1. 1.3 Reacting Masses and Volumes Reacting Gases
Kristin Page
IB SL Chemistry

2. Understandings
Avogadro’s law enables the mole ratio of reacting gases to be determined from volumes of the gases.
The molar volume of an ideal gas is a constant at specified temperature and pressure.
The molar concentration of a solution is determined by the amount of solute and the volume of solution.
A standard solution is one of known concentration.

3. Application & Skills
Calculation of reacting volumes of gases using Avogadro’s law.
Solution of problems and analysis of graphs involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas.
Solution of problems relating to the ideal gas equation.
Explanation of the deviation of real gases from ideal behavior at low temperature and high pressure.
Obtaining and using experimental values to calculate the molar mass of a gas from the ideal gas equation.
Solution of problems involving molar concentration, amount of solute and volume of solution.
Use of the experimental method of titration to calculate the concentration of a solution by reference to a standard solution.

4. Kinetic Theory of Gases
Gas particles have high energy are separated by a lot of space
Gas particles are move rapidly in straight lines but random directions
Gas particles collide with each other and with the container but do not lose energy
There is no attractive force between gas particles
These hold true for ideal gases

5. Ideal Gases Model of the behavior of real gases
Under “normal” conditions known as standard temperature and pressure (STP)
STP: P = 100 kPa and T = 273°K
Useful conversions: 100 kPa = 1atm and 273°K = 0°C
Gas particles have high energy are separated by a lot of space
Gas particles are move rapidly in straight lines but random directions
Gas particles collide with each other and with the container but do not lose energy
There is no attractive force between gas particles

6. Absolute Zero
When dealing with gases we need to use STP. This means that units of temperature must be in Kelvin (°K)
273°K = 0°C
The Kelvin scale starts at absolute zero (0°K)
It is not actually possible to reach absolute zero since this would be the point at which particles stopped all motion*
* Actually technically not true anymore but for our purposes we will stick with this. If you are curious go to zero

7. Avogadro’s Law
Relationship between amount of gas (n) and the volume (m3)
At constant temperature & pressure
𝑛∝𝑉
𝑛 1 𝑉 1 = 𝑛 2 𝑉 2
(m3)
Demonstration: Balloons, as I add more gas particles, volume increases. When I let gas out, volume decreases
This means that volumes can be used directly instead of moles in equations involving gases.
Ex: H2(g) + Cl2(g)  2HCl(g)

8. Avogadro’s Law Problems
Consider the following reaction for the synthesis of methanol: CO(g) + 2H2(g)CH3OH(g) (Assume same T & P)
What volume of H2 reacts exactly with 2.50dm3 of CO?
Mole ratio CO: H2 = 1:2
2.50 𝑑𝑚 3 𝐶𝑂 × 2 𝐻 𝐶𝑂 = 5.00 dm3 of H2 reacts
What volume of CH3OH is produced?
Mole ratio CO: CH3OH = 1:1
2.50 𝑑𝑚 3 𝐶𝑂 × 1 𝐶𝐻 3 𝑂𝐻 1 𝐶𝑂 =2.50 𝑑𝑚 3 𝐶𝐻 3 𝑂𝐻
Demonstration: Balloons, as I add more gas particles, volume increases. When I let gas out, volume decreases

9. Avogadro’s Law Problems
If 100 cm3 of oxygen reacts with 30cm3 of methane in the following reaction, how much oxygen will be left over at the end of the reaction? CH4(g) + 2O2(g)CO2(g)+H2O(g) (Assume same T & P)
Mole ratio CH4: O2 = 1:2
30 𝑐𝑚 3 𝐶 𝐻 4 × 2 𝑂 𝐶 𝐻 4 =60 𝑐𝑚 3 𝑂 2 𝑖𝑠 𝑢𝑠𝑒𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛
100 𝑐𝑚 3 −60 𝑐𝑚 3 =40 𝑐𝑚 3 𝑂 2 𝑖𝑠 𝑖𝑛 𝑒𝑥𝑐𝑒𝑠𝑠
Demonstration: Balloons, as I add more gas particles, volume increases. When I let gas out, volume decreases

