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

pV = constant p 1 V 1 = p 2 V 2 (at constant temp) Can you answer the questions that Mr Porter is giving you?

Do Now finish Q2 p71 p 1 V 1 = p 2 V 2 In a vertical cylinder. The initial pressure of the trapped air is 1.0 x 10 5 Pa. The volume of the air decreases from 860 cm 3 to 645 cm 3 The temperature is constant. Calculate the final pressure.

Learning today Pressure Law Describe qualitatively the effect of a change of temperature on the pressure of a gas at constant volume Charles’ Law Describe qualitatively the effect of a change of temperature on the volume of a gas at constant pressure

p 1 V 1 = p 2 V 2 In a Vertical cylinder. The initial pressure of the trapped air is 1.0 x 10 5 Pa. The volume of the air decreases from 860 cm 3 to 645 cm 3 The temperature is constant. Calculate the final pressure.

Graph y-axis labelled θ / 0 C[1] plots occupying at least half of grid on suitable scale start at about 50 0 C on the y - axis[1] all plots correct to ½ square[1] well judged single, smooth curve line, not ‘point-to-point’ [1] thin line [1]

Graph Mark scheme

The behaviour of gases- Pressure Law Gas Laws Gas Laws When we heat a gas at constant volume, what happens to the pressure? Why? Let’s do it!

The behaviour of gases When we heat a gas at constant volume, what happens to the pressure? Why? P α T (if T is in Kelvin)

The Pressure law- copy Gas Laws Gas Laws At constant volume, the pressure of a fixed mass of gas is proportional p α T or p/T = constant This is known as the Pressure law If T is in Kelvin

Explaining the behaviour of gases 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 an increase in pressure. Pressure Law

The Results P T K P T o C A value for absolute zero

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’ law When the conditions are changed V 1 /T 1 = V 2 /T 2 If 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, so the pressure remains constant. Charles Law

The Results V T K V T o C A value for absolute zero

Absolute Zero and the Kelvin Scale w Charles’ Law and the Pressure Law suggest that there is a lowest possible temperature that substances can go w This is called Absolute Zero w The Kelvin scale starts at this point and increases at the same scale as the Celsius Scale

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

Questions! Homework Graph question. Due Monday 12 th March

The equation of state By combining these three laws pV = constant V/T = constant p/T = constant We get pV/T = constant Or p 1 V 1 =p 2 V 2 T 1 T 2 Remember, T must be in Kelvin

The Results P V P 1/ V PV P

An example At the top of Mount Everest the temperature is around 250K, with atmospheric pressure around 3.3 x 10 4 Pa. At sea level these values are 300K and 1.0 x 10 5 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 example At the top of Mount Everest the temperature is around 250K, with atmospheric pressure around 3.3 x 10 4 Pa. At seas level these values are 300K and 1.0 x 10 5 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 m 3. Therefore at sea level p 1 = 1.0 x 10 5 Pa, V 1 = 0.83 m 3, T 1 = 300K.

An example At the top of Mount Everest the temperature is around 250K, with atmospheric pressure around 3.3 x 10 4 Pa. At seas level these values are 300K and 1.0 x 10 5 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 p 1 = 1.0 x 10 5 Pa, V 1 = 0.83 m 3, T 1 = 300K. At the top of Mount Everest p 2 = 3.3 x 10 4 Pa, V 2 = ? m 3, T 1 = 250K.

An example At the top of Mount Everest the temperature is around 250K, with atmospheric pressure around 3.3 x 10 4 Pa. At seas level these values are 300K and 1.0 x 10 5 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 p 1 = 1.0 x 10 5 Pa, V 1 = 0.83 m 3, T 1 = 300K. At the top of Mount Everestp 2 = 3.3 x 10 4 Pa, V 2 = ? m 3, T 1 = 250K. p 1 V 1 /T 1 = p 2 V 2 /T 2 (1.0 x 10 5 Pa x 0.83 m 3 )/300K = (3.3 x 10 4 Pa x V 2 )/250K V 2 = 2.1 m 3, This is the volume of 1kg of air on Everest Density = mass/volume = 1/2.1 = 0.48 kg.m -3.

pV= constant T

The equation of state of an ideal gas Experiment has shown us that pV = nR T p - pressure (Pa) V - volume (m 3 ) 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 0.1m 3 and temperature 25°C contains 3.20 x molecules. What is the pressure in the container? K.A.Tsokos “Physics for the IB Diploma” 5 th Edition

Sample question A container of hydrogen of volume 0.1m 3 and temperature 25°C contains 3.20 x molecules. What is the pressure in the container? # moles = 3.20 x /6.02 x = 0.53 K.A.Tsokos “Physics for the IB Diploma” 5 th Edition

Sample question A container of hydrogen of volume 0.1m 3 and temperature 25°C contains 3.20 x 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 10 4 N.m -2 K.A.Tsokos “Physics for the IB Diploma” 5 th Edition

An Ideal Gas w Is a theoretical gas that obeys the gas laws w And thus fit the ideal gas equation exactly

Real Gases w Real gases conform to the gas laws under certain limited conditions w But they condense to liquids and then solidify if the temperature is lowered w Furthermore, there are relatively small forces of attraction between particles of a real gas w This is not the case for an ideal gas