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Chapter 5 Gases A Gas- Uniformly fills any container Mixes completely with any other gas Can easily be compressed Exerts pressure on its surroundings

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The properties of a gas depends upon four variables- Pressure (P) –Is equal to force/unit area –Measured by a barometer –SI unit = Newtons/meter 2 = 1 Pascal (Pa) –1 standard atmosphere = 101,325 Pa –1 atm = 760 mm Hg = 760 torr Volume (V) of the gas = volume of the container Temperature (T) – measured in Kelvin Number of moles (n)

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A Torricellian Barometer

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A Simple Manometer

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The Gas Laws Boyles Law: At constant temperature, the volume of a gas is inversely proportional to the pressure V 1/P V = k x 1/P PV = k (where k is constant) P 1 V 1 = P 2 V 2 = constant (k) = PV (1 and 2 represents two states) Pressure x Volume = constant (T = constant) (Holds precisely only at very low pressures)

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A J-Tube Similar to the One Used by Boyle

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Plotting Data for Boyles Law

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Example: A sample of CO 2 gas is placed in a 125 mL flask where it has a pressure of 67.5 mm Hg. What pressure will it has when it is transferred to a 500 mL flask at the same temperature? P 1 = 67.5 mm HgP 2 = ? V 1 = 125 mLV 2 = 500 mL P 1 V 1 = P 2 V 2 P 2 = (P 1 V 1 )/V 2 = (67.5 mm Hg x 125 mL)/500 mL = 16.9 mm Hg

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Charles Law: At constant pressure (P), the volume of a gas is directly proportional to temperature of the gas and extrapolates to zero at zero Kelvin. V T V = bT Where, b = proportionality constant T = Kelvin temperature, V/T = b V 1 /T 1 = V 2 /T 2 = b

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Plots of V Versus T (ºC) for Several Gases

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Example: A sample of gas at 15 o C and 1 atm has a volume of 2.58 L. What volume will this gas occupy at 38 o C and 1 atm? T 1 = 15 o C + 273 = 288 K, V 1 = 2.58 L, T 2 = 38 o C + 273 = 311 K, V 2 =? V 1 /T 1 = V 2 /T 2 Solving for V 2 = V 1 T 2 / T 1 = (2.58 L x 311 K)/288 K = 2.79 L

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Avogadros Law: Equal volume of gases at the same temperature and pressure contain the same number of particles. At constant temperature and pressure, the volume is directly proportional to the number of moles of gas (at low pressures). V = an Where, V = the volume of the gas, n = the number of moles of gas particle, and a = proportionality constant V 1 /n 1 = V 2 /n 2 = a = constant At constant temperature and pressure, the volume of a gas is directly proportional to the number of moles of gas.

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Balloons Holding 1.0 L of Gas at 25º C and 1 atm

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Example: 12.2 L sample containing 0.50 mol oxygen gas at a pressure of 1 atm and a temperature of 25 o C. If all this O 2 were converted to ozone at the same temp. and pressure, what would be the volume of ozone? 3O 2 2O 3 (g) 0.50 mol O 2 x (2 mol O 3 )/(3 mol O 2 ) = 0.33 mol O 3 V 1 = 12.2 L,V 2 = ? n 1 = 0.50 mol,n 2 = 0.33 mol V 1 /n 1 = V 2 /n 2 Solving for V 2 = V 1 n 2 / n 1 = (12.2 L x 0.33 mol)/(0.50 mol) = 8.1 L

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The Ideal Gas Law: (An equation of state for a gas) Boyles law: V = k/P (at constant T and n) Charles law: V = bT (at constant P and n) Avogadros law: V = an (at constant T and P) The volume of gas depends on pressure, temperature, and number of moles of gas present, can be combined as follows: V = R(Tn/P); where, R = combined proportionality constant known as universal gas constant. PV = nRT (Holds closely at P < 1 atm) P = Pressure in atmV = Volume in liters n = number of molesT = Temperature in Kelvins R = Universal gas constant = 0.08206 L.atm K -1 mol -1

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Example: Nitrogen gas in an air bag at 850 mm Hg and 25 o C has a volume of 3.5 x 10 4 mL. How many moles of gas are in there? P = 850 mm Hg x 1 atm / 760 mm Hg = 1.118 atm T = 25 o C + 273 = 298 K V = 3.5 x 10 4 mL x 1L / 1000 mL = 35 L PV = nRT n = PV/RT =(1.118 atm x 35 L)/(0.0821 L.Atm K -1 mol -1 x 298K) = 1.6 moles N 2

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Example: A gas containing 0.35 mol argon at a temperature of 13 o C and a pressure of 568 torr is heated to 56 o C and pressure of 897 torr. Calculate the change in volume that occurs. P 1 = 568 torr x 1 atm/760 torr P 2 = 897 torr x 1 atm/760 torr = 0.747 atm = 1.18 atm V 1 = ? V 2 = ? T 1 = 13 o C + 273 = 286 K T 2 = 56 o C + 273 = 329 K n 1 = 0.35 mol n 2 = 0.35 mol P 1 V 1 = n 1 RT 1 V 1 = n 1 RT 1 /P 1 = (0.35 mol x 0.08206 L.atm.mol -1.K -1 x 286K) / (0.747 atm) = 11 L V 2 = n 2 RT 2 /P 2 = (0.35 mol x 0.08206 L.atm.mol -1.K -1 x 329K) / (1.18 atm) = 8.0 L V = V 2 - V 1 = 8.00 L – 11.0 L = -3 L (Volume decreases)