10. Molar Volume
According to Avogadro’s law, the volume of 1 mole of and ideal gas at STP is constant, this is known as the Molar Volume
Molar volume = 22.7 dm3 mol-1
Note: 1 dm3 = 1L =1000cm3
Calculate the number of moles in 250.cm3 of O2 at STP
250 𝑐𝑚 3 × 1 𝑑𝑚 𝑐𝑚 3 × 1 𝑚𝑜𝑙 22.7 𝑑𝑚 3 = 𝑚𝑜𝑙
Calculate the volume of mol CO2 at STP
0.135 𝑚𝑜𝑙 𝐶𝑂 2 × 𝑑𝑚 3 1 𝑚𝑜𝑙 =3.06 𝑑𝑚 3
Demonstration: Balloons, as I add more gas particles, volume increases. When I let gas out, volume decreases

11. Boyle’s Law Relationship between pressure and volume
At constant temperature, the volume of a fixed mass of an ideal gas is inversely proportional to its pressure
𝑃∝ 1 𝑉
Gases in smaller volume will have more collisions with the container and so higher pressure
This can also be written 𝑉 1 𝑃 1 = 𝑉 2 𝑃 2 where 𝑉 1 𝑎𝑛𝑑 𝑃 1 =𝑖𝑛𝑖𝑡𝑖𝑎𝑙 and 𝑉 2 𝑎𝑛𝑑 𝑃 2 =𝑓𝑖𝑛𝑎𝑙
Pressure chamber (vacuum) – inflate a marshmallow, balloon, bags of chips (decrease pressure increase volume)

12. Boyle’s Law Example 𝑃 2 = 𝑉 1 𝑃 1 𝑉 2 𝑃 2 = 1.50∙101 0.462 =328𝑘𝑃𝑎
To make an air horn, 1.50 dm3 of air at 101 kPa are compressed into a can with a volume of dm3. Assuming a constant temperature, what is the pressure on the compressed air?
𝑃 1 =101𝑘𝑃𝑎, 𝑉 1 =1.50 𝑑𝑚 3 , 𝑃 2 =? kPa, 𝑉 2 =0.462 𝑑𝑚 3 ,
𝑉 1 𝑃 1 = 𝑉 2 𝑃 2
𝑃 2 = 𝑉 1 𝑃 1 𝑉 2
𝑃 2 = 1.50∙ =328𝑘𝑃𝑎
Pressure chamber (vacuum) – inflate a marshmallow, balloon, bags of chips (decrease pressure increase volume)

13. Charles’s Law Relationship between volume and temperature
At constant pressure the volume of a fixed mass of ideal gas is directly proportional to its temperature (K)
𝑉∝𝑇 or 𝑉 1 𝑇 1 = 𝑉 2 𝑇 2
As temperature increase particles move more rapidly (higher kinetic energy) and collide with surface more causing an increase in volume
Temp. must be in Kelvin!
Balloon in boiling water bath, balloon in freezer

14. Charles’s Law Example
On hot days, you may have noticed that potato chip bags seem to “inflate”, even though they have not been opened. If I have a 250mL bag at a temperature of 19°C, and I leave it in my car which has a temperature of 45°C, what will the new volume of the bag be?
Temp. must be in Kelvin!
𝑇 1 =19+273=292𝐾 𝑇 2 =45+273=318 𝐾
𝑉 1 𝑇 1 = 𝑉 2 𝑇 2
𝑽 𝟐 = 𝑽 𝟏 𝑻 𝟐 𝑻 𝟏
𝑉 2 = 250𝑚𝐿∙318𝐾 292𝐾 =272 𝑚𝐿
Balloon in boiling water bath, balloon in freezer

