The Gas Laws
Leaving Certificate Chemistry
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
Boyle’s Law
Volume (V)
Boyle’s Law:
At a constant temperature, the pressure exerted by a gas
is inversely proportional to the volume occupied by that
gas.
Pressure (P)
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.
His law became known
as Charles’s Law
Jacques Charles
Charles’s Law
Volume (V)
Charles’s Law:
At a constant pressure, the temperature of a gas is
proportional to the volume occupied by that gas.
Temperature (T)
2002 Q. 4 (i)
2004 Q. 4 (d)
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, 6.02214179 ×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.
Gay-Lussac’s Law
Joseph Gay-Lussac – French was a
French chemist and physicist
Jacques Charles
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
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
• Celsuis
• Kelvin
K ( + 273 )
Celsuis (- 273 )
Changing units of measurement
•
a)
b)
c)
d)
Convert the following to the Kelvin:
-10oC
-272oC
25oC
100oC
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
P1V1
T1
=
P2V2
T2
is measured
in degrees
Kelvin:
PTemperature
,
V
and
T
are
initial
pressure,
volume
and 293 K
o
120 1 C
1=
20
+
273
=
Add 273 to of
convert
degrees Celsius to degrees Kelvin
temperature
the
gas
100 oC
=
100
+
273
=
373 K
P
later pressure,
2, V2oand
—50
C T2=are a—50
+
273volume
= and223 K
temperature of the gas
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
P1V1
T1
=
P2V2
T2
20 oC
=
20
+
273
293
100 oC
=
100
+
273
373
—50 oC
=
+
273
223
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)
P1V1
=
P2V2
T1
T2
101,000 × 500
300
=
101,325 × V2
273
101,000
3 500 × 273
453.5 cm×
300 × 101,325
= V2
P1
Q.200
V1
P2
=
T1
V2
T2
P1 = 101000Pa
V1 = 500cm3
T1 = 27+ 273= 300K
P2 = 101325 Pa
V2 = ? cm3
T2 = 0 + 273 = 273K
(101000)(500)
300
(101000)(500) (273)
(101,325)( 300)
=
=
(101,325)(V2)
273
V2
=
V2
P1
Q.201
V1
P2
=
T1
P1 = 975kPa
V1 = 30cm3
T1 = 25+ 273= 298K
(975)(30)
298
(975)(30) (273)
(100)( 298)
267.9614
V2
T2
P2 = 100kPa
V2 = ? cm3
T2 = 0 + 273 = 273K
=
=
=
(100)(V2)
273
V2
V2
Answer =267.9614cm3
P1
Q.202
V1
P2
=
T1
P1 = 101kPa
V1 = 1.2m3
T1 = 5+ 273= 278K
(101)(1.2)
278
(101)(1.2) (303)
(108)( 278)
V2
T2
P2 = 108 kPa
V2 = ? m3
T2 = 30+ 273 = 303K
=
=
1.2231=
(108)(V2)
303
V2
V2
Answer = 1.2231 m3
P1
Q.203
V1
P2
=
T1
P1 = 50,662.5 Pa
V1 = 283cm3
T1 = 25+ 273= 298K
V2
T2
P2 = 50,662.5 x2= 101325Pa
V2 = ? cm3
T2 =278 x 4 = 1192 K
(59,662.5)(298)
278
=
(101,325)(V2)
1112
(59662.5)(283) (1192)
(101,325)( 298)
566=
=
V2
V2
Answer = 566cm3
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.
Gas particle
The average kinetic energy of the
particles is proportional to the Kelvin
temperature.
no attraction
diameters
orare
repulsion
negligible
forces
All collisions are perfectly elastic.
Deviations from the ideal gas laws
• Two of the characteristics of ideal gases:
1) The gas molecules themselves occupy no appreciable
volume ( diameters are almost negligible in size)
2) 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.
Which will deviate most from real gas
behaviour?
• Consider the following gases – H2, HF, NH3 and F2
Least like
an ideal gas
(deviates
the most)
1. HF has hydrogen bonding between molecules - strong
attractive forces
2. 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)
3. F2 – has Van Der Walls forces attracting molecules –
very weak intermolecular forces
4. H2 – Van Der Waals forces are even less here(less
electrons present )– weakest attractive forces between
Most like an
molecules
ideal gas
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
Mandatory experiment
• Determination of the relative molecular mass
of a volatile liquid using the ideal gas equation
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
• Celsuis
• Kelvin
• Pa
• kPa
• cm3
• m3
K ( + 273 )
Celsuis (- 273 )
(÷ 1000 )
Pa (x 1000 )
m3 ( x 10 -6 )
cm3 ( x 10 6 )
Changing units of measurement
•
a)
b)
c)
d)
Convert the following to the Kelvin:
-10oC
-272oC
25oC
100oC
Convert the following to Pascals:
a)
100kPa
b)
1 x 10 3 kPa
c)
2 x 10 – 6 kPa
d)
1.01 x 105 KPa
Convert the following to cm3:
a)
1000m3
b)
1 m3
c)
2L
d)
10 x 10 4cm3
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
• PV = nRT
(200,000)(0 .000002) = n (8.31) (308)
(200,000)(0 .000002) = n
(8.31) (308)
0.4
2559.48
= n
0.000156281
= n ( the number of moles)
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
• PV = nRT
(200,000)(0 .0001) = n (8.31) (303)
(200,000)(0 .0001) = n
(8.31) (303)
20
2517.93
0.0079
= n
= n ( the number of moles)
Calculations using the ideal gas law
Q210. 1.236g of a volatile liquid gave 512cm3 of vapour
at 20oC and a pressure of 101325Pa. Calculate the
relative molecular mass
(The gas constant is 8.31JK-1 mol-1 )
PV = nRT
•
•
•
•
•
P = Pressure = 101.325kPa = 101325Pa
V = Volume = 512cm3 = .000512m3
n = number of moles = ?
