1. Leaving Certificate Chemistry
The Gas Laws
Leaving Certificate Chemistry

2. Boyle’s Law
Robert Boyle – Born in Lismore, Co. Waterford to the Earl of Cork
Referred to as the founder of modern chemistry
1662 – He studied the relationship between the pressure exerted by a gas and the volume (space) it occupies.
Robert Boyle

3. Boyle’s Law Boyle’s Law:
At a constant temperature, the pressure exerted by a gas is inversely proportional to the volume occupied by that gas.
Volume (V)
Pressure (P)

4. Charles’s Law
Jacques Charles – French inventor, scientist, mathematician, and balloonist
He launched the world's first (unmanned) hydrogen-filled balloon in August 1783
He was interested in the relationship between the temperature of a gas and the volume (space) it occupies.
Jacques Charles
His law became known as Charles’s Law

5. Charles’s Law Charles’s Law:
2002 Q. 4 (i)
Charles’s Law:
At a constant pressure, the temperature of a gas is proportional to the volume occupied by that gas.
2004 Q. 4 (d)
Volume (V)
Temperature (T)

6. Avogadro’s Law Amedeo Avogadro – Italian Scientist
In tribute to him, the number of elementary entities (atoms, molecules, ions or other particles) in 1 mole of a substance, ×1023, is known as the Avogadro constant.
Avogadro’s Law:
Equal volumes of all gases, at the same temperature and pressure, contain the same number of particles, or molecules.

7. Gay-Lussac’s Law
Joseph Gay-Lussac – French was a French chemist and physicist
Jacques Charles

8. N2(g) + 3H2(g) -----> 2NH3(g)
Gay-Lussac’s Law:
N2(g) + 3H2(g) -----> 2NH3(g)
1 litres
3 litres
2 litre
If all the reactants and products in a reaction are gases then
the ratio of the volumes of the gases can be expressed as simple whole numbers
The volume of gases taking part in a chemical reaction show simple whole number ratios to one another when those volumes are measured at the same temperature and pressure.

9. 2H2(g) + O2(g) -----> 2H2O(g)
What is the ratio of the volumes of gases in the reaction?
If all the reactants and products in a reaction are gases then the ratio of the volumes of the gases can be expressed as simple whole numbers

10. Converting units
Celsuis K ( )
Kelvin Celsuis (- 273 )

11. Changing units of measurement
Convert the following to the Kelvin:
-10oC
-272oC
25oC
100oC
My notes = normal pressure Nm -2 and this is equal to 1Pa
Normal pressure = Pa r 105 Pa or 100kPa
Standard temperature and pressure = 273 K and 1 x 10 5Pa

12. P1 V1 P2 V2 = T1 T2 The Combined Gas Law
The combined gas law is a gas law which combines Charles's law, Boyle's law, and Gay-Lussac's law P1 V1 P2 V2 = T1 T2
Temperature is measured in degrees Kelvin:
Add 273 to convert degrees Celsius to degrees Kelvin
P1, V1 and T1 are initial pressure, volume and temperature of the gas
20 oC = 20 +
273 =
293 K
100 oC =
100 +
273 =
373 K
P2, V2 and T2 are a later pressure, volume and temperature of the gas
—50 oC =
—50 +
273 =
223 K

13. P1 V1 T1 = P2 V2 T2 The Combined Gas Law
The combined gas law is a gas law which combines Charles's law, Boyle's law, and Gay-Lussac's law P1 V1 T1 = P2 V2 T2
20 oC = 20 +
273
293
100 oC =
100 +
273
373
—50 oC = +
273
223 13

14. (Standard temperature = 0 oC, standard pressure = 101,325 Pa)
If a definite mass of gas occupies 500 cm3 at a pressure of 101,000 Pa and a temperature of 27 oC, what is its volume in cm3 at s.t.p.?
(Standard temperature = 0 oC, standard pressure = 101,325 Pa) P1 V1 P2 V2 = T1 T2
101,000
× 500
101,325
× V2 =
300
273
101,000
× 500
× 273
453.5 cm3 = V2
300
× 101,325

