Well done!
Kavin (Nik) Supatravanij
Bangkok Patana School, 11B
Tuesday, 29 November 2011
Sunday, 27 November 2011
Past Paper Answers
Dear 11.1X,
1. Finish off steps 3 for the final page (mark scheme below)
2. Do step 4 for everything in Black pen or Red pen
3. Learn your formulae (see previous mail) – test yourself/test each other/etc
Cheers,
Mr B
P5 PPQs
08 November 2011
13:32
Getting the most out of Past Paper Questions
· Step 1 = Blue Pen: Exam Conditions
o Complete the questions under exam conditions in Blue Pen. Exam conditions means in a quiet environment away from any disturbances, without looking up anything in your books. Treat it as if this is the real test. Allow yourself 1 mark = 1 minute as a time limit. This step allows you to practice your exam technique but it doesn't teach you anything new. It's important, but not as important as...
· Step 2 = Black Pen: Open Book
o Open your books and review all questions. Make any corrections and additions in Black Pen. The aim here is to get 100%. Still not sure? Use the internet or phone your friends! This step is extremely important because this is how you will learn new stuff and improve your mark so make sure you take your time and do it carefully and thoroughly.
· Step 3 = Red Pen: Mark your work
o Using the mark scheme, mark your work and correct all mistakes in Red Pen. If you've done a really good job of step 2 your corrections should be minimal. Ask your teacher for help if unsure.
· Step 4 = Learn from your mistakes
o All the stuff in Blue Pen that was correct you already know - don't waste any time revising it.
o All the stuff in Black Pen is stuff you understand but can't yet remember. Use active revision techniques such as Q+A flashcards/bullet point notes/mind mapping/etc to help you learn it.
o All the stuff in Red Pen shows problems with your understanding - ask your teacher for help.
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Tuesday, 22 November 2011
Do you know your formulae?
Do you know your formulae?
28 February 2011
07:14
Unit 5 Formulae
Tell the person next to you…
· All of the formulae from this unit (there are 6!)
o eg. "The formula that links ρ, m and V is…"
· All of the quantities from this unit
o eg. "ρ = density"
· All of the units from this unit
o eg. "The units of density are g/cm3 or kg/m3"
Answers
Formulae
· ρ = m / V
· p = F / A
· ∆p = ρgh
· p1 / T1 = p2 / T2 Gay-Lussac's Law
· p1V1 = p2V2 Boyle's Law
· TK = ToC + 273
Quantities and Units
· ρ = density (kg/m3 or g/cm3)
· m = mass (kg or g)
· V = volume (m3 or cm3)
· p = pressure (Pa or N/m2)
· F = force (N)
· A = surface area (m2)
· Δp = change in pressure (Pa or N/m2)
· g = gravitational field strength (N/kg)
· h = height/depth of fluid (m)
· T = absolute temperature (K) (not oC!)
· ToC = temperature (oC)
5.19 Boyle's Law
5.19 Boyle's Law
28 October 2011
11:11
· 5.19 use the relationship between the pressure and volume of a fixed mass of gas at constant temperature:
p1V1 = p2V2
p1 = Pressure at the beginning [kPa, bar or atm]
V1 = Volume at the beginning [m3 or cm3]
p2 = Pressure at the end [kPa, bar or atm]
V2 = Volume at the end [m3 or cm3]
(Note: can use any units for V and p as long as they are the same at the beginning and end)
5.19 Boyle's Law demos
02 November 2011
20:01
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Fun with the vacuum pump!
· Marshmellows
· Food colouring in pipettes
· Surgical gloves
5.19 Ideal graph and conclusion
09 November 2011
15:15
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5.19 Questions
07 November 2011
14:52
PFY, p.36, Q.1a, 3 and 4
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Extension: PFY, p.36, Q.5.
Thursday, 17 November 2011
5.19 Experiment
5.19 Experiment 
07 November 2011
14:32
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· Change the pressure of a fixed mass of gas at a constant temperature
· Measure the volume
· Use the EXCEL spreadsheet to analyse your results
5.19 Blank EXCEL spreadsheet for Boyle's Law practical
07 November 2011
16:16
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Tuesday, 15 November 2011
5.18
5.17 Demo
02 November 2011
19:56
Cloud formation
· Place a little water in the bottom of a 1½ litre plastic bottle
· Squeeze a few times
· Introduce a small amount of smoke
· Squeeze and release several times
· When you squeeze, the cloud disappears; when you release, the cloud reforms
Explanation
· When the pressure increases the temperature increases and vica versa
· The smoke particles are nucleating sites on which the water can condense
ANSWERS
Collins) p1/T1 = p2/T2 => 3/293 = p2/(273+55) => p2 = 3.4 bar
a) The pressure decreases, as temperature decreases.
