.103 mol = 0.262
PCHCl3= 0.0262 x 92.68 atm
PCHCl3= 24.29 atm
Collecting a gas over water
Ptot = Pbar = Pgas + PH2O
Pgas = Pbar - PH2O
Sample Problem 11 Calculating the Amount of Gas Collected over Water
Acetylene (C2H2), an important fuel in welding, is produced in the laboratory when calcium carbide (CaC2) reacts with water:
CaC2(s) + 2H2O(l) C2H2(g) + Ca(OH)2(aq)
For a sample of acetylene that is collected over water, total gas pressure (adjusted to barometric pressure) is 738 torr and the volume is 523 mL. At the temperature of the gas (23oC), the vapor pressure of water is 21 torr. How many grams of acetylene are collected?
0.943 atm 0.523 L x n
C2H2 =
atm•L
mol•K 0.0821 x 296 K
= 0.0203 mol
0.0203 mol x 26.04 g C2H2
mol C2H2 = 0.529 g C2H2
SOLUTION:
P C2H2 = 738 torr – 21 torr = 717 torr 717 torr x
atm
760 torr
= 0.943 atm
grams ? PV = nRT n= PV
RT moles grams
Visualizing Molecular Motion
12-7 Kinetic Molecular Theory of Gases
• Particles are point masses in constant,
random, straight line motion.
• Particles are separated by great
distances.
• Collisions are rapid and elastic.
• No force between particles.
• Total energy remains constant.
Distribution of molecular speeds at three temperatures.
Kinetic Energy and Temperature
NAm = Mw:
Ek = m x u 2 1
2
Ek = mass x speed2 1
2
Ek = T 3
2
R
NA
m u 2 = T 3
2
R
NA
1
2
u 2 = T 3
2
R
NAm
1
2
u 2 = 3RT
Mw M
3RTu rms
M
3RTu rms
Distribution of Molecular Speeds – the effect
of mass and temperature
14-8 Gas Properties Relating to the
Kinetic-Molecular Theory
A. Diffusion
–Migration of molecules of
different substances due to
random molecular motion.
B. Effusion
–Escape of gas molecules
from their container through
a tiny pinhole.
Graham’s Law
rate of effusion of A
rate of effusion of B
(urms )A
(urms )B
3RT/MA
3RT/MB
MB
MA
rateA
rateB =
√Mw B
√Mw A
Applying Graham’s Law of Effusion
A mixture of helium (He) and methane (CH4) is placed in an effusion apparatus. Calculate the ratio of their effusion rates (He: CH4).
SOLUTION:
M of CH4 = 16.04 g/mol M of He = 4.003 g/mol
CH4
He rate
rate = √ 16.04
4.003 = 2.002 /
Sample Problem 12
A sample of hydrogen, H2, was found to effuse through a pinhole 5.2 times as rapidly as the same volume of unknown gas (at the same temperature and pressure). What is the molecular weight of the unknown gas?
Applying Graham’s Law of Effusion Sample Problem 13
rateH2
rateunk =
√Munk
√MH2
5.2 =
√Munk
√2 g/mol
2
27.04 =
Munk
2 g/mol
Munk = 54.08 g/mol
SOLUTION:
12-9 Real Gases: Deviation from
Ideal Behavior
Gas particles have finite volume
Vreal gas > Videal gas
Presence of attractive force between
gas molecules
Preal gas < Pideal gas
Compressibility factor
PV/nRT = 1 for ideal gas.
Deviations for real gases.
PV/nRT > 1 - molecular volume is significant.
PV/nRT < 1 – intermolecular forces of attraction.
Deviation from Ideal Behavior
The behaviour of real gases – compressibility
factor as a function of pressure at 0°C
Gases tend to behave ideally at high temperatures and low pressure.
Gases tend to behave nonideally at low temperature and high pressure.
Behavior of Gases
van der Waals Equation
P + n2a
V2 V – nb = nRT
a - intermolecular attraction
b - volume of gas molecules
adjusts
pressure up
adjusts
volume down
A 1.74 g sample of a compound that contains only carbon and hydrogen contains 1.44 g of carbon and 0.300 g of hydrogen. At STP 101 mL of the gas has a mass of 0.262 gram. What is its molecular formula?
QUIZ
104252
52
HC)H(C :formulaMolecular
229
58.1
g/mol 1.58L 0.101atm 00.1
K 273K mol
atm L0.0821)(262.0
PV
RT g=MW
RTMW
g PV
nRT PV
29 = masswith HC
5.20.120
0.2971
0.120
0.120
H mol 297.0H g 1.01
H mol 1H g 0.300 = atoms H mol ?
C mol 120.0C g 0.12
C mol 1C g 1.44 = atoms C mol ?
g
HC
Answer
3 pts
4 pts
1 pt
Summary
• Simple gas laws and the Ideal gas equation
• Determining MW and densities of gases using the ideal gas equation
• Chemical reactions involving gases
• Mixtures of gases
• Kinetic molecular theory
• Rate of effusion
• Ideal gas vs. Real gases