1. PRINCIPLES OF CHEMISTRY I CHEM 1211 CHAPTER 6
DR. AUGUSTINE OFORI AGYEMAN
Assistant professor of chemistry
Department of natural sciences
Clayton state university

2. CHAPTER 6
THE GASEOUS STATE

3. CHARACTERISTICS OF GASES
- Gases are composed of nonmetals
- Gases form homogeneous mixtures of each other
Air
mixture of 78% N2
21% O2
0.9% Ar
other substances (CO2, H2, Ne, Kr, He)

4. CHARACTERISTICS OF GASES
Gases at Ordinary Temperature and Pressure
Noble gases (monatomic gases)
He, Ne, Ar, Kr, Xe
Diatomic Gases
H2, N2, O2, F2, Cl2
Other Common Gases
propane, ammonia, carbon dioxide,
hydrogen sulfide, methane, carbon monoxide, sulfur dioxide

5. PRESSURE
- Gases exert pressure on any surface they come in contact with
- Pressure is force applied per unit area
A = area (m2)
F = force (newton, N = kg-m/s2)
P = pressure (N/m2 = pascal, Pa)

6. PRESSURE
F = m x a
m = mass (kg)
a = acceleration (m/s2)

7. THE GAS LAWS Four Variables Define the Physical State of any Gas
Amount (mole)
Temperature (K)
Volume (L)
Pressure (bar, Pa, mm Hg, torr, atm, psi)
1 bar = 105 Pa
1 atm = 760 mmHg
= 760 torr
= x 105 Pa
= kPa
= 14.7 psi

8. THE GAS LAWS mm Hg: millimeters mercury
atm: atmosphere (atmospheric pressure = 1atm)
Pa: Pascal
psi: pound per square inch (Ib/in2)
Pressure Instruments
barometers, manometers, gauges
760, 700, 650 mm Hg
- Considered to have 3 significant figures

9. BOYLE’S LAW - The volume of a fixed amount of a gas is inversely
proportional to the pressure applied to the gas
if the temperature is kept constant
PV = constant
P1V1 = P2V2
- P1 and V1 are the pressure and volume of a gas
at an initial set of conditions
- P2 and V2 are the pressure and volume of the same gas
under a new set of conditions
- The temperature and amount of gas remain constant

10. BOYLE’S LAW A sample of N2 gas occupies a volume of 3.0 L at
6.0 atm pressure . What is the new pressure if the gas is
allowed to expand to 4.8 L at constant temperature?
P1 = 6.0 atm V1 = 3.0 L
P2 = ? V2 = 4.8 L
P1V1 = P2V2
(6.0 atm)(3.0 L) = (P2)(4.8 L)
P2 = 3.8 atm

11. CHARLES’S LAW - The volume of a fixed amount of gas is directly
proportional to its absolute temperature
if the pressure is kept constant
- V1 and T1 are the volume and absolute temperature
of a gas at an initial set of conditions
- V2 and T2 are the volume and absolute temperature
of the same gas under a new set of conditions
- The pressure and amount of gas remain constant

12. CHARLES’S LAW A sample of Ar gas occupies a volume of 1.2 L at 125 oC
and a pressure of 1.0 atm. What is the new temperature, in
Celsius, if the volume of the gas is decreased to 1.0 L at the
same pressure?
V1 = 1.2 L T1 = 125 oC = 398 K
V2 = 1.0 L T2 = ?
T2 = 332 K = 59 oC

13. AVOGADRO’S LAW
- The volume of a gas maintained at constant temperature
and pressure is directly proportional to the
number of moles of the gas
n = number of moles of a gas

14. AVOGADRO’S LAW Avogadro’s Hypothesis
- Equal volumes of gases at the same temperature
and pressure contain equal numbers of molecules
At Standard Temperature and Pressure (STP)
1 mol of any gas (= x 1023 molecules)
occupies a volume of L
Conditions of STP
Temperature = 0 oC = 273 K = 32 oF
Pressure = 1.00 atm ( kPa or 100 kPa)

