1. Passing Gas Characteristics of Gases Gases expand to fill a container
Gases form homogeneous mixtures with other gases
Gases can be easily compressed
Readily measured properties of gas are its temperature,
volume, and pressure
5/19/2018

2. Kinetic Molecular Theory
1. Gases consist if large number of molecules that are in continuous, random motion
2. The volume of all molecules of the gas is negligible compared to the total volume
in which the gas is contained.
3. Attractive and repulsive forces between gas molecules are negligible
4. Energy can be transferred between molecules during collisions, but the average
kinetic energy of the molecules does not change with time, as long as the temperature
is constant. In other words collisions are perfectly elastic.
5. The average kinetic energy of the molecules is proportional to absolute temperature.
At any given temperature all gases have the same average kinetic energy
5/19/2018

3. Boy-Am I Under PressureF A
Pressure may be defined as a force exerted upon any surface:
Gravitational force
Mass = 1.06 kg
Area = 7.85 x 10-3 m2
Since F= mass x acceleration
F = 1.06 kg x 9.81 m/sec2 = 10.4 N where N is an SI unit called a Newton
with units of kg-m/s2
P = F A
10.4 N = =
1.32 x 103 N/m2 or pascals (Pa)
7.85 x 10-3 m2
5/19/2018

4. Atmospheric Pressure
P=F/A = (10,000 kg)(9.81 m/s2)/1 m2 = 1 x 105 Pa = kPa
5/19/2018

5. Atmospheric Pressure 760 mm Hg 760 torr 101.3 kPa 1 atm
Atmospheric pressure can be measured using a barometer
760 mm Hg
760 torr
101.3 kPa
1 atm
5/19/2018

6. Atmospheric Pressure
A manometer measures the pressures of encloses gases
For a given pressure difference, the difference in heights of the liquid levels in the two arms of the manometer is inversely proportional to the density of the liquid
A closed-tube manometer
measures pressures below
atmospheric pressure
An open-tube manometer
measures pressures near
atmospheric pressure
5/19/2018

7. Atmospheric Pressure
A manometer measures the pressures of encloses gases
A manometer is filled with dibutylphthalate (D = 1.05 g/ml) If the conditions
are such that h = 12.2 cm at atm, what is the pressure of the enclosed
gas in mm Hg (D =13.6 g/ml).
Patm = atm x
760 mm Hg =
733 mm Hg
1 atm
10 mm
Ph = cm x x
1.05 g
1 ml x =
9.4 mm Hg
1 cm
13.6 g ml
Because the pressure inside the manometer is less than atmospheric pressure, Pgas = P atm + Ph2
Pgas = = 742 mm Hg
5/19/2018

8. Boyle's Law
The volume of a fixed quantity of gas maintained at constant
temperature is inversely proportional to the pressure
5/19/2018

9. Boyle's Law Boyle’s Law: V  1/P, where (n, T constant) or PV = k
5/19/2018

10. Charles' Law Charles’s Law: V  T, where (n, P constant) or V/T = k
The volume of a fixed quantity of gas maintained at constant
pressure is inversely proportional to the temperature
5/19/2018

11. Gay-Lussac’s Law P1 = P2 volume and moles T1 T2 are constant (n, V)
5/19/2018

12. Examples
The pressure of nitrogen gas in a 12.0 L tank at 27°C is 2300 lb/in2. What
volume would the gas in the tank have at 1 atm pressure (14.7 lb/in2) if the
temperature remains unchanged?
Boyle’s Law: V  1/P, where (n, T constant) or PV = k
P1V1 = P2V2
(2300 lb/in2)(12.0 l) = (14.7 lb/in2)(V2)
V2 = 1880 L
The gas pressure in an aerosol can is 1.5 atm at 25° C. Assuming that the gas
inside obeys the ideal-gas equation what would be the pressure if the can were
heated to 450°C?
Gay-Lussac’s Law: P  T, where (n,V constant) or P/T = k
P P2 =
T T2
1.5 atm = P2 =
298 K = 723 K
5/19/2018
P2 = 3.6 atm

13. the Combined gas Law is:
Boyle’s Law: V  1/P, where (n, T constant) or PV = k
P1V1 = P2V2
Charles’s Law: V  T, where (n, P constant) or V/T = k
V V2 =
T T2
Gay-Lussac’s Law: P  T, where (n,V constant) or P/T = k
P P2 =
T T2
the Combined gas Law is:
P1V P2V2 =
T T2
5/19/2018

14. Combined Gas law
A quantity of helium gas occupies a volume of 16.5 L at 78°C and 45.6 atm.
What is its volume at STP?
The Combined gas Law is:
P1V P2V2 =
T T2
(45.6 atm)(16.5L) (1 atm)(V2) =
351 K K
V2 = 585 L
5/19/2018

