Topic 10 – Thermal physics

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)

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

The behaviour of gases When we heat a gas at constant volume, what happens to the pressure? Why? Let’s do it!

The behaviour of gases 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? Let’s do it!

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)

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 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 If T is in Kelvin

The equation of state By combining these three laws pV = constant V/T = constant p/T = constant We get pV/T = constant Or p 1 V 1 =p 2 V 2 T 1 T 2 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 10 4 Pa. At seas level these values are 300K and 1.0 x 10 5 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 example At the top of Mount Everest the temperature is around 250K, with atmospheric pressure around 3.3 x 10 4 Pa. At seas level these values are 300K and 1.0 x 10 5 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 m 3. Therefore at sea level p 1 = 1.0 x 10 5 Pa, V 1 = 0.83 m 3, T 1 = 300K.

An example At the top of Mount Everest the temperature is around 250K, with atmospheric pressure around 3.3 x 10 4 Pa. At seas level these values are 300K and 1.0 x 10 5 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 p 1 = 1.0 x 10 5 Pa, V 1 = 0.83 m 3, T 1 = 300K. At the top of Mount Everest p 2 = 3.3 x 10 4 Pa, V 2 = ? m 3, T 1 = 250K.

An example At the top of Mount Everest the temperature is around 250K, with atmospheric pressure around 3.3 x 10 4 Pa. At seas level these values are 300K and 1.0 x 10 5 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 p 1 = 1.0 x 10 5 Pa, V 1 = 0.83 m 3, T 1 = 300K. At the top of Mount Everestp 2 = 3.3 x 10 4 Pa, V 2 = ? m 3, T 1 = 250K. p 1 V 1 /T 1 = p 2 V 2 /T 2 (1.0 x 10 5 Pa x 0.83 m 3 )/300K = (3.3 x 10 4 Pa x V 2 )/250K V 2 = 2.1 m 3, This is the volume of 1kg of air on Everest Density = mass/volume = 1/2.1 = 0.48 kg.m -3.

pV= constant T

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

Moles! Hi Chris!

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 (N A ) which is equal to 6.02 x You need to learn this definition.

How big is 6 x ? Imagine the whole of the United states You are here!

How big is 6 x ? Imagine the whole of the United states covered in unpopped popcorn

How big is 6 x ? Imagine the whole of the United states covered in unpopped popcorn to a depth of six miles!

How big is 6 x ? Imagine the whole of the United states covered in unpopped popcorn to a depth of six miles! Count the grains and that is 6 x !

Moles! For example, Hydrogen (H 2 ) 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 )

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 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

Relative formula mass We can use the idea of moles and apply it to molecules using relative formula mass. C 2 H 5 OH RFM = (2 x 12) + (6 x 1) + (1 x 16) = 46 46g of ethanol = I mole of ethanol molecules

What about moles and gases?

Equal volumes Luckily, equal volumes of gas contain the same number of particles (at the same temperature and pressure)

Sample question A container of hydrogen of volume 0.1m 3 and temperature 25°C contains 3.20 x molecules. What is the pressure in the container? K.A.Tsokos “Physics for the IB Diploma” 5 th Edition

Sample question A container of hydrogen of volume 0.1m 3 and temperature 25°C contains 3.20 x molecules. What is the pressure in the container? # moles = 3.20 x /6.02 x = 0.53 K.A.Tsokos “Physics for the IB Diploma” 5 th Edition

Sample question A container of hydrogen of volume 0.1m 3 and temperature 25°C contains 3.20 x 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 10 4 N.m -2 K.A.Tsokos “Physics for the IB Diploma” 5 th Edition

Questions!