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How Do Gases Behave?

How Do Gases Behave?. What is a solid, liquid or gas?. Help Marvin the Martian understand what a solid, liquid and gas are! Draw what solids, liquids, gases look like Describe physical/chemical properties What would happen if we changed pressure? What would happen if we changed temperature?.

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How Do Gases Behave?

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  1. How Do Gases Behave?

  2. What is a solid, liquid or gas? • Help Marvin the Martian understand what a solid, liquid and gas are! • Draw what solids, liquids, gases look like • Describe physical/chemical properties • What would happen if we changed pressure? • What would happen if we changed temperature?

  3. What is Pressure? • Pressure = Force/Area • 1 atmosphere (atm) • 760 Torr • 760 mmHg • 1.01 Bar • 101,327 Pascal • 101.3 Kpa • 14.7 lbs/in2 • Measured with a barometer

  4. MANOMETER • Column of mercury to measure pressure. • h is how much lower or higher the pressure is than outside. • Pgas = Patm - h • Pgas = Patm + h h h

  5. What is Temperature? • Average Kinetic Energy (1/2 mv2) of an atom or molecule • Measured in Fahrenheit, Celsius or Kelvin (SI) • F = (C x 1.8) + 32 • K = C + 273 • 0 Kelvin: absolute zero (atom stops moving completely) • Is there a maximum temperature in the universe?

  6. Kinetic Molecular Theory • Theory explains why ideal gases behave the way they do. • Assumptions that simplify the theory, but don’t work in real gases. • The particles are so small we can ignore their volume. • The particles are in constant motion and their collisions cause pressure. • The particles do not affect each other, neither attracting or repelling. • The average kinetic energy is proportional to the Kelvin temperature. • The molecules move in straight path and all collisions are elastic

  7. What is an Ideal Gas? • An ideal gas or perfect gas is a hypothetical gas consisting of identical particles of: • Negligible volume • With no intermolecular forces • Atoms or molecules undergo perfectly elastic collisions with the walls of the container • Ideal gas law calculations are favored at low pressures and high temperatures. • Real gases existing in reality do not exhibit these exact properties, although the approximation is often good enough to describe real gases.

  8. What is Boyle’s Law? • In the mid 1600's, Robert Boyle studied the relationship between the pressure P and the volume V of a confined gas held at a constant temperature. • Boyle observed that the product of the pressure and volume are observed to be nearly constant. • The product of pressure and volume is exactly a constant for an ideal gas. • P * V = constant • This relationship between pressure and volume is called Boyle's Lawin his honor.

  9. BOYLE’S LAW V P (at constant T)

  10. Slope = k V 1/P (at constant T)

  11. 22.41 L atm O2 PV CO2 P (at constant T)

  12. 20.5 L of nitrogen at 25ºC and 742 torr are compressed to 9.8 atm at constant T. What is the new volume? P1V1=P2V2 (0.98 atm)(20.5 L) = (9.8 atm)(V2) 20.09 atm*L/9.8 atm= V2 2.1 L = V2 30.6 mL of carbon dioxide at 740 torr is expanded at constant temperature to 750 mL. What is the final pressure in kPa? (30.6mL)(740Torr)=(750 mL)(P2) 22,644 mL*Torr/750 mL= P2 30.2 Torr =P2 (30.2 Torr) (1 atm/760 Torr) (101.3 kPa/1 atm)= 3059.3 kPa/760= 4.03 kPa

  13. What is Charles’ Law? • The relationship between temperature and volume, at a constant number of moles and pressure, is called Charles and Gay-Lussac's Law in honor of the two French scientists who first investigated this relationship. • Charles did the original work, which was verified by Gay-Lussac. They observed that if the pressure is held constant, the volume V is equal to a constant times the temperature T, or: • V / T= constant

