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Gas Laws

Gas Laws. Definitions. Heat Energy transferred due to differences in temperature Temperature Measure of the average kinetic energy of particles composing a material Pressure Force per unit area Volume The amount of space a material occupies. Pressure vs. Volume: Boyles Law.

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Gas Laws

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  1. Gas Laws

  2. Definitions • Heat • Energy transferred due to differences in temperature • Temperature • Measure of the average kinetic energy of particles composing a material • Pressure • Force per unit area • Volume • The amount of space a material occupies

  3. Pressure vs. Volume: Boyles Law Pressure and volume are indirectly related Pressure and 1/volume are directly related

  4. Gas Law Problems: Boyles’ Law A balloon initially occupies 12.4 L at 1.00 atm. What will be the volume at 0.800 atm? A sample of gas expands from 10.0 L to 30.0 L. If the initial pressure was 1140 mm Hg, what is the new pressure?

  5. Using Boyles law Calculate: A balloon contains 40.0 L of a gas at 101.3 kPa. What is the volume of the gas when the balloon is brought to an altitude where the pressure is only 10.0 kPa? (Assume that the temperature remains constant.)

  6. Volume vs. Temperature: Charles Law • Temperature and volume are directly related • At constant pressure • Absolute zero • Theoretically, point at which all motion stops. • The temperature at which volume of a gas becomes zero when the plot of temperature vs. volume is extrapolated.

  7. Gas Law Problems: Charles’Law 1. Calculate the decrease in temperature when 2.00 L at 20.0°C is compressed to 1.00 L. 2. A gas occupies 900.0 mL at a temperature of 27.0°C. What is the volume at 132.0°C?

  8. Pressure vs. Temperature: Gay-Lussac’s Law • Temp and pressure are directly related • Absolute zero • The temperature at which the pressure of a gas becomes zero when a plot of pressure versus temperature for a gas is extrapolated • -273C = 0 K or K = C + 273

  9. Gas Law Problems: Gay-Lussac’s • Consider a container with a volume of 22.4 L filled with a gas at 1.00 atm at 273 K. What will be the new pressure if the temperature increases to 298K?

  10. Gas Law Problems: Gay-Lussac’s 2. A container is initially at 47 mm Hg and 77K (liquid nitrogen temperature.) What will the pressure be when the container warms up to room temperature of 25C?

  11. Gas Law Problems: Gay-Lussac’s 3. A thermometer reads a pressure of 248 Torr at 0.0C. What is the temperature when the thermometer reads a pressure of 345 Torr?

  12. Combined Gas Law • STP(standard temperature and pressure) • T = 273K (which is 0°C) • P = 1 atm, 101.3 kPa, 760 mmHg, 760 torr • V = 22.4 L (one mole of gas) • Formula P1V1 = P2V2 T1 T2

  13. Combined Gas Law 10.0 cm3 volume of a gas measured 75.6 kPa and 60.0C is to be corrected to correspond to the volume it would occupy at STP.

  14. Ideal Gas Law The Kinetic Molecular Theory describes IDEAL gases • Most gases approximate ideal gases and the model works great for most gases. • Some REAL gases need adjustments • Real gases experience electrostatic attractions. • Real gases have volume

  15. Ideal Gas Law • PV = nRT • P = pressure-measured in atm, kPa, mmHg, torr • V = volume - usually measured in L • T = temperature in K • n = number of moles (“mol”) • R = “universal gas constant” • Calculate R if pressure is in atm: (most common) (1 atm)(22.4 L) = 0.0821 atm*L/K*mol (273 K)(1mol)

  16. Ideal Gas Law • Calculate R if pressure is in kPa: (most common) (101.3 kPa)(22.4 L) = 8.312 kPa*L/K*mol (273 K)(1mol) • Calculate R if pressure is in torr or mm Hg: (rarely used) (760 torr)(22.4 L) = 62.36 torr*L/K*mol (273 K)(1mol) (760 mmHg)(22.4 L) = 62.36 mmHg*L/K*mol (273 K)(1mol)

  17. Ideal Gas Law • Calculate the temperature of 0.500 moles of a gas occupying a volume of 20.0 L with a pressure of 99.9 kPa. (99.9 kPa)(20 L) = 480.75 K (8.312 kPa*L/K*mol)(.5 mol)

  18. Calculate the moles of a gas occupying a volume of 500.0 mL with a temperature of 25.5C and a pressure of 755 torr. (755 torr)(.5 L) = 0.0203 mol (298.5 K)(62.36 torr*L/K*mol)

  19. 22.4 L • Calculate the volume of 1.00 mol of H2 at STP STP = 0C 273 K and 1.00 atm • P = 1 atm • V = ? • n = 1 mol • R = 0.0821 L-atm/mol-K • T = 273K PV = nRT nRT/P = V (1 mol)(.0821L-atm/mol-K)(273 K) = 1 atm That number should look familiar! It is the volume of one mole of any gas at STP. The number from the mole map comes from the ideal gas law.

  20. End

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