Preparation - Chapter 5.1a, Chapter 5.1b, Chapter 5.1c, Chapter 5.1d, Chapter 5.1e, Chapter 5.1f
Participation - Assignment P.167 Q10, 12, 14, 16 and P.185 Q28-31
Presentation - Blackboard, Student's Work
Possession
> Pressure is force per unit are, p = F/A
> Gas pressure is due to collision of molecules on the wall of container.
> A balloon maintain its volume when the gas pressure inside equals the external pressure outside.
> Bourdon gauge measures gas pressure
> Atmospheric pressure ~ 100 000 Pa
> Draw a labelled diagram and describe the following experiments. State the precautions and explain the data analysis (what graph to plot? What the graph should look like?)
> Boyle's law, pV = constant for a fix mass of gas at constant temperature.
> Pressure-temperature relationship, p/T for a fixed mass of gas with constant volume.
> Volume-temperature relationship, V/T for a fixed mass of gas under constant pressure.
> General gas law pV = nRT where n is the number of mole
> n = N/NA where NA is called Avogadro's number 6.02 x 10^23
> IDEAL GAS always obey general gas law
> Solve problems of mixing gases (P.163)
> Graphs of gas laws (P.164) - what the slope and intercept(if any) depends on?
Chapter 5.1a (5:57) - 116
Chapter 5.1b (3:59) - 101
Chapter 5.1c (5:00) - 94
Chapter 5.1d (3:38) - 88
Chapter 5.1e (3:58) - 81
Chapter 5.1f (1:32) - 64
(a) If V is fixed (e.g. using a glass container), when Temperature T increases by 20%, the pressure P will …
A.increase by 20%
B.decrease by 20%
C.increase by 16.7%
D.decrease by 16.7%
Answer : A (increases by 20%)
Pressure and Temperature are directly proportional.
When temperature increases by 20%, pressure will also increase by 20%.
(b) If temperature T is constant, when P increases by 20%, how would V change?
A.increase by 20%
B.decrease by 20%
C.increase by 16.7%
D.decrease by 16.7% <ANS>
Answer : D (decreases by 16.7%)
Pressure and volume are inversely proportional.
When pressure increases by 20%, 1/Pressure decreases, and it decreases by 16.7%.
For example, P = 100, then P' = 120. So, 1/P = 0.01 while 1/P' = 0.00833,
so precentage change is (P'-P)/P = (0.00833 - 0.01)/(0.01) = -0.167 = -16.7%
Also note that percentage change is NOT (P'-P)/P' !!
Preparation - Chapter 5.2a, Chapter 5.2b, Chapter 5.2c, Chapter 5.2d
Participation - Assignment P.167 Q5, 9-11 and P.185 Q17, 20-27 except 24
Presentation - Blackboard, Student's Work
Possession
> Brownian motion (of smoke particle) is a result of frequent collision by gas molecules in random motion
> IDEAL gas assumption (P.170) - no intermolecular force, negligible gas molecule volume
> Kinetic theory equation pV = 1/3 (Nmc^2), refer to P.171
> Calculate TOTAL moleuclar KE = 3RT/2
> Calculate average molecular KE = 3RT/2NA, so absolute temperature and average molecular KE are directly proportional
> Calculate INTERNAL energy of a gas
> Explain gas laws using kinetic theory (NOT kinetic theory equation)
> PV = constant. When external pressure increases, gas volume is reduced, gas pressure increases as particle collision becomes more frequent. Application of PV = constant to solve problems.
> P/T = constant. When gas temperature increases, gas molecules move faster, the collision on the container wall becomes more frequent and more violent (change in momentum per collision increases), so gas pressure increases.
> V/T = constant. When gas temperature increases, gas molecules move faster, the collision on the container wall becomes more frequent and more violent (change in momentum per collision increases), so gas pressure becomes greater than the external pressure. The gas expands. After volume is increased, the collision frequency decreases and gas pressure also decreases until it is the same as the external pressure again.
Chapter 5.2a (4:18) - 81
Chapter 5.2b (2:30) - 73
Chapter 5.2c (2:37) - 61
Chapter 5.2d (8:42) - 63