Presentation on theme: "Boyle, Charles and Gay-Lussac"— Presentation transcript:
1Boyle, Charles and Gay-Lussac Gas Laws Day 2Boyle, Charles and Gay-Lussac
2The Gas LawsThe gas laws apply to ideal gases, which are described by the kinetic theory in the following five statements.Gas particles do not attract or repel each other.Gas particles are much smaller than the spaces between them.
3The Gas LawsGas particles are in constant, random motion.No kinetic energy is lost when gas particles collide with each other or with the walls of their container.All gases have the same kinetic energy at a given temperature.
4Boyle’s Law: Pressure and Volume Robert Boyle ( ), an English scientist, used a simple apparatus pictured to compress gases.
5Boyle’s Law: Pressure and Volume After performing many experiments with gases at constant temperatures, Boyle had four findings.a) If the pressure of a gas increases, itsvolume decreases proportionately.b) If the pressure of a gas decreases, itsvolume increases proportionately.
6Boyle’s Law: Pressure and Volume c) If the volume of a gas increases, itspressure decreases proportionately.d) If the volume of a gas decreases, itspressure increases proportionately.By using inverse proportions, all four findings can be included in one statement called Boyle’s law.
7Boyle’s Law: Pressure and Volume Boyle’s law states that the pressure and volume of a gas at constant temperature are inversely proportional.The equation for Boyle’s lawP1V1=P2V2When Temperature and moles are unchanged.Click box to view movie clip.
8Applying Boyle’s LawA sample of compressed methane has a volume of 648 mL at a pressure of 503 kPa.To what pressure would the methane have to be compressed in order to have a volume of 216 mL?Examine the Boyle’s law equation. You need to find P2, the new pressure, so solve the equation for P2.P1V1=P2V2V2V2
9Substitute known values and solve. Applying Boyle’s LawSubstitute known values and solve.P1V1=P2V2(503kPa)(648mL)=P2= 1510kPa216mL
10Charles’s LawWhen the temperature of a sample of gas is increased and the volume is free to change, the pressure of the gas does not increase. Instead, the volume of the gas increases in proportion to the increase in Kelvin temperature. This observation is Charles’s law, which can be stated mathematically as follows.V2V1=T2T1
11Charles’s Law Topic 13 Gases: Basic Concepts Click box to view movie clip.
12Applying Charles’s Law A weather balloon contains 5.30 kL of helium gas when the temperature is 12°C.At what temperature will the balloon’s volume have increased to 6.00 kL?Start by converting the given temperature to kelvins.
13Applying Charles’s Law Next, solve the Charles’s law equation for the new temperature, T2.V2T2T1T1T2V1=T2V1V1T1
14Applying Charles’s Law Then, substitute the known values and compute the result.T1V2(285K)(6.00kL)T2==V1(5.30kL)=323KFinally, convert the Kelvin temperature back to Celsius.New Temperature = 323 – 273 = 50oC
15Gay Lussac’s LawThis law represents the relationship between pressureand temperature… you will see it is very similar to theCharles law, and calculations are similar.P2P1=T2T1
16practiceA gas system has an initial temperature of 135°C with the pressure unknown. When the temperature changes to °C the pressure is found to be 1.67 atm. What was the initial pressure in atm?First solve Gay-Lussac’s law for P1then plug in the variables
17practiceP2T1T1P1=T2T1Change temp to kelvin… T1= 135C +273= 408KT2= C = 49.9K(1.67atm)(408K)==13.65atm49.9K