15. P1 V1 P2 V2
Q.200 = T1 T2
P1 = Pa P2 = Pa
V1 = 500cm3 V2 = ? cm3
T1 = = 300K T2 = = 273K
(101000)(500) = (101,325)(V2)
(101000)(500) (273) = V2
(101,325)( 300)
= V2

16. = Q.201 P1 = 975kPa P2 = 100kPa V1 = 30cm3 V2 = ? cm3T1 T2
P1 = 975kPa P2 = 100kPa
V1 = 30cm3 V2 = ? cm3
T1 = = 298K T2 = = 273K
(975)(30) = (100)(V2)
(975)(30) (273) = V2
(100)( 298)
= V2
Answer = cm3

17. P1 V1 P2 V2
Q.202 = T1 T2
P1 = 101kPa P2 = 108 kPa
V1 = 1.2m3 V2 = ? m3
T1 = = 278K T2 = = 303K
(101)(1.2) = (108)(V2)
(101)(1.2) (303) = V2
(108)( 278)
1.2231= V2
Answer = m3

18. P1 V1 P2 V2
Q.203 = T1 T2
P1 = 50,662.5 Pa P2 = 50,662.5 x2= Pa
V1 = 283cm3 V2 = ? cm3
T1 = = 298K T2 =278 x 4 = K
(59,662.5)(298) = (101,325)(V2)
( )(283) (1192) = V2
(101,325)( 298)
566 = V2
Answer = 566cm3

19. The kinetic theory of gases
The kinetic theory of gases attempts to explain properties of gases
Assumptions of the kinetic theory:
2009 Q. 10 (a)
Gases are made up of particles whose diameters are negligible
There are no attractive or repulsive forces between gas particles.
Particles are in constant random motion colliding with each other.
A perfectly elastic collision is defined as one in which there is no loss of kinetic energy in the collision. An inelastic collision is one in which part of the kinetic energy is changed to some other form of energy in the collision. Any macroscopic collision between objects will convert some of the kinetic energy into internal energy and other forms of energy, so no large scale impacts are perfectly elastic
The average kinetic energy of the particles is proportional to the Kelvin temperature.
Gas particle
no attraction or repulsion forces
diameters are negligible
All collisions are perfectly elastic.

20. Deviations from the ideal gas laws
Two of the characteristics of ideal gases:
The gas molecules themselves occupy no appreciable volume ( diameters are almost negligible in size)
The gas molecules have no attraction or repulsion for each other
Real gas molecules, however, do have an appreciable volume and do attract one another – especially under conditions of low temperature and high pressure
So no gas is completely ideal, and some are less ideal than others.
At high pressures, and low volumes, the intermolecular distances can become quite short, and attractive forces between molecules becomes significant
Neighboring molecules exert an attractive force, which will minimize the interaction of molecules with the container walls. And the apparent pressure will be less than ideal (PV/RT will thus be less than ideal).
As pressures increase, and volume decreases, the volume of the gas molecules becomes significant in relationship to the container volume
In an extreme example, the volume can decrease below the molecular volume, thus PV/RT will be higher than ideal (V is higher)
At high temperatures, the kinetic energy of the molecules can overcome the attractive influence and the gasses behave more ideal
At higher pressures, and lower volumes, the volume of the molecules influences PV/RT and its value, again, is higher than ideal

21. Which will deviate most from real gas behaviour?
Consider the following gases – H2, HF, NH3 and F2
HF has hydrogen bonding between molecules - strong attractive forces
NH3 – has hydrogen bonding between molecules (but electronegativity difference between N and H is less so attractive forces between molecules will be weaker than in 1)
F2 – has Van Der Walls forces attracting molecules – very weak intermolecular forces
H2 – Van Der Waals forces are even less here(less electrons present )– weakest attractive forces between molecules
Least like an ideal gas
(deviates the most)
Most like an ideal gas

