topic 10 – thermal physics
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Topic 10 – Thermal physics. Topic 9 TEST, Thursday October 1 st. 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;. - PowerPoint PPT PresentationTRANSCRIPT
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Topic 10 – Thermal physics
Topic 9 TEST, Thursday October 1st.
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The kinetic theory of gases and the gas laws
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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;
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Ideal gas assumptions
• The particles of gas (atoms or molecules) obey Newton’s laws of motion.
You should know these by now!
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Ideal gas assumptions
• The particles in a gas move with a range of speeds
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Ideal gas assumptions
• The volume of the individual gas particles is very small compared to the volume of the gas
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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)
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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?
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Ideal gas assumptions
• The duration of a collision is small compared to the time between collisions.
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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)
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Pressure – A reminder
What is origin of the pressure of a gas?
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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
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The behaviour of gases
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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?
Let’s do it!
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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)
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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!
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The behaviour of gases
When we compress (reduce the volume) a gas at constant temperature, what happens to the pressure? Why?
pV = constant
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The behaviour of gases
When we heat a gas a constant pressure, what happens to its volume? Why?
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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)
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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)
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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
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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 lawIf T is in Kelvin
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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
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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
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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?
“Physics”, Patrick Fullick, Heinemann
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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.
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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.
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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.
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pV = constantT
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The equation of state
Experiment has shown us that
pV = nRT
Where n = number of moles of gas and R = Gas constant
(8.31J.K-1.mol-1) Remember, T must be in Kelvin
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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
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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
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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
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Questions!
Page 181Questions 2, 4, 6, 9
Page 182Questions 12, 13, 17.