b) As the temperature decreases, the average kinetic energy of the particles decreases, meaning that they move around with less speed, and with less energy. This means that they hit the walls of the rigid container with less force, and since the surface area of the container is still the same (rigid container), according to the p = F/A law, the pressure decreases with temperature. (pressure is prop. to temperature)
5.18 Gay-lussac's law
28 October 2011
11:11
· 5.18 use the relationship between the pressure and Kelvin temperature of a fixed mass of gas at constant volume:
p1 / T1 = p2 / T2
p1 = Pressure at the beginning [kPa, bar or atm ]
T1 = Absolute temperature at the beginning [K]
p2 = Pressure at the end [kPa, bar or atm]
T2 = Absolute temperature at the end [K]
(Note: the units of temperature must be Kelvin, not oC! The units of pressure can be any, as long as the same at the beginning and the end)
5.18 Ideal graph and conclusion
09 November 2011
15:15
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5.18 Question
07 November 2011
15:08
Collins, p.116
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a. If we cool the gas in a rigid, sealed tin can, what happens to the pressure inside the can? (1 mark)
b. Explain your answer to part a. by using the Kinetic Theory (4 marks)
Sunday, 13 November 2011
5.17
5.17 starter 
02 November 2011
20:01
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Why do the eggs get sucked into the bottles?!
Explanation
· The burning paper in the bottle heats the air in the bottle
· When the egg gets placed on top, the oxygen supply in the bottle is rapidly depleted and the paper goes out
· The bottle is sealed by the egg and now has a constant volume of gas inside
· The hot gas in the bottle now starts to cool which reduces the pressure inside the bottle
· The pressure outside the bottle remains unchanged and so there is now an unbalanced force on the egg which accelerates the egg into the bottle
5.17
28 October 2011
11:11
· 5.17 describe the qualitative relationship between pressure and Kelvin temperature for a gas in a sealed container
Instructions
· Launch the application on this website: http://phet.colorado.edu/en/simulation/gas-properties
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· Put 5 pumps of gas in
· Set volume as the Constant Parameter
· Heat to 1000K
· Watch what happens to the Pressure
Conclusion
· If you increase the temperature, you increase the pressure
5.18 Gay-lussac's law
5.18 Gay-lussac's law
28 October 2011
11:11
· 5.18 use the relationship between the pressure and Kelvin temperature of a fixed mass of gas at constant volume:
p1 / T1 = p2 / T2
p1 = Pressure at the beginning [kPa, bar or atm ]
T1 = Absolute temperature at the beginning [K]
p2 = Pressure at the end [kPa, bar or atm]
T2 = Absolute temperature at the end [K]
(Note: the units of temperature must be Kelvin, not oC! The units of pressure can be any, as long as the same at the beginning and the end)
5.18 Ideal graph and conclusion
09 November 2011
15:15
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Saturday, 12 November 2011
5.16 Answers
5.16 Answers
28 October 2011
11:11
Answers
1. What variable remains constant for this experiment?
Volume
2. Explain in terms of the particles what happened to the pressure when the temperature increased
When the temperature is increased the particles have a greater average KE and therefore hit the walls of the container with more force and more frequently. This increases the pressure.
3. Is the temperature proportional to the average speed? Justify your answer
No; the graph is not a straight line
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4. Is the temperature proportional to the average kinetic energy of the particles? Justify your answer
Yes; the graph of temperature against (average speed of particles)2 is a straight line.
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NB: m, the mass of the particles is a constant and so will only affect the gradient of the graph, not the shape of the graph
5. Why is the word 'average' used?
The particles in the container have got a range of speeds and therefore a range of KEs. Some particles will be moving faster and some slower but, on average, T α KE.
You can view this in the programme. Click on "measurement tool", "energy histograms".
Thursday, 10 November 2011
5.14
5.14
28 October 2011
11:10
· 5.14 describe the Kelvin scale of temperature and be able to convert between the Kelvin and Celsius scales
Converting Centigrade to Kelvin
TK = ToC + 273
Converting Kelvin to Centigrade
ToC = TK - 273
TK = Temperature in Kelvin [K]
ToC = Temperature in Degrees Centigrade [oC]
5.14 Questions
02 November 2011
18:29
· Collins p.118
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ANSWERS
1. Kinetic Theory says that the higher the temperature, the more energy the particles in that system will have. Therefore, in order for something to have an 'absolute zero' temperature, it must similarly have absolutely zero energy. -273 Celsius is regarded to be this temperature (or 0 Kelvin), though it has not been ever found in the universe.
2. (a) Celsius -> Kelvin
i) 20'C + 273 = 293K
ii) 150'C + 273 = 423K
iii) 1000'C +273 = 1273K
(b) Kelvin -> Celsius
i) 300K - 273 = 27'C
ii) 650 - 273 = 377'C
iii) 1000 - 273 = 727'C
5.13
5.13 Starter
02 November 2011
18:17
· How can you fit a giraffe, 2 dogs and a swan into a standard laboratory beaker?!
5.13 Starter 2
02 November 2011
18:17
· Use particle theory to explain why the gas in the balloon contracts
Explanation
· The temperature of the gas inside the balloon decreases so the average speed of the particles decreases
· Consequently the gas particles collide with the walls of the balloon with less force and less collisions per second
· Because the walls of the container are flexible, the volume decreases
5.13 Charles' law
28 October 2011
11:10
· 5.13 understand that there is an absolute zero of temperature which is –273oC
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Open the Charles' law interactive experiment
· Adjust the temperature
· What’s the relationship between temperature and volume?