R = gas constant = 8.31JK-1 mol-1
T = temperature = 20oC = 293K
Calculations using the ideal gas law
• PV = nRT
(101325)(0 .000512) = n (8.31) (293)
(101325)(0 .000512) = n
(8.31) (293)
51.8784
2424.83
0.0213
= n
= n ( the number of moles)
Calculations using the ideal gas law
261e) Calculate the relative molecular mass of the volatile liquid
X Relative Molecular Mass
Mass of X in g
Moles of X
÷ Relative Molecular Mass
0.0213 moles x RMM = 1.236g
RMM = 1.236/ 0.0213
RMM = 58.0282
RMM = 58
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 = 101300Pa
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
• PV = nRT
(101300)(0 .0002) = n (8.31) (298)
(101300)(0 .0002) = n
(8.31) (298)
20.26
2476.38
= n
0.0082
= n ( the number of moles)
Calculations using the ideal gas law
261e) Calculate the relative molecular mass of the volatile liquid
X Relative Molecular Mass
Mass of X in g
Moles of X
÷ Relative Molecular Mass
0.0082 moles x RMM = 1.0 g
RMM = 1.0/ 0.0082
RMM = 121.9512
Calculations using the ideal gas law
Q212. 1.1g of a volatile liquid occupied 3.15 x 10 -4 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 10 -4 m3
n = number of moles = ?
R = gas constant = 8.31JK-1 mol-1
T = temperature = 300K
Calculations using the ideal gas law
• PV = nRT
(100000)(3.15 x 10 -4 m3 ) = n (8.31) (300)
(100000)(0 .000315) = n
(8.31) (300)
31.5
2493
= n
0.0126
= n ( the number of moles)
Calculations using the ideal gas law
X Relative Molecular Mass
Mass of X in g
Moles of X
÷ Relative Molecular Mass
0.0126moles x RMM = 1.1 g
RMM = 1.1/ 0.0126
RMM = 87.3016
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 = 101000Pa
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
X Relative Molecular Mass
Mass of X in g
Moles of X
÷ Relative Molecular Mass
moles x 58 = 0.29 g
Moles = 0.29 / 58
Moles = 0.005
Calculations using the ideal gas law
• PV = nRT
(101000)V = (0.005)(8.31)(293)
V = (0.005)(8.31)(293)
101000
V = 0.0001m3
Calculations using the ideal gas law
To go from m3 to litres multiply by
103
• Volume has to be in litres...
0.0001m3 x 10 3 = 0.1 L
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 = 100000Pa
V = Volume = 720cm3 = .00072m3
n = number of moles = ?
R = gas constant = 8.31JK-1 mol-1
T = temperature = 10oC = 283K
Calculations using the ideal gas law
• PV = nRT
(100000)(0 .00072) = n (8.31) (283)
(100000)(0 .00072) = n
(8.31) (283)
0.0306
= n ( the number of moles)
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 100000Pa.
(The gas constant is 8.31JK-1 mol-1 )
PV = nRT
• P = Pressure = 100000Pa
• 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
X Relative Molecular Mass
Mass of X in g
Moles of X
÷ Relative Molecular Mass
moles x 44 = 2500 g
Moles = 2500 / 44
Moles = 56.8182
Calculations using the ideal gas law
• PV = nRT
(100000)V = (56.8182)(8.31)(288)
V = (0.005)(8.31)(293)
101000
V = 1.3598m3
Calculations using the ideal gas law
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?
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 10 -6 m3
n = number of moles = ?
R = gas constant = 8.31JK-1 mol-1
T = temperature = 370K
Calculations using the ideal gas law
• PV = nRT
(100000)(0.000095 ) = n (8.31) (370)
(100000)(0 .000095) = n
(8.31) (370)
0.0031
= n ( the number of moles)
Calculations using the ideal gas law
261e) Calculate the relative molecular mass of the volatile liquid
X Relative Molecular Mass
Mass of X in g
Moles of X
÷ Relative Molecular Mass
0.0031 moles x RMM = 0.275 g
RMM = 0.275/ 0.0031
RMM = 88.7097
To the nearest whole number – RMM = 89
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 = 0.000328m3
n = number of moles = ?
R = gas constant = 8.31JK-1 mol-1
T = temperature = 373K
Calculations using the ideal gas law
• PV = nRT
(98000)(0.000328 ) = n (8.31) (373)
(98000)(0 .000328) = n
(8.31) (373)
0.0104
= n ( the number of moles)
Calculations using the ideal gas law
261e) Calculate the relative molecular mass of the volatile liquid
X Relative Molecular Mass
Mass of X in g
Moles of X
÷ Relative Molecular Mass
0.0104 moles x RMM = 1.43 g
RMM = 1.43/ 0.0104
RMM = 137.5
To the nearest whole number – RMM = 138
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 = 100000Pa
V = Volume = 0.000315m3
n = number of moles = ?
R = gas constant = 8.31JK-1 mol-1
T = temperature = 300K
Calculations using the ideal gas law
• PV = nRT
(100000)(0.000315 ) = n (8.31) (300)
(100000)(0 .000315) = n
(8.31) (300)
0.0126
= n ( the number of moles)
Calculations using the ideal gas law
261e) Calculate the relative molecular mass of the volatile liquid
X Relative Molecular Mass
Mass of X in g
Moles of X
÷ Relative Molecular Mass
0.0126 moles x RMM = 1.1 g
RMM = 1.1/ 0.0126
RMM = 87.3016
To the nearest whole number – RMM = 87
Calculations using the ideal gas law