1. Chapter 5
Gases

2. Elements that exist as gases at 250C and 1 atmosphere
5.1

3. 5.1

4. Physical Characteristics of Gases
Gases assume the volume and shape of their containers.
Gases are the most compressible state of matter.
Gases will mix evenly and completely when confined to the same container.
Gases have much lower densities than liquids and solids.
5.1

5. Force Pressure = Area Units of Pressure 1 pascal (Pa) = 1 N/m2
Barometer
Pressure =
Units of Pressure
1 pascal (Pa) = 1 N/m2
1 atm = 760 mmHg = 760 torr
1 atm = 101,325 Pa
5.2

6. 10 miles 0.2 atm 4 miles 0.5 atm Sea level 1 atm
5.2

7. h = pressure difference
5.2

8. As P (h) increases
V decreases
5.3

9. Boyle’s Law P a 1/V Constant temperature P x V = constant
Constant amount of gas
P x V = constant
P1 x V1 = P2 x V2
5.3

10. Boyle’s Law : P - V Relationship
Pressure is inversely proportional to volume
Or pressure is directly proportional to 1/V
P = or V = or PV=k
If n (number of moles) and T are constant
P1V1 = k P2V2 = k
Then : P1V1 = P2V2 k V k P

11. P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL V2 = 154 mL P1 x V1
A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL?
P1 x V1 = P2 x V2
P1 = 726 mmHg
P2 = ?
V1 = 946 mL
V2 = 154 mL
P1 x V1 V2
726 mmHg x 946 mL
154 mL =
P2 =
= 4460 mmHg
5.3

13. Charles Law - V - T- Relationship
Volume proportional to T (in K) : V = kT
n and P must be constant
V/T = k or T1 V1 T2 = V2 V1 T1 V2 = T2 or
Temperatures must be expressed in Kelvin .

14. Charles Law Problem
A sample of carbon monoxide, a poisonous gas, occupies 3.20 L at 125 oC. Calculate the temperature (oC) at which the gas will occupy 1.54 L if the pressure remains constant.
V1 = 3.20 L T1 = 125oC = 398 K
V2 = 1.54 L T2 = ?
T2 = 192 K oC = K =
oC = -81oC

15. Constant same for all gases
Avogadro’s Law
Constant temperature
Constant pressure
V a number of moles (n)
V = constant x n
Constant same for all gases
V1/n1 = V2/n2
5.3

16. 4NH3 + 5O2 4NO + 6H2O 1 mole NH3 1 mole NO At constant T and P
Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many liters of NO are obtained from 10 liters of ammonia at the same temperature and pressure? How many liters of oxygen are required?
4NH3 + 5O NO + 6H2O
1 mole NH mole NO
At constant T and P
1 volume NH volume NO
5.3

17. Ideal Gas Equation 1 Boyle’s law: V a (at constant n and T) P
Charles’ law: V a T (at constant n and P)
Avogadro’s law: V a n (at constant P and T)
V a nT P
V = constant x = R nT P
R is the gas constant
PV = nRT
5.4

18. PV = nRT PV (1 atm)(22.414L) R = = nT (1 mol)(273.15 K)
The conditions 0 0C and 1 atm are called standard temperature and pressure (STP).
Experiments show that at STP, 1 mole of an ideal gas occupies L.
PV = nRT
R = PV nT =
(1 atm)(22.414L)
(1 mol)( K)
R = L • atm / (mol • K)
5.4

19. Problems for Chapter 5
Pages
1-5, 8, 13-16, 18, 20, 22, 24-30, 32, 34, 36, 38, 40, 42, 44, 48, 50, 52, 60, 62, 65, 66, 72, 76, 80

20. PV = nRT nRT V = P 1.37 mol x 0.0821 x 273.15 K V = 1 atm V = 30.6 L
What is the volume (in liters) occupied by 49.8 g of HCl at STP?
T = 0 0C = K
P = 1 atm
PV = nRT
n = 49.8 g x
1 mol HCl
36.45 g HCl
= 1.37 mol
V =
nRT P
V =
1 atm
1.37 mol x x K
L•atm
mol•K
V = 30.6 L
5.4

21. What is the pressure exerted by 168 g of argon in a 38
What is the pressure exerted by 168 g of argon in a 38.0 L container at 27oC?

