Gases

1. Gases

2. -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.
Kinetic Molecular Theory

3. Kinetic Molecular Theory

4. Gas molecules have high translational energy. They move randomly
in all directions, in straight lines
Kinetic Molecular Theory

5. Kinetic Molecular Theory

6. Molecular Motion and States of Matter
3 types of molecular motion:
-Solids cannot move through space so they can only vibrate.
-Liquids and gases are free to move around in space so they can have all three modes of motion.
Vibrational motion –
molecules vibrating (in all
three phases)
Translational motion –
molecules moving from one
place to another (in liquids
and gases)
Rotational motion –
molecule is rotating or the
parts of the molecule are
rotating (gas and liquid
phase)

7. Molecular Motion and States of Matter
Molecular vibrations:

8. Pressure
Volume
Temperature
ALL related to one
another.
If you change one,
you change the
other.
Three quantities involving gases

9. 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
Gases: Pressure

10. Volume (V) is measured in Litres (L)
Gases: Volume

11. 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
Gases: Temperature

12. 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)
Gases: Relationships

13. 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
Gases: Boyle’s Law

14. Boyle’s Law: Pressure and volume are inversely proportional
PiVi = PfVf
Gases: Boyle’s Law

15. Charles’ Law: Volume and temperature are directly proportional
Vi = Vf
Ti Tf
Gases: Charles’ Law

16. Gay-Lussac’s Law: Pressure and temperature are directly proportional
Pi = Pf
Ti Tf
Gases: Gay-Lussac’s Law

17. COMBINED GAS LAW
PiVi = PfVf
Ti Tf
Since pressure, volume, and temperature are all related, they
can all be combined together:
Gases: Combined Gas Law

18. 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:
STP and SATP

19. 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
Gases: Calculations
Practice: Page 457 #19-21

20. 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
Gases: Dalton’s Law of Partial Pressures

21. Gay-Lussac:
Mole ratios are the same as volume ratios
Avogadro’s Law:
Equal volumes of all ideal gases at the same temperature and pressure contain the
same number of molecules.
The volume of a gas is directly related to the number of moles of gas.
Gases: Avogadro’s Law

22. 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
Gases: Avogadro’s Law

23. Ideal Gas Law formula:
PV = nRT
Pressure (kPa)
Volume (L)
Number of moles (mol)
Universal gas constant
8.314 kPa∙L
mol∙K
Temperature (K)
Gases: Avogadro’s Law

24. 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
Gases: Stoichiometry

25. 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
Gases: Stoichiometry

26. 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 596, Table 1)
Ptotal = PH2O + PH2
(101.8 kPa) = 3.78 kPa + PH2
Dalton’s Law of Partial Pressures
PH2
= 98.0 kPa
Gases: Stoichiometry

27. 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
Gases: Stoichiometry

28. 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
Gases: Stoichiometry

29. 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
Gases: Stoichiometry