The Gas Laws Do Now read pages 70-71. The Gas Laws What happens if the Pressure and Volume are...

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The Gas Laws

Do Now read pages 70-71

The Gas Laws

• What happens if the Pressure and Volume are changed and constant temperature

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)

Pressure – A reminder

What is origin of the pressure of a gas?

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

The behaviour of gases – Boyles Law

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

Let’s do it!

The behaviour of gases

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

pV = constant

The Boyle’s laws – copy

We have found experimentally that;

At constant temperature, the pressure of a fixed mass of gas is inversely proportional to its volume.

If the volume halves the pressure doubles

p α 1/V or pV = constant

This is known as Boyle’s law

Explaining the behaviour of gases

When we compress (reduce the volume) a gas at constant temperature, the pressure increases. Why?

Explaing the behaviour of gases

When we compress (reduce the volume) a gas at constant temperature, the pressure increases. Why?

A smaller volume increases the likelihood of a particle colliding with the container walls.

Boyle’s Law

The Gas Laws

Do Q1-3 page 71

The behaviour of gases- Pressure Law

http://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!

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)

Boyle’s Law

States that the pressure of a fixed mass of gas is inversely proportional to its volume at constant temperature

P 1/V or PV = constant

When the conditions are changed P1V1 = P2V2

What to do

• A column of trapped dry air in a sealed tube by the oil

• The pressure on this volume of air can be varied by pumping air in or out of the oil reservoir to obtain different pressures

• Wait to allow the temperature to return to room temperature

Charles’ Law

States that the volume of a fixed mass of gas is directly proportional to its absolute temperature at constant pressure

V T or V/T = constant

When the conditions are changed V1/T1 = V2/T2

The ExperimentTap 1

Tap 2 Tap 3

Water reservoir

Fixed massof gas

Mercury in U tube

What to do

Fill the mercury column with mercury using the right hand tube (tap 1 open, tap 2 closed)

With tap 1 open drain some mercury using tap 2, then close tap 1 and 2. To trap a fixed mass of gas

Fill the jacket with water (make sure tap 3 is closed)

and then

Change the temperature of the water by draining some water from tap 3 and adding hot water

Equalise the pressure by leveling the columns using tap 2

Read the volume from the scale

The Results

V

T K

V

T oCA value forabsolute zero

The Results

P

V

P

1/ V

PV

P

The Charles’ Law copy

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 Charles’ lawIf T is in Kelvin

Explaing the behaviour of gases

When we heat a gas a constant pressure, the volume increases. Why?

Explaining the behaviour of gases

When we heat a gas a constant pressure, the volume increases. Why?

Increasing the volume reduces the chance of particles colliding with the container walls, opposing the effect of the particles increased kinetic energy.

Charles Law

Explaing the behaviour of gases

When we heat a gas a constant pressure, the volume increases. Why?

Explaining the behaviour of gases

When we heat a gas a constant pressure, the volume increases. Why?

Increasing the volume reduces the chance of particles colliding with the container walls, opposing the effect of the particles increased kinetic energy.

Charles Law

The Pressure law

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

Explaining the behaviour of gaseshttp://phet.colorado.edu/sims/ideal-gas/gas-properties.jnlp

When we heat a gas at constant volume, the pressure increases. Why?

Explaining the behaviour of gases

When we heat a gas at constant volume, the pressure increases. Why?

Increased average kinetic energy of the particles means there are more collisions with the container walls in a period of time and the collisions involve a greater change in momentum.

Pressure Law

Absolute Zero and the Kelvin Scale Charles’ Law and the Pressure Law suggest

that there is a lowest possible temperature that substances can go

This is called Absolute Zero The Kelvin scale starts at this point and

increases at the same scale as the Celsius Scale

Therefore -273oC is equivalent to 0 K ∆1oC is the same as ∆1 K To change oC to K, add 273 To change K to oC, subtract 273

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 exampleAt 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

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 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.

                                                                                    

                                                                                                            

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 of an ideal gas

Experiment has shown us that

pV = nR T

• p - pressure (Pa)• V - volume (m3)• n - number of mols• R - molar gas constant ( 8.31 J mol-1 K-1) • T - Temperature (K)

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

• 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

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

An Ideal Gas

Is a theoretical gas that obeys the gas laws

And thus fit the ideal gas equation exactly

Real Gases

Real gases conform to the gas laws under certain limited conditions

But they condense to liquids and then solidify if the temperature is lowered

Furthermore, there are relatively small forces of attraction between particles of a real gas

This is not the case for an ideal gas

Questions!

Questions Lots of questions.

Homework questions due 24th

January