22. PV = nRT n, V and R are constant nR V = P T = constant P1 T1 P2 T2 =
Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)?
PV = nRT
n, V and R are constant nR V = P T
= constant
P1 = 1.20 atm
T1 = 291 K
P2 = ?
T2 = 358 K P1 T1 P2 T2 =
P2 = P1 x T2 T1
= 1.20 atm x
358 K
291 K
= 1.48 atm
5.4

23. Practice Exercise 5.4
A gas initially at 4.0 L, 1.2 atm, and 66oC undergoes a change so that its temperature and pressure become 42oC and 1.7 atm. What is its final volume?

24. d is the density of the gas in g/L
Density (d) Calculations
m is the mass of the gas in g m V = PM RT
d =
M is the molar mass of the gas
Molar Mass (M ) of a Gaseous Substance
dRT P
M =
d is the density of the gas in g/L
5.4

25. Example A chemist has synthesized a greenish-yellow compound of Cl and O and finds its density to be 7.71g/L at 36oC and 2.88 atm. Calculate the molar mass and determine its molecular formula.
A 2.10 L vessel contains 4.65 g of gas at 1.00 atm and 27.0oC. What is its density and molar mass?

26. C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)
Gas Stoichiometry
What is the volume of CO2 produced at 370 C and 1.00 atm when 5.60 g of glucose are used up in the reaction:
C6H12O6 (s) + 6O2 (g) CO2 (g) + 6H2O (l)
g C6H12O mol C6H12O mol CO V CO2
1 mol C6H12O6
180 g C6H12O6 x
6 mol CO2
1 mol C6H12O6 x
5.60 g C6H12O6
= mol CO2
0.187 mol x x K
L•atm
mol•K
1.00 atm =
nRT P
V =
= 4.76 L
5.5

27. Ideal Gas Law and Reaction Stoichiometry
Sodium azide (NaN3) is used in some air bags in automobiles. Calculate the volume of nitrogen gas generated at 21 oC and 823 mm Hg by the decomposition of 60.0 g of NaN3 .
2 NaN3 (s) Na (s) + 3 N2 (g)

28. EH Assignment
Due March ?
Unit 7
Minimum score
Sec 1 80
Sec 2 90
Sec 3
Sec 4 85
Sec 5

29. Dalton’s Law of Partial Pressures
V and T are constant P1 P2
Ptotal = P1 + P2
5.6

30. Dalton’s Law of Partial Pressures
Definition: In a mixture of gases, each gas contributes to the total pressure the amount it would exert if the gas were present in the container by itself. This is called the partial pressure of the gas.
To obtain a total pressure, add all of the partial pressures: Ptotal = p1+p2+p3+...

31. Dalton’s Law of Partial Pressures
A 2.00 L flask contains 3.00 g of CO2 and g of helium at a temperature of 17.0 oC.
What are the partial pressures of each gas, and the total pressure?
T = 17 oC = 290 K
nCO2 = 3.00 g CO2 x 1 mole CO2 / g CO2
= mole CO2
pCO2 = nCO2RT/V
pCO2 =
pCO2 = atm
( mol CO2) ( L atm/mol K) ( 290 K)
(2.00 L)

32. Dalton’s Law Problem - cont.
nHe = 0.10 g He / 1 mole He / g He
= mole He
pHe = nHeRT/V
pHe =
pHe = 0.30 atm
PTotal = pCO2 + pHe = atm atm
PTotal = 1.11 atm
(0.025 mol) ( L atm / mol K) ( 290 K )
( 2.00 L )

33. PA = nART V PB = nBRT V XA = nA nA + nB XB = nB nA + nB PT = PA + PB
Consider a case in which two gases, A and B, are in a container of volume V.
PA =
nART V
nA is the number of moles of A
PB =
nBRT V
nB is the number of moles of B
XA = nA
nA + nB
XB = nB
nA + nB
PT = PA + PB
PA = XA PT
PB = XB PT
Pi = Xi PT
5.6

34. Pi = Xi PT PT = 1.37 atm 0.116 8.24 + 0.421 + 0.116 Xpropane =
A sample of natural gas contains 8.24 moles of CH4, moles of C2H6, and moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)?
Pi = Xi PT
PT = 1.37 atm
0.116
Xpropane = =
Ppropane = x 1.37 atm
= atm
5.6

35. Dalton’s Law is used for calculations when a gas is collected over water. Whenever water (or any liquid) is placed in a closed container some water will evaporate and reach a state of equilibrium where the partial pressure of the water remains constant. This pressure is called the equilibrium vapor pressure of the liquid. It depends only on the temperature.

