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