Post on 20-Jan-2021
Warm Up Questions:1. Define temperature.
2. What is heat?
3. Draw and label the table on the board.
- include solid, liquid and gas as well as the latent heat of fusion and vaporization.
3.2 – Modelling a gas
Moles!
Boo!
Moles!
Equal masses of different elements will contain different numbers of atoms (as atoms of different elements have different masses)
Moles!
It is sometimes useful for physicists and chemists (but we don’t care about them) to compare the number of atoms or molecules in an amount of substance.To do this we use the idea of moles.
A chemist
Moles!
• One mole of a substance contains the same number of molecules/atoms as in 12 grams of carbon-12.
• This number (of atoms or molecules) is known as the Avogadro constant (NA) which is equal to 6.02 x 1023
You need to learn this definition.
How big is 6.02 x 1023?
Imagine the floor of this classroom covered in unpopped popcorn
How big is 6.02 x 1023?
Imagine the all the floors in the school covered in unpopped popcorn
I hope you’re going to clear it
up!
How big is 6.02 x 1023?
Imagine the whole of Warsaw covered in unpopped popcorn
How big is 6.02 x 1023?
Imagine the whole of Poland covered in unpopped popcorn
How big is 6.02 x 1023?
Imagine the whole of Europe covered in unpopped popcorn
You are here!
How big is 6.02 x 1023?
Imagine the whole of Europe covered in unpopped popcorn to a depth of 12km!
You are here!
How big is 6.02 x 1023?
Imagine the whole of Europe covered in unpopped popcorn to a depth of 12 km! Count the grains and that is 6 x 1023!
600000000000000000000000
That’s how big 6.02 x 1023 is!
Moles!
• One mole of a substance contains the same number of molecules/atoms as in 12 grams of carbon-12.
• This number (of atoms or molecules) is known as the Avogadro constant (NA) which is equal to 6.02 x 1023
You need to learn this definition.
Moles!
For example, Hydrogen (H2) has a relative molecular mass of 2, so 2 grams of hydrogen (one mole) contains the same number of molecules as atoms in 12g of carbon-12 (6.02 x 1023)
Moles!
It follows therefore that 7g of lithium (atomic mass 7), 20g neon (atomic mass 20) or 39 g potassium (atomic mass 39) all contain the same number of atoms (1 mole or 6.02 x 1023 atoms)
Moles!
• The number of moles of a substance can thus be found by dividing the mass of substance by its relative atomic or molecular mass
n = mass/RAM
Example
How many moles of sulphur atoms are there in 80g of sulphur? How many grams of carbon would have the same number of atoms?
Example
How many moles of sulphur atoms are there in 80g of sulphur? How many grams of carbon would have the same number of atoms?
N = mass/RAM = 80/32 = 2.5 moles
Example
How many moles of sulphur atoms are there in 80g of sulphur? How many grams of carbon would have the same number of atoms?
N = mass/RAM = 80/32 = 2.5 moles
Mass of carbon = RAM x n = 12 x 2.5 = 30 g
The kinetic theory of gases and the gas laws
Kinetic theory/ideal gas
We can understand the behaviour of gases using a very simple model, that of an “ideal” gas.
The model makes a few simple assumptions;
Ideal gas assumptions
• The particles of gas (atoms or molecules) obey Newton’s laws of motion.
You should know these by now!
Ideal gas assumptions
• The particles in a gas move with a range of speeds
Ideal gas assumptions
• The volume of the individual gas particles is very small compared to the volume of the gas
Ideal gas assumptions
• The collisions between the particles and the walls of the container and between the particles themselves are elastic (no kinetic energy lost)
Ideal gas assumptions
• There are no forces between the particles (except when colliding). This means that the particles only have kinetic energy (no potential)
Do you remember what internal energy is?
Ideal gas assumptions
• The duration of a collision is small compared to the time between collisions.
Pressure – A reminder
Pressure is defined as the normal (perpendicular) force per unit area
P = F/A
It is measured in Pascals, Pa (N.m-2)
Pressure – A reminder
What is origin of the pressure of a gas?
Pressure – A reminder
Collisions of the gas particles with the side of a container give rise to a force, which averaged of billions of collisions per second macroscopically is measured as the pressure of the gas
Change of momentum
Let’s start by investigating gas behaviour
3.2 Pressure law practical
When we heat a gas at constant volume, what happens to the pressure? Why?
Let’s do it!
3.2 Boyle’s law practical
When we compress (reduce the volume) a gas at constant temperature, what happens to the pressure? Why?
Let’s do it!
HOMEWORK
• Finish BOTH investigations for 11th December.
Don’t forget – Specific latent
heat questions in next Thursday 4th
December
The behaviour of gaseshttp://phet.colorado.edu/sims/ideal-gas/gas-properties.jnlp
When we heat a gas at constant volume, what happens to the pressure? Why?
P α T (if T is in Kelvin)
The behaviour of gases
When we compress (reduce the volume) a gas at constant temperature, what happens to the pressure? Why?
pV = constant
The behaviour of gases
When we heat a gas a constant pressure, what happens to its volume? Why?