15. Considering all three gas laws
THE IDEAL GAS LAW
Considering all three gas laws
V α 1/P V α T V α n
PV = nRT

16. R is the ideal gas constant
THE IDEAL GAS LAW
R is the ideal gas constant
= L-atm/mol.K
= J/mol-K
= m3-Pa/mol-K
= cal/mol-K
= L-torr/mol-K

17. RELATING THE GAS LAWS
PV = nRT
If n is constant

18. RELATING THE GAS LAWS
A 1.00-L container is filled with mole of CO gas at
35.0 oC. Calculate the pressure, in atmospheres, exerted
by the gas in the container
PV = nRT
P = ? V = 1.00 L n = mol
T = 35.0 oC = 308 K R = atm.L/mol.K
(P)(1.00 L) = (0.500 mol)( atm.L/mol.K)(308 K)
P = 12.6 atm

19. RELATING THE GAS LAWS
A balloon filled with helium initially has a volume of
1.00 x 106 L at 25 oC and a pressure of 752 mm Hg.
Determine the volume of the balloon after a certain
time when it encounters a temperature of
-33 oC and a pressure of 75.0 mm Hg
P1 = 752 mm Hg V1 = 1.00 x 106 L T1 = 25 oC = 298 K
P2 = 75.0 mm Hg V2 = ? T2 = -33 oC = 240 K

20. GAS DENSITIES AND MOLAR MASS
- Density (d) = mass/volume (m/V)
- Rearrange the ideal gas equation and multiply both
sides by molar mass (M) to give
M x n = mass (m) and m/V = d

21. Calculate the density of carbon dioxide gas
GAS DENSITIES AND MOLAR MASS
Calculate the density of carbon dioxide gas
at 0.45 atm and 252 K
Density (d) = 0.96 g/L

22. Molar mass = 29.0 g/mol GAS DENSITIES AND MOLAR MASS
Calculate the average molar mass of dry air
if its density is 1.17 g/L at 21 oC and torr
Molar mass = 29.0 g/mol

23. Consider the following reaction
REACTION STOICHIOMETRY
Consider the following reaction
4Al(s) + 3O2(g) → 2Al2O3(s)
Calculate the mass of aluminum that would react completely with 2.00 L of pure oxygen gas at STP

24. REACTION STOICHIOMETRY
1 mol O2 at STP = 22.4 L

25. REACTION STOICHIOMETRY
- At the same conditions
volumes of gases combine in the same proportions as the coefficients of the chemical equation
Example
3H2(g) + N2(g) → 2NH3(g)
This implies that
- 3 mol H2 reacts with 1 mol N2 to produce 2 mol NH3
- 3 L H2 reacts with 1 L N2 to produce 2 L NH3

26. DALTON’S LAW OF PARTIAL PRESSURES
- The total pressure exerted by a mixture of gases is the sum of
the partial pressures of the individual gases present
- The partial pressure is the pressure that a gas in a mixture of
gases would exert if it were present alone under the same
conditions

27. DALTON’S LAW OF PARTIAL PRESSURES
- Ptotal is the total pressure of a gaseous mixture
- P1, P2, P3,…. are the partial pressures of the individual gases
- ntotal is the total number of moles of a gaseous mixture
- n1, n2, n3,….are the number of moles of the individual gases

28. The total pressure by a mixture of He, Ne, and Ar gases is
DALTON’S LAW OF PARTIAL PRESSURES
The total pressure by a mixture of He, Ne, and Ar gases is
3.50 atm. Find the partial pressure of Ar if the partial pressures of He and Ne are 0.50 atm and 0.75 atm, respectively
Ptotal = 3.50 atm
P1 = 0.50 atm
P2 = 0.75 atm
P3 = ?
Ptotal = P1 + P2 + P3
P3 = Ptotal - (P1 + P2)
= 3.50 atm - (0.50 atm atm) = 2.25 atm

