Warm Up Questions: 1. Define temperature. 3. Draw and label the table on the board. well as the latent heat of fusion and vaporization.
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1 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.
2 3.2 Modelling a gas
3 Moles! Boo!
4 Moles! Equal masses of different elements will contain different numbers of atoms (as atoms of different elements have different masses)
5 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
6 Moles! You need to learn this definition. 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
7 23 How big is 6.02 x 10? Imagine the floor of this classroom covered in unpopped popcorn
8 23 How big is 6.02 x 10? Imagine the all the floors in the school covered in unpopped popcorn I hope you re going to clear it up!
9 23 How big is 6.02 x 10? Imagine the whole of Warsaw covered in unpopped popcorn
10 23 How big is 6.02 x 10? Imagine the whole of Poland covered in unpopped popcorn
11 23 How big is 6.02 x 10? Imagine the whole of Europe covered in unpopped popcorn You are here!
12 23 How big is 6.02 x 10? Imagine the whole of Europe covered in unpopped popcorn to a depth of 12km! You are here!
13 23 How big is 6.02 x 10? Imagine the whole of Europe covered in unpopped popcorn to a depth of 12 km! Count the grains and that is 6 x 1023! That s how big 6.02 x 1023 is!
14 Moles! You need to learn this definition. 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
15 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)
16 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)
17 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
18 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?
19 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
20 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
21 The kinetic theory of gases and the gas laws
22 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;
23 Ideal gas assumptions The particles of gas (atoms or molecules) obey Newton s laws of motion. You should know these by now!
24 Ideal gas assumptions The particles in a gas move with a range of speeds
25 Ideal gas assumptions The volume of the individual gas particles is very small compared to the volume of the gas
26 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)
27 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?
28 Ideal gas assumptions The duration of a collision is small compared to the time between collisions.
29 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)
30 Pressure A reminder What is origin of the pressure of a gas?
31 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
32 Let s start by investigating gas behaviour
33 3.2 Pressure law practical When we heat a gas at constant volume, what happens to the pressure? Why? Let s do it!
34 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!
35 HOMEWORK Finish BOTH investigations for 11th December.
36 Don t forget Specific latent heat questions in next Thursday 4th December
37 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)
38 The behaviour of gases When we compress (reduce the volume) a gas at constant temperature, what happens to the pressure? Why? pv = constant
39 The behaviour of gases When we heat a gas a constant pressure, what happens to its volume? Why?
40 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)
41 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)
42 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?).
43 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
44 The gas laws At constant pressure, the volume of a fixed mass of gas is proportional to its temperature; VαT or V/T = constant This is known as Charle s law If T is in Kelvin
45 The gas laws At constant volume, the pressure of a fixed mass of gas is proportional to its temperature; pαt or p/t = constant This is known as the Pressure lawif T is in Kelvin (Gay-Lussac s law)
46 The equation of state By combining these three laws pv = constant V/T = constant p/t = constant We get pv/t = constant Or Remember, T must be in Kelvin p 1V 1 = p 2V 2 T1 T2
47 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
48 An example At 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.
49 An example At 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.
50 An example At 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.
51 pv = constant T
52 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
53 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
54 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
55 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
56 Average kinetic energy of a particle (ideal gas) Ēk = (3/2)kBT = 3RT/2NA where Ēk = average kinetic energy kb = Boltzmann s constant T = temperature in Kelvin R = Gas constant 8.31 J.K-1.mol-1 NA = Number of particles
57 3.2 Ideal gas questions
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