15. Avogadro's Law Ar N2 H2 Avogadro’s Law: V  n, where (R,T constant)
Volume L
Pressure atm
Temperature °C
Mass of gas g
# of particles x 1022
Volume L
Pressure atm
Temperature °C
Mass of gas g
# of particles x 1022
Volume L
Pressure atm
Temperature °C
Mass of gas g
# of particles x 1022
The volume of a fixed quantity of gas maintained at constant pressure
and temperature is directly proportional to the number of moles of gas
5/19/2018

16. PV = nRT Ideal Gas Law R = 0.0821 L-atm/K mol = 62.36 L mmHg/K mol
Avogadro’s Law: V  n, where (R,T constant) or V = kn
Charles’s Law: V  T, where (n, P constant) or V/T = k
Boyle’s Law: V  1/P, where (n, T constant) or PV = k
Since these Are True
then:
R = L-atm/K mol
= L mmHg/K mol
= L torr/K mol
= 8.31 L kPa/K mol
= 8.31 J/K mol nT
V  P nT
V = R P
PV = nRT
5/19/2018

17. Example
What volume would 3.5g of Methane occupy at STP?
5/19/2018

18. Ideal Gas Law Molar Mass and Gas Densities
The ideal gas law may be used to determine the density or the molecular
mass of a gas n P
Since PV = nRT, then: = V RT
Let M = the molar mass which is grams per one mole of a substance.
n M
P M = V RT
If we multiply both sides of the expression by M, then the top left-hand side
of the expression becomes: or g/L which is density (d)
moles
grams x
liter
mole
P M
dRT d = or M = RT P
5/19/2018

19. Ideal Gas Law
Determine the molar mass of a gas if g of gas completely fills a 936 ml
Container at 735 torr….
2.889g
( l-atm/mol-K) (304K)
dRT
.936 L M = P
1 atm
735
760 M
= 79.7 g/mol
5/19/2018

20. Ideal Gas Law STOICHIOMETRY Volumes of Gases in Chemical Reactions
The industrial synthesis of nitric acid involves the reaction of nitrogen dioxide gas
with water:
3NO2(g) + H2O(l)  2HNO3(aq) + NO(g)
1. How grams of nitric acid can be prepared using 540 L of NO2 at a pressure
of 5.00 atm and a temperature of 295 K? PV
(5.00 atm)( 540 liters)
2 mol HNO g
nNO2 = =
= 4700 g HNO3 RT
3 mol NO mol
( l-atm/mol-K)(295 K)
STOICHIOMETRY
5/19/2018

21. Ideal Gas Law Gas Mixtures and Partial pressure
The ideal gas law may be used to determine the partial pressure of a mixture of gases
Since PV = nRT, then:
n1RT
This would be true of a single gas or mixture of gases P1
n2RT
n3RT = P2 = P3 = V V V
According to Dalton’s law of partial pressures, the total pressure of a mixture of gas equals the sum of
the pressures that each gas would exert if it were present alone.
Therefore Pt = (P1 + P2 + P3 +...)V = nRT or Pt V =
(n1+ n2 + n3+ …) RT
5/19/2018

22. Ideal Gas Law
What pressure, in atm, is exerted by a mixture of 2.00 g of H2 and 8.00 g of
N2 at 273 K in a 10 L vessel?
2.00 g H2
1 mol
= mol H2
2.02 g
8.00 g N2
1 mol
= mol N2
28.0 g
Pt (10.0 L) =
(0.990 mol mol)
( l-atm/mol-K)(273 K) Pt
= atm
5/19/2018

23. Ideal Gas Law Partial Pressures and Mole Fractions
The mole fraction, X, of any component of a mixture is simple the ratio of the total number of moles in the mixture
Moles of component 1
Mole fraction of component 1 = X1 =
Total moles of mixture
We can relate the mole fractions to partial pressures because
n1RT P1 n1 = V = = X1
ntRT Pt nt V
Therefore
P1 = X1Pt
5/19/2018

24. Ideal Gas Law
A study of the effects of certain gases on plant growth requires a synthetic atmosphere
composed of 1.5 mol percent CO2, 18.0 mol percent O2, and 80.5 mol percent Ar. (a)
Calculate the partial pressure of O2 in the mixture if the total pressure of the atmosphere
is to be 745 mm Hg. (b) If this atmosphere is to be held in a 120 L space at 295 K,
how many moles of O2 are needed?
Converting mole percent to pressure in mm Hg
Since P1 = X1Pt
P02 = 0.18 (745 mm Hg) = 134 mm Hg
Converting mm Hg pressure atmospheres
P02 = 134 mm Hg
1 atm
= atm
760 mm Hg
Solving for moles of O2 using the ideal gas equation PV
(0.176 atm)( 120L)
nO2 = =
= mol RT
( l-atm/mol-K)(295 K)
5/19/2018

