Topic 10 – Thermal physics

Topic 10 – Thermal physics

Topic 9 TEST, Thursday October 1st.

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;

Ideal gas assumptions

• The particles of gas (atoms or molecules) obey Newton’s laws of motion.

You should know these by now!


• The particles in a gas move with a range of speeds


• The volume of the individual gas particles is very small compared to the volume of the gas


• The collisions between the particles and the walls of the container and between the particles themselves are elastic (no kinetic energy lost)


• 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?


• 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)


What is origin of the pressure of a gas?


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

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!

http://phet.colorado.edu/sims/ideal-gas/gas-properties.jnlp

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)

http://phet.colorado.edu/sims/ideal-gas/gas-properties.jnlp


When we compress (reduce the volume) a gas at constant temperature, what happens to the pressure? Why?

Let’s do it!


When we compress (reduce the volume) a gas at constant temperature, what happens to the pressure? Why?

pV = constant


When we heat a gas a constant pressure, what happens to its volume? Why?


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 lawIf 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 = constantV/T = constantp/T = constant

We get pV/T = constant

Or p1V1 = p2V2

T1 T2

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 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

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.

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.

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.

pV = constantT

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

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

Sample question


# moles = 3.20 x 1023/6.02 x 1023 = 0.53


Sample question


# 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


Questions!

Page 181Questions 2, 4, 6, 9

Page 182Questions 12, 13, 17.

Topic 10 – Thermal physics

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