36. PT = PO + PH O 2KClO3 (s) 2KCl (s) + 3O2 (g) 5.6
Bottle full of oxygen gas and water vapor
2KClO3 (s) KCl (s) + 3O2 (g)
PT = PO + PH O 2
5.6

37. Relative Humidity pressure of water in air Rel Hum = x 100%
Example : the partial pressure of water at 15oC is 6.54 mm Hg, what is the relative humidity?
Rel Hum =(6.54 mm Hg/ mm Hg )x100%
= 51.1 %
What if the temperature is increased to 25oC?
pressure of water in air
maximum vapor pressure of water

38. Collection of Hydrogen Gas over Water - Vapor Pressure
2 HCl(aq) + Zn(s) ZnCl2 (aq) + H2 (g)
Calculate the mass of hydrogen gas collected over water if 156 ml of gas is collected at 20oC and 769 mm Hg.
PTotal = p H2 + pH2O pH2 = PTotal - pH2O
pH2 = 769 mm Hg mm Hg
= 752 mm Hg = atm
T = 20oC = 293 K
V = L

39. Kinetic Molecular Theory of Gases
A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume.
Gas molecules are in constant motion in random directions. Collisions are perfectly elastic. Pressure is caused by collisions of molecules with the walls of the container.
Gas molecules exert neither attractive nor repulsive forces on one another.
The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy
5.7

40. Kinetic theory of gases explains the gas laws and properties of gases
Compressibility of Gases
Boyle’s Law - Why does a decrease in volume cause an increase in pressure?
Charles’ Law – Why does an increase in temperature cause an increase in pressure?
5.7

42. Velocity and Energy Kinetic Energy = 1/2mu2
All gas molecules have the same average kinetic energy at the same temperature.
Therefore heavier molecules must have slower average speeds.
Consider samples of N2 and He gas at the same temperature. Which has the higher average kinetic energy? Which has the higher average speed?

43. Next EH assignment due Mar 12
Molecular Mass and Molecular Speeds
Problem: Calculate the molecular speeds of the molecules of hydrogen,
methane, and carbon dioxide at 300K!
Compare speeds and kinetic energies.
Next EH assignment due Mar 12
Section 7 – Properties of Gases

44. Molecular Mass and Molecular Speeds - III
Molecule H CH CO2
Molecular
Mass (g/mol)
Kinetic Energy
(J/molecule)
6.213 x x x
Velocity
(m/s) 1,

45. Diffusion vs. Effusion
Diffusion - One gas mixing into another gas, or gases, of which the molecules are colliding with each other, and exchanging energy between molecules.
Effusion - A gas escaping from a container into a vacuum.
Graham’s Law

46. NH3 (g) + HCl(g) = NH4Cl (s)
HCl = g/mol NH3 = g/mol
RateNH3/ RateHCl = 1.463

48. Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties.
NH4Cl
NH3
17 g/mol
HCl
36 g/mol
5.7

49. Apparatus for studying molecular speed distribution
5.7

50.  3RT urms = M The distribution of speeds of three different gases
at the same temperature
The distribution of speeds
for nitrogen gas molecules
at three different temperatures
urms =
3RT M 
5.7

51. Deviations from Ideal Behavior
1 mole of ideal gas
Repulsive Forces
PV = nRT
n = PV RT
= 1.0
Attractive Forces

52. For a gas to be an ideal gas
its molecules must have
zero volume and there must
be no forces between the
molecules.

53. Gases show the greatest deviation from ideal behavior at 1
Gases show the greatest deviation from ideal behavior at 1. High pressures 2. Low temperatures Why?

54. Effect of intermolecular forces on the pressure exerted by a gas.
5.8

55. The van der Waals Equation
n2a V2
P (V-nb) = nRT
Gas a b
atm L2
mol2 L
mol He Ne Ar Kr Xe H N O Cl CO NH
H2O

57. Determination of the Gas Constant R
THIS WEEK
EXPERIMENT #9
Determination of the Gas Constant R