25. Ideal Gas Law Collecting Gases Over Water P total = P gas + PH2O
5/19/2018

26. Ideal Gas Law Collecting Gases Over Water
Suppose the 2.00 L of oxygen is collected over water. The temperature of the water
and gas is 26°C and the pressure is 750 mm Hg. How many moles of O2 are collected?
What volume would the O2 gas collected occupy when dry, at the same temperature
and pressure?
(a) Keep in mind that water vapor may be dealt with in Dalton’s partial pressure law just like any
other gas.
P total = P gas + PH2O
PO2 = P total - PH2O
PO2 = = 725 mm Hg PV
(725 mm Hg/760 Hg)(0.200 L)
nO2 = =
= 7.77 x 10-3 moles O2 RT
( l-atm/mol-K)(299 K)
(b) P1V1 = P2V2
(725 mm Hg)(0.200 L)1 = (750 mm Hg)(V2)
V2= L
5/19/2018
Note: ( the vapor pressure of water at 26 °C is 25 mm Hg., Appendix B)

27. Kinetic Molecular Theory
How does KMT explain PV=k, P/T = k
At higher temperatures a greater fraction of molecules is moving
at greater speeds. This results in a higher average kinetics energy
Root-mean-square (rms) speed
Indicates the speed of a particle
possessing average kinetic energy
( )such that  = ½ m2
5/19/2018

28. Kinetic Molecular Theory
If two molecules which differ in mass have the same kinetic energy, then because
Ek = ½ m  2 is true, the velocities of the particles must be different; lighter particles must move
with a higher average speed.
5/19/2018

29. Kinetic Molecular Theory
Molecules Effusion and Diffusion
Calculate the rms speed () of a N2 molecules at 25 °C
Keep in mind that R needs to fit the specific circumstances you are dealing with.
Since we are dealing with particles possessing mass and traveling at some velocity,
R = 8.31 J/K-mol and J = kg-m2/s2, therefore R = 8.31 kg-m2/s2-K-mol
3(8.31 kg-m2/s2-K-mol)( 298K)
 =
28 g
1 kg
1000g
 = 5.15 x 102 m/s
5/19/2018

30. Kinetic Molecular Theory
Molecules Effusion and Diffusion
Diffusion
Effusion
5/19/2018

31. Kinetic Molecular Theory
Graham’s Law of Effusion
The effusion rate of a gas is inversely proportional to the square root of its molar mass r1 1
3RT/ M1 M2 = = = r2 2
3RT/ M2 M1 r1 M2 = r2 M1 or
T1 = √ M1
T M2
5/19/2018

32. Kinetic Molecular Theory
Graham’s Law of Effusion
If an unknown gas effuses at a rate that is only times that of O2 at the same temperature,
what is the molar mass of the unknown gas? rx M2 =
rO2 M1
32 g/mol
0.468 = M1 M1
= 146 g/mol
5/19/2018

33. Kinetic Molecular Theory
Diffusion and Mean Free Path
Diffusion speeds are slower than molecular speeds
The average distance traveled by a molecule between collisions is called
the mean free path. The higher the density of the gas the smaller the mean
free path (the shorter the distance between collisions
5/19/2018

34. Deviations From Ideal Behavior
10 atm
PV=nRT deviates at high pressure
5/19/2018

35. Deviations From Ideal Behavior
Note: as the temperature increases, the PV/RT behavior of
gases approaches ideal behavior
Ideal Gas
PV=nRT deviates at low temperature
5/19/2018

36. Deviations From Ideal Behavior
The Van der Waals Equation
n2a
P +
(V-nb) = nRT V2
The correction for pressure, n2a/V2, takes
into account the intermolecular attractions
between molecules.
(V-nb) is a correction for the finite volume
of the gas molecules The Van der Waal constant, b, varies with each gas and is the actual volume occupied by the gas molecules and increases with an increase in the mass or complexity of the molecule
5/19/2018

37. Deviations From Ideal Behavior
The Van der Waals Equation
If mol of an ideal gas were confined to L at 0.0°C, it would exert a pressure
of atm. Use the Van der Waals equation and the constants in Table 10.3 to estimate the pressure exert by mol of Cl2(g) in 22.4 L at 0.0 °C
n2a
P +
(V-nb) = nRT V2
nRT
n2a
Solving for pressure: P = -
V-nb V2
(1.000 mol)( l-atm/mol-K)(273.2 K)
(1.000 mol)(6.49 L2-atm)/mol2
P = -
22.4 L -(1.000 mol)( l/mol)
(22.4 L)2
P = atm
5/19/2018