15. Gay-Lussac’s Law ( )
1800’s studied gases using hot air balloons
When volume is constant, pressure of the gas is directly proportional to temperature
𝑝∝𝑇 or 𝑝 1 𝑇 1 = 𝑝 2 𝑇 2
Pressure in Nm-2 (Pa)
Temp in K
Demonstrations: Crushing Can, Egg in a Bottle
Large Erlenmeyer flask, small amount of water, heat on hot plate. Balloon on top – balloon inflates, put flask in ice water bath, balloon inflates inside the jar

16. Combined Gas Law
All 3 gas laws can be combined into one equation known as the combined gas law
For a fixed amount of gas 𝒑 𝟏 𝑽 𝟏 𝑻 𝟏 = 𝒑 𝟐 𝑽 𝟐 𝑻 𝟐
This equation can technically be used for ANY gas law problem since the constant variable will cancel out.
Note: all units must be the same on both sides of the equation and temperature must be in Kelvin!

17. Combined Gas Law Example
If the volume of an ideal gas collected at 0°C and 100 kPa is 50.0cm3, what would be the volume at 60°C and 108 kPa?
𝑇 1 =0+273=273𝐾, 𝑝 1 =100𝑘𝑃𝑎, 𝑉 1 =50.0 𝑐𝑚 3 , 𝑇 2 =60+273=333𝐾, 𝑝 2 =108 𝑘𝑃𝑎, 𝑉 2 =?
𝒑 𝟏 𝑽 𝟏 𝑻 𝟏 = 𝒑 𝟐 𝑽 𝟐 𝑻 𝟐 so 𝑽 𝟐 = 𝒑 𝟏 𝑽 𝟏 𝑻 𝟐 𝑻 𝟏 𝒑 𝟐
𝑽 𝟐 = 𝟏𝟎𝟎∙𝟓𝟎.𝟎∙𝟑𝟑𝟑 𝟐𝟕𝟑∙𝟏𝟎𝟖 =𝟓𝟔.𝟓 𝒄𝒎 𝟑

18. Ideal Gas Equation 𝒑𝑽=𝒏𝑹𝑻
Now we can add the gas equations with Avogadro’s Law to get the ideal gas equation
𝒑𝑽=𝒏𝑹𝑻
R = gas constant = 8.31 J K-1mol-1 (Data Book p.2)
Must use the following units:
p (Pa)
V (m3)
T (K)
Conversion Factors
1 Pa = 1 J m-3
1 dm3 = 1 x 10-3 m3 (Data book)

19. Ideal Gas Equation Practice
An ideal gas occupies 590 cm3 at 120°C and 202 kPa. What amount of gas (in moles) is present?
𝒑𝑽=𝒏𝑹𝑻
𝑝=202 𝑘𝑃𝑎∙ 1000 𝑃𝑎 1 𝑘𝑃𝑎 =202,000 𝑃𝑎
𝑉=590 𝑐𝑚 3 ∙ 𝑚 𝑐𝑚 3 =5.9× 10 −4 𝑚 3
R = 8.31 J K-1mol-1
𝑇= =393 𝐾
𝑛= 202,000𝑃𝑎×(5.9× 10 −4 ) 𝑚 𝐽 𝐾 −1 𝑚𝑜𝑙 −1 × 393𝐾 =0.036 𝑚𝑜𝑙

20. Ideal Gas Equation Practice
A gas has a density of 1.24 g dm-3 at 0°C and 1.00 x 105 Pa. Calculate its molar mass.
𝑺𝑻𝑷 𝑮𝒊𝒗𝒆𝒏
𝑀𝑜𝑙𝑎𝑟 𝑉𝑜𝑙𝑢𝑚𝑒=22.7 𝑑𝑚 3 𝑚𝑜𝑙 −1
1.24 𝑔 𝑑𝑚 3 ∙ 22.7 𝑑𝑚 3 𝑚𝑜𝑙 = 28.1𝑔 𝑚𝑜𝑙