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Chapter 13 13.1 Pressure - force exerted over an area

Chapter 13 13.1 Pressure - force exerted over an area (pressure = force x area) (F in Newtons, area in m 2 ) (1 N-m 2 = 1 Pascal = 1 Pa) Barometer - device for measuring atmospheric pressure (Torricelli) Manometer - device for measuring pressure in a container Units of Pressure

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Chapter 13 13.1 Pressure - force exerted over an area

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  1. Chapter 13 13.1 Pressure- force exerted over an area (pressure = force x area) (F in Newtons, area in m2) (1 N-m2 = 1 Pascal = 1 Pa) Barometer- device for measuring atmospheric pressure (Torricelli) Manometer- device for measuring pressure in a container Units of Pressure 1 atm = 760.0 mm Hg = 760.0 torr = 101, 325 Pa = 101.3 kPa Example: 525 mm Hg x 1 atm/ 760 mm Hg = 0.691 atm Problem: convert 105.2 kPa to atm Problem: Convert 225,400 Pa to mm Hg

  2. Figure 13.2:A glass tube is filled with mercury and inverted in a dish of mercury at sea level.

  3. Figure 13.3: Gas pressure = atmospheric pressure – h.

  4. Figure 13.3: Gas pressure = atmospheric pressure + h.

  5. 13.2 Pressure & Volume: Boyle’s Law Pressure Volume Temperature # of Particles of Gas If temperature and the # of particles of Gas are held constant, then pressure and volume are indirectly proportional. P x V = k P1V1 = P2V2 Graph Example:A sample of neon to be used in a neon sign has a volume of 1.51 L at a pressure of 635 torr. Calculate the volume of the gas after it is pumped into the glass tubes of the sign, where it shows a pressure of 785 torr. V2 = (V1 x P1)/P2 = (1.51 L x 635 torr)/785 torr = 1.22 L Problem: V = 53.2 mL @ 785 mm Hg; V = ? @ 700 mm Hg V2 = (V1 x P1)/P2 = (53.2 mL x 785 mm Hg)/700 mm Hg = 59.7 mL

  6. Table 13.1

  7. Figure 13.4: A J-tube similar to the one used by Boyle.

  8. Figure 13.5: A plot of P versus V from Boyle’s data.

  9. Figure 13.6: Illustration of Boyle’s law.

  10. 13.3 Volume & Temperature : Charles’s Law Pressure Volume Temperature # of Particles of Gas If pressure and # of particles of gas are held constant, then temperature and volume are directly proportional. (Note: Temperature MUST be in Kelvin) V/T = k V1/T1 = V2/T2 Graph Example: 13.3, p. 406 V2 = (V1 x T2)/T1 = (23 cm3 x 291 K)/301 K = 22 cm3 Problem: If a 45.0 mL sample of gas at 26.5oC is heated to 55.2oC, what is the new volume of the gas sample at constant pressure? V2 = (V1 x T2)/T1 = (45.0 mL x 328.2K)/299.5 K = 49.3 mL

  11. Figure 13.7: Plots of V (L) versus T (°C) for several gases.

  12. Figure 13.8: Plots of V versus T using the Kelvin scale for temperature.

  13. 13.4 Volume & Moles: Avogadro’s Law Pressure Volume Temperature # of Particles of Gas If pressure and temperature are held constant, then volume and # of particles of gas are directly proportional. V/n = k V1/n1 = V2/n2 Example:Consider two samples of nitrogen gas (composed of N2 molecules). Sample 1 contains 1.5 mol of N2 and has a volume of 36.7 L @ 25oC and 1 atm. Sample 2 has a volume of 16.5 L at 25oC and 1 atm. Calculate the # of moles of N2 in sample 2. n2 = (n1 x V2)/V1 = (1.5 mol x 16.5 L)/36.7 L = 0.67 mol Problem: If 0.214 mol of argon gas occupies a volume of 652 mL at a particular temperature and pressure, what volume would 0.375 mol of argon occupy under the same conditions? V2 = 1.14 L

  14. Figure 13.9: The relationship between volume V and number of moles n.

  15. 13.5 Ideal Gas Law Pressure Volume Temperature # of particles of gas R- universal gas constant R = (1.000 atm)(22.4 liters) = (1 mol)(273K) PV = nRT Example: Problem:

  16. Combined Gas Law- if # of particles of gas is held constant, then the other quantities vary in this relationship: P1V1/T1 = P2V2/T2 Example: Problem:

  17. 13.6 Dalton’s Law of Partial Pressures Partial pressure- the pressure that a gas would have if it was alone in a container Total pressure exerted is the sum of the partial pressures. PT = p1 + p2 + … + pn Example: Problem:

  18. Figure 13.10: When two gases are present, the total pressure is the sum of the partial pressures of the gases.

  19. Figure 13.11: The total pressure of a mixture of gases depends on the number of moles of gas particles present.

  20. Figure 13.12: The production of oxygen by thermal decomposition.

  21. Postulates of KMT.

  22. Figure 13.13: (a) A gas confined in a cylinder with a movable piston. (b) The temperature of the gas is increased at constant pressure Pext.

  23. Figure 13.14: A gas sample is compressed.

  24. 13.9 The Implications of the Kinetic Molecular Theory The average kinetic energy of gas particles is directly proportional to the Kelvin temperature. (I.e., the warmer the gas, the faster the particles move) 13.10 Real Gases Gas Laws do not work accurately when temperature is low and pressure is high. (Gas  liquid) 13.11 Gas Stoichiometry 1 mole of any gas occupies 22.4 liters of volume at 0oC and 1 atm 0oC and 1 atm are called Standard temperature and Pressure (STP) (22.4 liters is called molar volume) Example: Problem:

  25. (not in textbook) Graham’s Law- compares the rate of effusion (movement of a gas through a pinhole) of one gas to the rate of effusion of another gas; they are indirectly proportional to the square root of the their respective molar masses. Rate gas 1 = (molar mass gas 2)1/2 Rate gas 2 (molar mass gas 1)1/2

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