22. Ideal gases
Approximations about the behaviour of gases can be made by using the ideal gas formula:
PV = nRT
Temperature ( K)
Pressure (Pa)
Gas constant
R = 8.31JK-1mol-1
Volume (m3)
Number of moles of the gas

23. Mandatory experiment
Determination of the relative molecular mass of a volatile liquid using the ideal gas equation

24. Determination of the RMM of a volatile liquid
A known amount of a volatile liquid is vaporised at a known temperature, pressure and in a container of known volume
PV = nRT
By doing this experiment we can find n ( number of moles of the gas) and then use this to find the relative molecular mass of the gas

25. Converting units Celsuis K ( + 273 ) Kelvin Celsuis (- 273 )
Pa (÷ 1000 )
kPa Pa (x 1000 )
cm3 m3 ( x )
m3 cm3 ( x 10 6 )

26. Changing units of measurement
Convert the following to the Kelvin:
-10oC
-272oC
25oC
100oC
Convert the following to Pascals:
100kPa
1 x 10 3 kPa
2 x 10 – 6 kPa
1.01 x 105 KPa
Convert the following to cm3:
1000m3
1 m3 2L
10 x 10 4cm3
My notes = normal pressure Nm -2 and this is equal to 1Pa
Normal pressure = Pa r 105 Pa or 100kPa
Standard temperature and pressure = 273 K and 1 x 10 5Pa

27. Calculations using the ideal gas law
How many moles of nitrogen gas are present in 200cm3 of the gas at 35oC and a pressure of 200kPa? (The gas constant is 8.31JK-1 mol-1 )
PV = nRT
P = Pressure = 200kPa = 200,000Pa
V = Volume = 200cm3 = .0002m3
n = number of moles = ?
R = gas constant = 8.31JK-1 mol-1
T = temperature = 35oC = 308K
Calculations using the ideal gas law

28. Calculations using the ideal gas law
PV = nRT
(200,000)( ) = n (8.31) (308)
(200,000)( ) = n
(8.31) (308)
= n
= n ( the number of moles)
Calculations using the ideal gas law

29. Calculations using the ideal gas law
Q209. How many moles of nitrogen gas are present in 100cm3 of the gas at 30oC and a pressure of 200kPa? (The gas constant is 8.31JK-1 mol-1 )
PV = nRT
P = Pressure = 200kPa = 200,000Pa
V = Volume = 100cm3 = .0001m3
n = number of moles = ?
R = gas constant = 8.31JK-1 mol-1
T = temperature = 30oC = 303K
Calculations using the ideal gas law

30. Calculations using the ideal gas law
PV = nRT
(200,000)( ) = n (8.31) (303)
(200,000)( ) = n
(8.31) (303)
20 = n
= n ( the number of moles)
Calculations using the ideal gas law

31. Calculations using the ideal gas law
Q g of a volatile liquid gave 512cm3 of vapour at 20oC and a pressure of Pa. Calculate the relative molecular mass (The gas constant is 8.31JK-1 mol-1 )
PV = nRT
P = Pressure = kPa = Pa
V = Volume = 512cm3 = m3
n = number of moles = ?
R = gas constant = 8.31JK-1 mol-1
T = temperature = 20oC = 293K
Calculations using the ideal gas law

32. Calculations using the ideal gas law
PV = nRT
(101325)( ) = n (8.31) (293)
(101325)( ) = n
(8.31) (293)
= n
= n ( the number of moles)
Calculations using the ideal gas law

33. Calculations using the ideal gas law
261e) Calculate the relative molecular mass of the volatile liquid
Moles of X
Mass of X in g
X Relative Molecular Mass
÷ Relative Molecular Mass
moles x RMM = 1.236g
RMM = 1.236/
RMM =
RMM = 58
Calculations using the ideal gas law

34. Calculations using the ideal gas law
Q211. What is the relative molecular mass to the nearest whole number if 1.00g occupies a volume of 200cm3 of vapour at 25oC and a pressure of 101.3kPa. (The gas constant is 8.31JK-1 mol-1 )
PV = nRT
P = Pressure = 101.3kPa = Pa
V = Volume = 200cm3 = .0002m3
n = number of moles = ?
R = gas constant = 8.31JK-1 mol-1
T = temperature = 25oC = 298K
Calculations using the ideal gas law