· Plot a graph of V against T
· Take a screen shot of the graph
5.13 results and conclusion
28 October 2011
11:10
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Conclusion
· Volume is directly proportional to absolute (Kelvin) temperature
· V α T
Wednesday, 9 November 2011
5.11 (REDONE)
Instructions for Objective 5.11 
THE DIAGRAM SHOULD SHOW ARROWS OF RANDOM LENGTH AND DIRECTION BETWEEN EACH AIR PARTICLE AND THE SMOKE PARTICLE. (THESE SHOW THE 'COLLISIONS' RESPONSIBLE FOR BROWNIAN MOTION).
2. Explain what is meant by Brownian Motion of smoke particles in air and how it provides evidence for air particles (4 marks)
-> Brownian motion explains the phenomenon of visible smoke particles apparently 'jiggling' about, and traveling in random directions. These smoke particles are pushed in these random directions by the random collisions with small, fast moving air particles; this provides evidence for air particles as without the air particles, the smoke particles would not be 'jiggling about' but BE STATIONARY, AS THE SMOKE PARTICLES ARE SOLID, AND THEREFORE HAVE NO INTRINSIC KINETIC ENERGY TO MOVE ABOUT (ONLY THE ATOMS VIBRATE, BUT THIS DOES NOT AFFECT THE MOVEMENT OF THE ENTIRE SMOKE PARTICLE)
3. What change would you expect to see in the movement of the smoke particles if the air was cooled down? Why? (2 marks)
-> I would expect the smoke particles to move less erratically, as the air particles would have less kinetic energy, meaning that there would be less collisions with the smoke particles to move them about (collision theory). The smoke particles would therefore still be moving, simply not as much.
1. 5.11 Starter. Watch the video and think about the question. No need to type anything.
2. 5.11. Watch the videos and animations for the 3 models of Brownian Motion (for Model 3 you need to open the attached). Think about the questions. No need to type anything.
3. 5.11 explained. Check your understanding with the model answers.
4. 5.11 Questions. Forward this e-mail to your blog and complete the questions.
5. Answers to step 2 will be sent separately. Don’t look at them until you’ve done the work!
5.11 Starter
02 November 2011
16:58
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· You're looking at smoke particles in air under a microscope
· They appear to be jiggling about
· Why?
· (Don't worry if you can't work this out straight away - Albert Einstein was the bloke who eventually explained what's happening here!)
5.11
28 October 2011
11:10
· 5.11 understand the significance of Brownian motion
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Model 1
· What does the red puck represent?
· What do the metal balls represent?
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Model 3
· What do the "smoke" particles look like?
· Why are they moving?
· What do the "air" particles look like?
5.11 explained
28 October 2011
11:10
Model 1
· What does the red puck represent?
o The large, visible smoke particle
· What do the metal balls represent?
o The small, not visible air particles
Model 2
· What do the small red particles represent?
o The small, not visible air particles
· What does the large blue particle represent?
o The large, visible smoke particle
· What does the view on the left of the screen represent?
o The view through the microscope lense
· Why can‘t you see the red particles in this view?
o They are too small to see
Model 3
· What do the "smoke" particles look like?
o They are the 5 large, sand coloured particles
· Why are they moving?
o Small, fast moving air particles are colliding with the smoke particles and making them move
· What do the "air" particles look like?
o They are the numerous, small, white particles
5.11 Questions
02 November 2011
17:21
1. Draw the path of a smoke particle in air (3 marks)
-> Random path (or 'random walk') similar to:
THE DIAGRAM SHOULD SHOW ARROWS OF RANDOM LENGTH AND DIRECTION BETWEEN EACH AIR PARTICLE AND THE SMOKE PARTICLE. (THESE SHOW THE 'COLLISIONS' RESPONSIBLE FOR BROWNIAN MOTION).
2. Explain what is meant by Brownian Motion of smoke particles in air and how it provides evidence for air particles (4 marks)
-> Brownian motion explains the phenomenon of visible smoke particles apparently 'jiggling' about, and traveling in random directions. These smoke particles are pushed in these random directions by the random collisions with small, fast moving air particles; this provides evidence for air particles as without the air particles, the smoke particles would not be 'jiggling about' but BE STATIONARY, AS THE SMOKE PARTICLES ARE SOLID, AND THEREFORE HAVE NO INTRINSIC KINETIC ENERGY TO MOVE ABOUT (ONLY THE ATOMS VIBRATE, BUT THIS DOES NOT AFFECT THE MOVEMENT OF THE ENTIRE SMOKE PARTICLE)
3. What change would you expect to see in the movement of the smoke particles if the air was cooled down? Why? (2 marks)
-> I would expect the smoke particles to move less erratically, as the air particles would have less kinetic energy, meaning that there would be less collisions with the smoke particles to move them about (collision theory). The smoke particles would therefore still be moving, simply not as much.
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