1. GASES

2. GASES
Kinetic Molecular Theory of Gases
-The volume of an individual gas molecule is negligible compared to the volume of the
container holding the gas. This means that individual gas molecules, with virtually no
volume of their own, are extremely far apart and most of the container
is “empty” space
-There are neither attractive nor repulsive forces between gas molecules
-Gas molecules have high translational energy. They move randomly in all directions, in
straight lines
-When gas molecules collide with each other or with a container wall, the collisions are
perfectly elastic. This means that when gas molecules collide, somewhat like billiard
balls, there is no loss of kinetic energy
-The average kinetic energy of gas molecules is directly related to the temperature. The
greater the temperature, the greater the average kinetic energy.

3. GASES
Gas molecules have high translational energy. They move randomly in
all directions, in straight lines

4. GASES
Pressure
Volume
Temperature
ALL related to one
another.
If you change one,
you change the
other.

5. GASES: Pressure
Pressure (P) is measured in kilopascals (kPa)
Pressure = Force
Area
The unit Pa is the same as N/m2
For example:
The standard atmospheric pressure at 0ºC is 101.3kPa
How to measure pressure in pascals (Pa)?
Converting common units: 760 mm Hg = 760 Torr = 1 atm = 101.3 kPa

6. GASES: Volume
Volume (V) is measured in Litres (L)

7. GASES: Temperature
Temperature (T) is measured in Kelvin (K)
These are the same degrees as ºC, but:
0ºC = 273.15K
To convert Celcius to Kelvin,
use the following formula:
T (in K) = ºC + 273

8. GASES: Relationships
Boyle’s Law: Pressure and volume are inversely proportional
PiVi = PfVf
Charles’ Law: Volume and temperature are directly proportional
Vi = Vf
Ti Tf
Gay-Lussac’s Law: Pressure and temperature are directly proportional
Pi = Pf
Ti Tf
(assuming constant temperature)
(assuming constant pressure)
(assuming constant volume)

9. GASES: Relationships
P
1
V
If a sample of gas
at initial conditions
has an increase of
pressure applied
to it, its volume
decreases
proportionally
Boyle’s Law: Pressure and volume are inversely proportional
PiVi = PfVf

10. GASES: Relationships
Boyle’s Law: Pressure and volume are inversely proportional
PiVi = PfVf

11. GASES: Relationships
Charles’ Law: Volume and temperature are directly proportional
Vi = Vf
Ti Tf

12. GASES: Relationships
Gay-Lussac’s Law: Pressure and temperature are directly proportional
Pi = Pf
Ti Tf

13. GASES: Relationships
COMBINED GAS LAW
PiVi = PfVf
Ti Tf
Since pressure, volume, and temperature are all related, they
can all be combined together:

14. GASES
STANDARD TEMPERATURE and PRESSURE (STP)
Pressure = 101.3 kPa Temperature = 273K (0°C)
STANDARD AMBIENT TEMPERATURE and PRESSURE (SATP)
Pressure = 100 kPa Temperature = 298K (25°C)
One of two conditions will be used for gas calculations:

15. GASES: Calculations
Sandra is having a birthday party on a mild winter’s day. The weather
changes and a higher pressure (103.0 kPa) cold front (-25°C) rushes into
town. The original air temperature was -2°C and the pressure was 100.8
kPa. What will happen to the volume of the 4.2 L balloons that were tied to
the front of the house?
Given:
Pi = 100.8 kPa Pf = 103.0 kPa
Vi = 4.2 L Vf = ?
Ti = -2°C = 271K Tf = -25°C = 248K
PiVi = PfVf
Ti Tf
(100.8 kPa) (4.2 L) = (103.0 kPa) Vf
(271 K) (248K)
Vf = 3.76 L
Therefore the volume of the balloons will decrease to 3.8 L

16. GASES: Calculations
Practice: Page 457 #19-21

17. GASES: Dalton’s Law
Dalton’s Law of Partial Pressures:
The total pressure of a mixture of gases is the sum of
the pressures of each of the individual gases
Ptotal = P1 + P2 + P3 + P4 + P5 +… + Pn
Practice: Page 460 #22-23

18. GASES: Ideal Gas Law
Gay-Lussac:
Mole ratios are the same as volume ratios
Avogadro’s Hypothesis:
Equal volumes of all ideal gases at the same
temperature and pressure contain the same number
of molecules

19. GASES: Ideal Gas Law
For example:
2H2(g) + O2(g)  2H2O(g)
2 mol + 1 mol  2 mol
2 volumes + 1 volume  2 volume
So if you react 2L of hydrogen gas with 1L of oxygen
gas, you will get…
2L of water vapour!!!