35. Calculations using the ideal gas law
PV = nRT
(101300)( ) = n (8.31) (298)
(101300)( ) = n
(8.31) (298)
= n
= n ( the number of moles)
Calculations using the ideal gas law

36. Calculations using the ideal gas law
261e) Calculate the relative molecular mass of the volatile liquid
Moles of X
Mass of X in g
X Relative Molecular Mass
÷ Relative Molecular Mass
moles x RMM = 1.0 g
RMM = 1.0/
RMM =
Calculations using the ideal gas law

37. Calculations using the ideal gas law
Q g of a volatile liquid occupied 3.15 x m3 at 300K and a pressure of 105Pa. Calculate the relative molecular mass (The gas constant is 8.31JK-1 mol-1 )
PV = nRT
P = Pressure = 105 Pa
V = Volume = 3.15 x m3
n = number of moles = ?
R = gas constant = 8.31JK-1 mol-1
T = temperature = 300K
Calculations using the ideal gas law

38. Calculations using the ideal gas law
PV = nRT
(100000)(3.15 x m3 ) = n (8.31) (300)
(100000)( ) = n
(8.31) (300)
= n
2493
= n ( the number of moles)
Calculations using the ideal gas law

39. Calculations using the ideal gas law
Moles of X
Mass of X in g
X Relative Molecular Mass
÷ Relative Molecular Mass
0.0126moles x RMM = 1.1 g
RMM = 1.1/
RMM =
Calculations using the ideal gas law

40. Calculations using the ideal gas law
Q213. A small quantity of a volatile organic solvent propanone evaporates at room temperature and pressure. Use the ideal gas equation to calculate the volume, in litres of propanone vapour formed when 0.29g of liquid propanone evaporated taking temperature of 20oC and a pressure of 101kPa. (The gas constant is 8.31JK-1 mol-1 )
PV = nRT
P = Pressure = 101kPa = Pa
V = Volume = ?
n = number of moles = ? ( but mass is known)
R = gas constant = 8.31JK-1 mol-1
T = temperature = 293 K
Calculations using the ideal gas law

41. Calculations using the ideal gas law
Moles of X
Mass of X in g
X Relative Molecular Mass
÷ Relative Molecular Mass
moles x 58 = 0.29 g
Moles = 0.29 / 58
Moles = 0.005
Calculations using the ideal gas law

42. Calculations using the ideal gas law
PV = nRT
(101000)V = (0.005)(8.31)(293)
V = (0.005)(8.31)(293)
101000
V = m3
Calculations using the ideal gas law

43. To go from m3 to litres multiply by 103
Volume has to be in litres...
0.0001m3 x 10 3 = 0.1 L

44. Calculations using the ideal gas law
Q214. Assuming ideal behaviour how many moles of carbon dioxide are present in 720cm3 at 10oC and a pressure of 105 Pa. (The gas constant is 8.31JK-1 mol-1 )
PV = nRT
P = Pressure = Pa
V = Volume = 720cm3 = m3
n = number of moles = ?
R = gas constant = 8.31JK-1 mol-1
T = temperature = 10oC = 283K
Calculations using the ideal gas law

45. Calculations using the ideal gas law
PV = nRT
(100000)( ) = n (8.31) (283)
(100000)( ) = n
(8.31) (283)
= n ( the number of moles)
Calculations using the ideal gas law

46. Calculations using the ideal gas law
Q215. Carbon dioxide is stored under pressure in liquid form in a fire extinguisher. 2.5 Kilograms of carbon dioxide are released into the air as a gas on teh discharge of the fire extinguoisher. What volume does this gas occupy at a temperature of 288K and a pressure of Pa. (The gas constant is 8.31JK-1 mol-1 )
PV = nRT
P = Pressure = Pa
V = Volume = ?
n = number of moles = ? ( but mass is known)
R = gas constant = 8.31JK-1 mol-1
T = temperature = 288K
Calculations using the ideal gas law