29. DALTON’S LAW OF PARTIAL PRESSURES
For gas 1 in a mixture of gases with n1 moles
Mole fraction (x1) = n1/ntotal
O2 is 21% of air
Mole fraction of O2 (x1) = 0.21
Total atmospheric pressure = 1 atm
Partial pressure of O2 = P1 = (0.21)(1 atm) = 0.21 atm
= (0.21)(760 mm Hg) = 160 mm Hg

30. EXPERIMENT TO COLLECT GAS OVER WATER
- Magnesium reacts with HCl to produce hydrogen gas
according to the following equation
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
- The H2 gas is collected in a gas collection tube initially filled
with water and inverted in a water pan
- The volume of H2 gas is measured by raising or lowering the
tube such that the water levels in the tube and pan are the same

31. EXPERIMENT TO COLLECT GAS OVER WATER
For this condition
The pressure inside the tube is equal to the
atmospheric pressure outside
Ptotal = Pgas + Pwater
Ptotal = total atmospheric pressure
Pgas = pressure of gas in tube
Pwater = vapor pressure of water at experimental temperature

32. KINETIC MOLECULAR THEORY OF GASES
- Gases consist of large numbers of small molecules that
are in continuous random motion
- Attractive and repulsive forces between gas molecules
are negligible
- The combined volume of all the molecules of a gas is
negligible relative to the total volume in which the
gas is contained

33. KINETIC MOLECULAR THEORY OF GASES
- Collisions between molecules are perfectly elastic
(the average kinetic energy of the colliding molecules does
not change at constant temperature)
- The average kinetic energy of the molecules is proportional
to the absolute temperature (molecules of all gases have the
same average kinetic energy at a given temperature)

34. AVERAGE KINETIC ENERGY
m = mass of an individual molecule
u = root-mean-square (rms) velocity at a given temperature

35. AVERAGE KINETIC ENERGY
Root-Mean-Square Velocity
- Square-root of the average of the squares
- The lower the molar mass of a gas the higher the u

36. GRAHAMS LAW OF EFFUSION
Diffusion
- The spread of one substance throughout space or
throughout a another substance
Effusion
- Escape of gas molecules through a tiny hole
into an evacuated space

37. GRAHAMS LAW OF EFFUSION
- The rate of effusion of a gas is inversely proportional to the
square root of its molar mass
- r1 and r2 are the effusion rates of gases 1 and 2 respectively
- M1 and M2 are the molar masses of gases 1 and 2 respectively

38. GRAHAMS LAW OF EFFUSION
- The rate of effusion of a gas is directly proportional to the
rms velocity of the molecules
- The time it takes for a gas to effuse is inversely proportional
to the rate of effusion

39. GRAHAMS LAW OF EFFUSION
It took 4.5 minutes for 1.0 L helium to effuse through a porous
barrier. How long will it take for 1.0 L Cl2 gas to effuse under
identical conditions?
t2 = 19 minutes

40. REAL GASES Mean Free Path
- The average distance traveled by a molecule between collisions
- Real Gases deviate from ideal gas behavior at high pressures
- Deviation is very small at low pressures (usually below 10 atm)
- Deviation increases with decreasing temperature
- Deviation is significant near the temperature at which a given
gas changes to the liquid state

41. REAL GASES - Molecules of real gases have finite volumes
and attract one another
- van der Waals equation

42. REAL GASES
a = constant
(measure of how strongly gas molecules attract each other
b = constant
(measure of the small but finite volume occupied by gas molecules)
n2a/V2
accounts for the attractive forces and adjusts pressure upwards nb
accounts for the finite volume occupied by molecules

43. REAL GASES Calculate the pressure exerted by 0.2500 mol N2 gas in a
5.000 L container at 22.5 oC using both the ideal gas law
and the van der Waals equation. Compare the results.
PV = nRT
P = nRT/V
= ( mol)( L-atm/mol-K)(295.5 K)/(5.000 L)
= atm

44. REAL GASES