  14. CHARLES LAW He CH4 H2O V (L) H2 -273.15ºC T (ºC)

  15. Examples What would the final volume be if 247 mL of gas at 22ºC is heated to 98ºC , if the pressure is held constant? 247 ml/295 K = X ml/371 K 91,637 mL * K = 295 X * K 91,637 mL * K/295 K= X 310 mL = X If the volume of oxygen at 21 °C is 785 L, at what temperature would oxygen occupy 804 L? 785 L/294 K = 804 L/X K 785 X = 236,376 X = 236,376/785 X= 301 K = 28 °C

  16. Combined Gas Law • Combining Charles’s Law and Boyle’s Law in a single statement: • P1V1/T1 = P2V2/T2 • 39.8 mg of caffeine gives 10.1 mL of nitrogen gas at 23°C and 746 mmHg. What is the volume of nitrogen at 0°C and 760 mmHg? • First change temperature to Kelvin • V1 = 10.1mL P1 = 746 mmHg K1 = 296 K • V2 = ? P2 = 760 mmHg K2 = 273 K 10.1 * 746/296 = V2 * 760/273 V2 = 9.14 mL

  17. Other Gas Laws • Gay-Lussac Law • At constant volume, pressure and absolute temperature are directly related. • P/T = k (constant) • Avogadro’s Law • At constant temperature and pressure, the volume of gas is directly related to the number of moles. • V /n= k (n is the number of moles)

  18. Gas Law Summary What equation would we get if we combined them all?

  19. What is the Ideal Gas Law? • Combining Boyle’s Law, Charles’ law & Avogadro’s Law we derive the Ideal Gas Law: • P V = n R T • P = Pressure (atm) • V = Volume (L) • n = # moles (0.0821 L atm /mol K) • R = Gas Constant • T = Temperature (K) • Ideal gas law calculations are favored at low pressures and high temperatures • Tells you about a gas NOW. • The other laws tell you about a gas when it changes.

  20. Let’ Try It! • Example: • If we had 1.0 mol of gas at 1.0 atm of pressure at 0°C (STP), what would be the volume? • PV = nRT • V = nRT/P • V = (1.0 mol)(0.0821 L atm/mol K)(273 K)/(1.0 atm) • V = 22.41 L • 1 mole of ANY gas at STP will occupy 22.4 Liters of volume

  21. Gas Density and Molar Mass • D = m/V • Let M stand for molar mass • M = m/n • n = m/M • PV = nRT • PV = (m/M) RT • P = mRT/VM = (m/V)(RT/M) • P = d RT/M • PM/RT = d (density)

  22. Examples • What is the density of ammonia at 23ºC and 735 torr? • Units must be: atm, K • 735 torr(1 atm/760 torr) = 0.967 atm • 23 + 273 = 296 K • Molar mass of NH3 = 17.0 g d = 0.967 * 17.0 g (0.0821 L* atm/mol * K)(296 K) d = 0.676 g / L

  23. Gases and Stoichiometry • Reactions happen in moles • At Standard Temperature and Pressure (STP, 0ºC and 1 atm) 1 mole of gas occupies 22.42 L. • If not at STP, use the ideal gas law to calculate moles of reactant or volume of product.

  24. Examples • Consider the following reaction: • Suppose you heat 0.0100 mol of potassium chlorate, KClO3, in a test tube. How many liters of oxygen can you produce at 298 K and 1.02 atm? • Break it into 2 problems, one involving stoichiometry and the other using the ideal gas law

  25. 0.0100 mol KClO3 X 3 mol O2/2 mol KClO3 = 0.0150 mol O2 Now that you have the moles of oxygen use the ideal gas law to calculate the volume: V = nRT/P 0.0150 mol x 0.0821 L * atm (K * mol) x 298 K 1.02 atm V = 0.360 L

  26. Using the following reaction • Calculate the mass of sodium hydrogen carbonate necessary to produce 2.87 L of carbon dioxide at 25ºC and 2.00 atm. n = PV/RT = (2.00 atm)(2.87 L) (0.0821 L*atm/K*mol)(298 K) n= 0.235 mol CO2 0.235 mol CO2 (1 mol NaHCO3) ( 84.0 g) (1 mol CO2 ) (1 mol NaHCO3) 19.7 g NaHCO3