47. Calculations using the ideal gas law
Moles of X
Mass of X in g
X Relative Molecular Mass
÷ Relative Molecular Mass
moles x 44 = 2500 g
Moles = 2500 / 44
Moles =
Calculations using the ideal gas law

48. Calculations using the ideal gas law
PV = nRT
(100000)V = ( )(8.31)(288)
V = (0.005)(8.31)(293)
101000
V = m3
Calculations using the ideal gas law

49. Check your learning
Describe some assumptions of the Kinetic gas theory
What is an ideal gas?
Give two reasons why real gases deviate from ideal behaviour
Under what conditions do gases depart from ideal behaviour?
What is the ideal gas law?

50. Calculations using the ideal gas law
Q216(d) In an experiment to measure the relative molecular mass of a volatile liquid was vaporised at 97oC. The volume occupied was found to be 95cm3 and a pressure of 105Pa. Calculate the number of moles and relative molecular mass (The gas constant is 8.31JK-1 mol-1 )
PV = nRT
P = Pressure = 1 x 105Pa
V = Volume = 95cm3 = 95 x m3
n = number of moles = ?
R = gas constant = 8.31JK-1 mol-1
T = temperature = 370K
Calculations using the ideal gas law

51. Calculations using the ideal gas law
PV = nRT
(100000)( ) = n (8.31) (370)
(100000)( ) = n
(8.31) (370)
= n ( the number of moles)
Calculations using the ideal gas law

52. Calculations using the ideal gas law
261e) Calculate the relative molecular mass of the volatile liquid
Moles of X
Mass of X in g
X Relative Molecular Mass
÷ Relative Molecular Mass
moles x RMM = g
RMM = 0.275/
RMM =
To the nearest whole number – RMM = 89
Calculations using the ideal gas law

53. Calculations using the ideal gas law
Q217. In order to determine the relative molecular mass of a volatile liquid a conical flask was filled with vapour of the liquid. The results of the experiment were summarised as follows ( on sheet). Calculate the relative molecular mass. ( To the neasreast whole number) (The gas constant is 8.31JK-1 mol-1 )
PV = nRT
P = Pressure = 98000Pa
V = Volume = 328cm3 = m3
n = number of moles = ?
R = gas constant = 8.31JK-1 mol-1
T = temperature = 373K
Calculations using the ideal gas law

54. Calculations using the ideal gas law
PV = nRT
(98000)( ) = n (8.31) (373)
(98000)( ) = n
(8.31) (373)
= n ( the number of moles)
Calculations using the ideal gas law

55. Calculations using the ideal gas law
261e) Calculate the relative molecular mass of the volatile liquid
Moles of X
Mass of X in g
X Relative Molecular Mass
÷ Relative Molecular Mass
moles x RMM = 1.43 g
RMM = 1.43/
RMM = 137.5
To the nearest whole number – RMM = 138
Calculations using the ideal gas law

56. Calculations using the ideal gas law
Q218. In order to determine the relative molecular mass of a volatile liquid a conical flask was filled with vapour of the liquid. The results of the experiment were summarised as follows ( on sheet). Calculate the relative molecular mass. ( To the nearest whole number) (The gas constant is 8.31JK-1 mol-1 )
PV = nRT
P = Pressure = Pa
V = Volume = m3
n = number of moles = ?
R = gas constant = 8.31JK-1 mol-1
T = temperature = 300K
Calculations using the ideal gas law

57. Calculations using the ideal gas law
PV = nRT
(100000)( ) = n (8.31) (300)
(100000)( ) = n
(8.31) (300)
= n ( the number of moles)
Calculations using the ideal gas law

58. Calculations using the ideal gas law
261e) Calculate the relative molecular mass of the volatile liquid
Moles of X
Mass of X in g
X Relative Molecular Mass
÷ Relative Molecular Mass
moles x RMM = 1.1 g
RMM = 1.1/
RMM =
To the nearest whole number – RMM = 87
Calculations using the ideal gas law