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Boyle, Charles and Gay-Lussac

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1 Boyle, Charles and Gay-Lussac
Gas Laws Day 2 Boyle, Charles and Gay-Lussac

2 The Gas Laws The 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.

3 The Gas Laws Gas 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.

4 Boyle’s Law: Pressure and Volume
Robert Boyle ( ), an English scientist, used a simple apparatus pictured to compress gases.

5 Boyle’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, its volume decreases proportionately. b) If the pressure of a gas decreases, its volume increases proportionately.

6 Boyle’s Law: Pressure and Volume
c) If the volume of a gas increases, its pressure decreases proportionately. d) If the volume of a gas decreases, its pressure increases proportionately. By using inverse proportions, all four findings can be included in one statement called Boyle’s law.

7 Boyle’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 law P1V1=P2V2 When Temperature and moles are unchanged. Click box to view movie clip.

8 Applying Boyle’s Law A 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. P1 V1 = P2 V2 V2 V2

9 Substitute known values and solve.
Applying Boyle’s Law Substitute known values and solve. P1 V1 = P2 V2 (503kPa) (648mL) = P2 = 1510kPa 216mL

10 Charles’s Law When 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. V2 V1 = T2 T1

11 Charles’s Law Topic 13 Gases: Basic Concepts
Click box to view movie clip.

12 Applying 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.

13 Applying Charles’s Law
Next, solve the Charles’s law equation for the new temperature, T2. V2 T2 T1 T1 T2 V1 = T2 V1 V1 T1

14 Applying Charles’s Law
Then, substitute the known values and compute the result. T1 V2 (285K) (6.00kL) T2= = V1 (5.30kL) =323K Finally, convert the Kelvin temperature back to Celsius. New Temperature = 323 – 273 = 50oC

15 Gay Lussac’s Law This law represents the relationship between pressure and temperature… you will see it is very similar to the Charles law, and calculations are similar. P2 P1 = T2 T1

16 practice A 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 P1 then plug in the variables

17 practice P2 T1 T1 P1 = T2 T1 Change temp to kelvin… T1= 135C +273= 408K T2= C = 49.9K (1.67atm)(408K) = = 13.65atm 49.9K

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