20. GASES: Ideal Gas Law
Ideal Gas Law formula: (Most IMPORTANT formula)
PV = nRT
Pressure (kPa)
Volume (L)
Number of moles (mol)
Universal gas constant
8.314 kPa∙L
mol∙K
Temperature (K)

21. GASES: CALCULATIONS
Sulfuric acid reacts with iron metal to produce gas and an iron
compound. What volume of gas is produced when excess sulfuric
acid reacts with 40.0g of iron at 18°C and 100.3 kPa?
Given: mFe= 40.0g T = 18.0°C = 291K P = 100.3 kPa
STEP 1: Write the balanced chemical equation
Fe(s) + H2SO4(aq)  H2(g) + FeSO4(aq)
n = m/M
= (40.0g) / (55.85g/mol)
= 0.716 mol Fe
STEP 2: Use molar ratios to solve for amount of product made (stoichiometry!)
1 mol H2 = x
1 mol Fe 0.716 mol Fe
x = 0.716 mol H2
STEP 3: Use the ideal gas law to solve for the volume

22. GASES: CALCULATIONS
Sulfuric acid reacts with iron metal to produce gas and an iron
compound. What volume of gas is produced when excess sulfuric
acid reacts with 40.0g of iron at 18°C and 100.3 kPa?
Given: mFe= 40.0g T = 18.0°C = 291K P = 100.3 kPa
STEP 3: Use the ideal gas law to solve for the volume
PV = nRT
V = nRT
P
= (0.716 mol x 8.314 kPa∙L/mol∙K x 291 K)
(100.3 kPa)
= 17.3 L
Therefore 17.3 L of hydrogen gas are produced
Practice: Page 506 #30-34

23. GASES: CALCULATIONS
A student reacts magnesium with excess dilute hydrochloric acid to
produce hydrogen gas. She uses 0.15g of magnesium metal. What
volume of dry hydrogen does she collect over water at 28°C and 101.8
kPA?
Given: mMg= 0.15g T = 28.0°C = 301K P = 98.0 kPa
?
Pressure of water vapour at 28°C = 3.78 kPa
(from page 507, Table 12.3)
Ptotal = PH2O + PH2
(101.8 kPa) = 3.78 kPa + PH2
Dalton’s Law of Partial Pressures
PH2
= 98.0 kPa

24. GASES: CALCULATIONS
A student reacts magnesium with excess dilute hydrochloric acid to
produce hydrogen has. She uses 0.15g of magnesium metal. What
volume of dry hydrogen does she collect over water at 28°C and 101.8
kPA?
Given: mMg= 0.15g T = 28.0°C = 301K P = 98.0 kPa
STEP 1: Write the balanced chemical equation
Mg(s) + 2HCl(aq)  MgCl2(ag) + H2(g)
n = m/M
= (0.15g) / (24.31g/mol)
= 0.0062 mol
STEP 2: Use molar ratios to solve for amount of product made (stoichiometry!)
1 mol H2 = x
1 mol Mg 0.0062 mol Mg
x = 0.0062 mol H2
STEP 3: Use the ideal gas law to solve for the volume

25. GASES: CALCULATIONS
STEP 3: Use the ideal gas law to solve for the volume
PV = nRT
V = nRT
P
= (0.0062 mol x 8.314 kPa∙L/mol∙K x 301 K)
(98.0 kPa)
= 0.16 L
Therefore the student collects 0.16 L of dry hydrogen
Practice: Page 511 #37-39
A student reacts magnesium with excess dilute hydrochloric acid to
produce hydrogen gas. She uses 0.15g of magnesium metal. What
volume of dry hydrogen does she collect over water at 28°C and 101.8
kPA?
Given: mMg= 0.15g T = 28.0°C = 301K P = 101.8 kPa