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Learning outcomes

Recall, explain and use the following relationships A change in pressure of a fixed mass of gas at constant volume is caused by a change in temperature of the gas A change in volume occupied by a fixed mass of gas at constant pressure is caused by a change in temperature of the gas

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Learning outcomes

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  1. Recall, explain and use the following relationships A change in pressure of a fixed mass of gas at constant volume is caused by a change in temperature of the gas A change in volume occupied by a fixed mass of gas at constant pressure is caused by a change in temperature of the gas A change in pressure of a fixed mass of gas at constant temperature is caused by a change in volume of the gas Apply gas laws to various situation. Learning outcomes

  2. 8.3 How is the kinetic model used to explain the behaviour of gases ? Motion of molecules and pressure • When randomly moving gas molecules hit the wall of a container, they exert forces on the wall • Since pressure is defined as force per unit area, the average force per unit area gives rise to gas pressure Pressure exerted by a gas is caused by molecular collisions with the container

  3. 8.3 How is the kinetic model used to explain the behaviour of gases ? Pressure-temperature Relationship in Gases • For a mixed mass of gas at constant volume, when the temperature of the gas in the container increases, the molecules move faster and hit the walls more frequently and with greater force • The average force per unit area increases. • This causes the pressure to increase • Therefore, for a fixed mass of gas at constant volume, gas pressure increases with gas temperature and vice versa

  4. 8.3 How is the kinetic model used to explain the behaviour of gases ? Volume-temperature Relationship in Gases • As we have seen earlier, for a fixed mass of gas at constant volume, when the temperature of the gas in the container increases, the molecules move faster and hit the walls more frequently and with greater force • If we want the pressure to remain constant, we have to reduce the number of collisions per unit time with the container • This can be done by increasing the volume of the container • Therefore, the volume of a fixed mass of gas at constant pressure increases with temperature and vice versa

  5. 8.3 How is the kinetic model used to explain the behaviour of gases ? Pressure-Volume Relationship in Gases • For a fixed mass of gas at constant temperature, the average speed/kinetic energy of the molecules remains the same • If we halve the volume of the container, the number of gas molecules per unit volume in the container will be doubled Half the volume, double the pressure

  6. 8.3 How is the kinetic model used to explain the behaviour of gases ? Pressure-Volume Relationship in Gases • The number of molecules hitting the wall per unit time will also be doubled • Consequently the gas pressure will be doubled • Therefore for a fixed mass of gas at constant temperature, gas pressure increases when volume decreases, and vice versa Half the volume, double the pressure

  7. P V Boyle’s Law

  8. The pressure and volume of a gas are inversely related at constant mass & temp P V Boyle’s Law PV = k

  9. V T Charles’ Law

  10. The volume and absolute temperature (K) of a gas are directly related at constant mass & pressure V T Charles’ Law

  11. P T Gay-Lussac’s Law

  12. The pressure and absolute temperature (K) of a gas are directly related at constant mass & volume P T Gay-Lussac’s Law

  13. P1V1 T1 P2V2 T2 = Combined Gas Law T PV = k

  14. A gas occupies 473 cm3 at 36°C. Find its volume at 94°C. CHARLES’ LAW GIVEN: V1 = 473 cm3 T1 = 36°C = 309K V2 = ? T2 = 94°C = 367K T V WORK: V1/T1 = V2/T2 (473 cm3) /(309 K)=V2/ (367 K) V2 = 562 cm3

  15. A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa if temperature remains constant. BOYLE’S LAW GIVEN: V1 = 100. mL P1 = 150. kPa V2 = ? P2 = 200. kPa P V WORK: P1V1 = P2V2 (150.kPa)(100.mL)=(200.kPa)V2 V2 = 75.0 mL

  16. A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP. COMBINED GAS LAW P T V GIVEN: V1=7.84 cm3 P1=71.8 kPa T1=25°C = 298 K V2=? P2=101.325 kPa T2=273 K WORK: P1V1/T1 = P2V2/T2 (71.8 kPa)(7.84 cm3) /(298 K) =(101.325 kPa)V2 / (273 K) V2 = 5.09 cm3

  17. A gas’ pressure is 765 torr at 23°C. At what temperature will the pressure be 560. torr? (1 torr appox 133 Pa) GAY-LUSSAC’S LAW GIVEN: P1 = 765 torr T1 = 23°C = 296K P2 = 560. torr T2 = ? P T WORK: P1/T1 = P2/T2 (765 torr)/(309K) = (560. torr) /T2 T2 = 226 K = -47°C

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