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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 ).
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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
pV = constant • p1V1 = p2V2 (at constant temp) Can you answer the questions that Mr Porter is giving you?
Do Now finish Q2 p71p1V1 = p2V2 In a vertical cylinder. The initial pressure of the trapped air is 1.0 x 105 Pa. The volume of the air decreases from 860 cm3 to 645 cm3 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
p1V1 = p2V2 In a Vertical cylinder. The initial pressure of the trapped air is 1.0 x 105 Pa. The volume of the air decreases from 860 cm3 to 645 cm3 The temperature is constant . • Calculate the final pressure.
Graph • y-axis labelled θ / 0C [1] • plots occupying at least half of grid on suitable scale start at about 50 0C 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]
The behaviour of gases- Pressure Law http://phet.colorado.edu/sims/ideal-gas/gas-properties.jnlpGas Laws 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)
The Pressure law- copyGas 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 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 an increase in pressure. Pressure Law
P P T oC T K A value for absolute zero The Results
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 • V1/T1 = V2/T2 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
V V T oC T K A value for absolute zero The Results
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 tooC, subtract 273
Questions! Homework Graph question. Due Monday 12th March
The equation of state By combining these three laws pV = constant V/T = constant p/T = constant We get pV/T = constant Or p1V1 = p2V2 T1 T2 Remember, T must be in Kelvin
P P PV V 1/ V P The Results
An example At 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 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? 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 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. 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.
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