  27. Dalton’s Law • The total pressure in a container is the sum of the pressure each gas would exert if it were alone in the container. • The total pressure is the sum of the partial pressures. • PTotal = P1 + P2 + P3 + P4 + P5 ... • For each P = nRT/V

  28. Dalton's Law • PTotal = n1RT + n2RT + n3RT +... V VV • In the same container R, T and V are the same. • PTotal = (n1+ n2 + n3+...)RT V • PTotal = (nTotal)RT V

  29. The Mole Fraction • Ratio of moles of the substance to the total moles. • symbol is Greek letter chi c • Because pressure of a gas is proportional to moles, for fixed volume and temperature then, • c1 = n1= P1nTotalPTotal

  30. Calculating the Partial Pressure and Mole Fraction of a Gas Mixture • A 1.00 L sample of dry air at 25°C and 786 mmHg contains 0.925 g N2, plus other gases including oxygen, argon and carbon dioxide. • What is the partial pressure (in mmHg) of N2 in the air sample? • What is the mole fraction and mole percent of N2 in the mixture?

  31. Convert grams into moles 0.925 g N2 x (1 mol N2/28.0g N2) =0.0330 mol N2 • Substitute into ideal gas law PN2 = nN2RT/V =0.0330mol x 0.0821 L*atm/K*mol x 298 1.00 L =0.807 atm = 613 mmHg

  32. The mole fraction of N2 in the air is = PN2/P = 613 mmHg/786 mmHg =0.780 • Mole percent equals mole fraction x 100 =0.780 x 100 = 78% • Air contains 78.0 mole percent of N2

  33. Vapor Pressure • Water evaporates! • When that water evaporates, the vapor has a pressure. • Gases are often collected over water so the vapor pressure of water must be subtracted from the total pressure. • Vapor pressure varies by temperature and must be given in the problem or in a table.

  34. Hydrogen gas is produced by the reaction of hydrochloric acid, HCl, on zinc metal: 2HCl (aq) + Zn (s) —> ZnCl2 (aq) + H2 (g) • The gas is collected over water. If 156 mL of gas is collected at 19°C and 769 mmHg total pressure, what is the mass of hydrogen collected?

  35. First find the Partial Pressure. The vapor pressure of water at 19°C is 16.5 mmHg P = PH2 + PH2O PH2 = P - PH2O PH2 =769 – 16.5 = 752 mmHg • Use the ideal gas law to find the moles of hydrogen collected. P: 752 mmHg x (1 atm/760 mmHg) = 0.989 atm V: 156 mL x (1 L/1000 mL) = 0.156 L T: 19 + 273 = 292 K R: 0.0821 L*atm/K*mol n: ?

  36. Solve for moles: n = PV/RT = 0.989 x 0.156/0.0821 x 292 =0.00644 mol H2 • Convert moles to grams: 0.00644 mol H2 x (2.02g/1 mol H2) =0.0130 g H2

  37. What’s Diffusion and Effusion? • Only a few physical properties of gases depends on the identity of the gas. • Diffusion - The rate at which two gases mix. • Effusion - The rate at which a gas escapes through a pinhole into a vacuum.

  38. What is Graham’s Law? • We know that Kinetic energy = 1/2 mv2 • If two bodies of unequal mass have the same kinetic energy, which moves faster? • The lighter one! • Thus, for two gases at the same temperature, the one with lower molecular mass will diffuse/effuse faster. • The rate of effusion/diffusion of a gas is inversely proportional to the square root of its mass.

  39. Calculate the ratio of effusion rates of molecules of carbon dioxide and sulfur dioxide from the same container and at the same temperature and pressure: Rate of effusion of CO2 = √Mm SO2 Rate of effusion of SO2 √Mm CO2 = √64.1/44.0 = 1.21 • In other words, carbon dioxide effuses 1.21 times faster than sulfur dioxide.

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