The behaviour of gases
When we heat a gas a constant pressure, what happens to its volume? Why?
V α T (if T is in Kelvin)
Explaining the behaviour of gases
In this way we are explaining the macroscopic behaviour of a gas (the quantities that can be measured like temperature, pressure and volume) by looking at its microscopic behaviour (how the individual particles move)
Note
• Real gases do behave approximately as ideal gases at high temperatures and low pressures (why?)
• At high pressures and low temperatures real gases do NOT behave ideally (why?).
The gas laws
We have found experimentally that;At constant temperature, the pressure of a fixed mass of gas is inversely proportional to its volume.
p α 1/V or pV = constant
This is known as Boyle’s or Mariotte’s law
The gas laws
At constant pressure, the volume of a fixed mass of gas is proportional to its temperature;
V α T or V/T = constant
This is known as Charle’s law If T is in Kelvin
The gas laws
At constant volume, the pressure of a fixed mass of gas is
proportional to its temperature;
p α T or p/T = constant
This is known as the Pressure law (Gay-Lussac’s law)
If T is in Kelvin
The equation of state
By combining these three laws
pV = constantV/T = constantp/T = constant
We get pV/T = constant
Or p1V1 = p2V2 T1 T2
Remember, T must be in Kelvin
An example
At the top of Mount Everest the temperature is around 250K, with atmospheric pressure around 3.3 x 104 Pa. At sea level these values are 300K and 1.0 x 105 Pa respectively. If the density of air at sea level is 1.2 kg.m-3, what is the density of the air on Mount Everest?
“Physics”, Patrick Fullick, Heinemann
An exampleAt the top of Mount Everest the temperature is around 250K, with atmospheric
pressure around 3.3 x 104 Pa. At seas level these values are 300K and 1.0 x 105 Pa respectively. If the density of air at sea level is 1.2 kg.m-3, what is the density of the air on Mount Everest?
Take 1kg of air at sea level
Volume = mass/density = 1/1.2 = 0.83 m3.
Therefore at sea level
p1 = 1.0 x 105 Pa, V1 = 0.83 m3, T1 = 300K.
An exampleAt the top of Mount Everest the temperature is around 250K, with atmospheric
pressure around 3.3 x 104 Pa. At seas level these values are 300K and 1.0 x 105 Pa respectively. If the density of air at sea level is 1.2 kg.m-3, what is the density of the air on Mount Everest?
Therefore at sea level p1 = 1.0 x 105 Pa, V1 = 0.83 m3, T1 = 300K.
At the top of Mount Everestp2 = 3.3 x 104 Pa, V2 = ? m3, T1 = 250K.
An exampleAt the top of Mount Everest the temperature is around 250K, with atmospheric
pressure around 3.3 x 104 Pa. At seas level these values are 300K and 1.0 x 105 Pa respectively. If the density of air at sea level is 1.2 kg.m-3, what is the density of the air on Mount Everest?
Therefore at sea level p1 = 1.0 x 105 Pa, V1 = 0.83 m3, T1 = 300K.At the top of Mount Everest p2 = 3.3 x 104 Pa, V2 = ? m3, T1 = 250K.
p1V1/T1 = p2V2/T2
(1.0 x 105 Pa x 0.83 m3)/300K = (3.3 x 104 Pa x V2)/250K
V2 = 2.1 m3,
This is the volume of 1kg of air on Everest
Density = mass/volume = 1/2.1 = 0.48 kg.m-3.
pV = constantT
The equation of state
Experiment has shown us that
pV = nR T
Where n = number of moles of gas and R = Gas constant
(8.31J.K-1.mol-1) Remember, T must be in Kelvin
Sample question
• A container of hydrogen of volume 0.1m3 and temperature 25°C contains 3.20 x 1023 molecules. What is the pressure in the container?
K.A.Tsokos “Physics for the IB Diploma” 5th Edition
Sample question
• A container of hydrogen of volume 0.1m3 and temperature 25°C contains 3.20 x 1023 molecules. What is the pressure in the container?
# moles = 3.20 x 1023/6.02 x 1023 = 0.53
K.A.Tsokos “Physics for the IB Diploma” 5th Edition
Sample question
• A container of hydrogen of volume 0.1m3 and temperature 25°C contains 3.20 x 1023 molecules. What is the pressure in the container?
# moles = 3.20 x 1023/6.02 x 1023 = 0.53
P = RnT/V = (8.31 x 0.53 x 298)/0.1 = 1.3 x 104 N.m-2
K.A.Tsokos “Physics for the IB Diploma” 5th Edition
Average kinetic energy of a particle (ideal gas)
• Ēk = (3/2)kBT = 3RT/2NA
where Ēk = average kinetic energykB = Boltzmann’s constantT = temperature in KelvinR = Gas constant 8.31 J.K-1.mol-1
NA = Number of particles
3.2 Ideal gas questions