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Gas Stoichiometry The conditions 0 o C and 1 atm called Standard Temperature and Pressure (STP) Suppose we have 1 mol of an ideal gas at 0 o C and 1 atm, the volume of the gas is given by ideal gas law, V = nRT / P = (1 mol x 0.08206 L.atm.mol -1.K -1 x 273K) / (1 atm) = 22.42 L Molar Volume of an ideal gas at STP

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A Mole of Any Gas Occupies a Volume of Approximately 22.4 L at STP

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Example: A sample of N 2 gas has a volume of 1.75 L at STP. How many moles of N 2 are present? 1.75 L N 2 x (1 mol N 2 )/22.42 L N 2 = 7.81 x 10 -2 mol N 2 OR PV = nRT n = PV/RT = (1 atm x 1.75 L)/(0.08206 L.atm.mol -1.K -1 x 273 K) = 7.81 x 10 -2 mol N 2

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Molar Mass By using ideal gas law we can calculate the molar mass (molecular weight) of a gas from its measured density. PV = nRT n = grams of gas/molar mass = m/molar mass PV = (m/molar mass)RT P = mRT/V(molar mass) m/V is the gas density d in units of grams per liter, Thus, P = dRT/molar mass molar mass = dRT/P If density of a gas at a given temperature and pressure is known, its molar mass can be calculated.

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Example: The density of a gas was measured at 1.50 atm and 27 o C and found to be 1.95 g/L. Calculate the molar mass of the gas. P = 1.50 atm T = 27 o C + 273 = 300K d = 1.95 g/L Molar mass = dRT/P = = 32.0 g/mol

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Daltons Law of Partial Pressures For a mixture of gases in a container, the total pressure exerted is the sum of the pressure that each gas would exert if it were alone. P total = P 1 + P 2 + P 3 + --- (P 1, P 2, P 3 --- represent partial P) Partial pressure of each gas can be calculated from ideal gas law, P 1 = n 1 RT/V, P 2 = n 2 RT/V, P 3 = n 3 RT/V P total = P 1 + P 2 + P 3 + ---- = n 1 RT/V + n 2 RT/V + n 3 RT/V = (n 1 +n 2 +n 3 +---)RT/V = n total (RT/V) For a mixture of ideal gas, the total number of moles of particles is important, not the identity or composition of the involved gas.

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Partial Pressure of Each Gas in a Mixture

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Partial Pressure Example: 46 L He at 25 o C and 1.0 atm and 12 L O 2 at 25 o C and 1.0 atm were pumped into a tank with a volume of 5.0 L. Calculate the partial pressure of each gas and the total pressure in the tank at 25 o C. PV = nRT n = PV/RT 1.0 atm x 46 L 0.08206 L.atm.mol -1.K -1 x (25+273) K 1.0 atm x 12 L 0.08206 L.atm.mol -1.K -1 x 298K

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…Partial Pressure (example) continued… Again, PV = nRT P = nRT /V 1.9 mol x 0.08206 L.atm.mol -1.K -1 x 298K 5.0 L 0.49 mol x 0.08206 L.atm.mol -1.K -1 x 298K 5.0 L P total = P He + P O2 = (9.3 + 2.4) atm = 11.7 atm

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Mole Fraction The ratio of the number of moles of a given component in a mixture to the total number of moles in the mixture. Mole fraction = 1 n = P(V/RT) [# of moles proportion to pressure]

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…Mole Fraction continued… 1 1 1 x P Total 2 2 x P Total

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Example: The partial pressure of oxygen was observed to be 156 torr in air with a total atomospheric pressure of 743 torr. Calculate the mole fraction of O 2 present. Mole fraction of O 2, O2 = P O2 /P Total = 156 torr/743 torr = 0.210

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Example: The mole fraction of N 2 in the air is 0.7808. Calculate the partial pressure of N 2 in air when the atomospheric pressure is 760 torr. The partial pressure of N 2, P N2 = N2 x P Total = 0.7808 x 760 torr = 593 torr

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The Kinetic Molecular Theory of Gases Postulates for an Ideal Gas: 1.The particles are so small that the volume of the particles can be assumed to be negligible (zero). 2.The particles are in constant motion. The collision of the particles with the walls of the container are the cause of the pressure exerted by the gas. 3.The particles are assumed to exert no forces on each other, they are assumed neither to attract nor to repel each other. 4.The average kinetic energy of a collection of gas particles is assumed to be directly proportional to the Kelvin temperature of the gas.

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The Effects of Decreasing the Volume of a Sample of Gas at Constant Temperature

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The Effects of Increasing the Temperature of a Sample of Gas at Constant Volume

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The Effects of Increasing the Temperature of a Sample of Gas at Constant Pressure

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The Effects of Increasing the Number of Moles of Gas Particles at Constant Temperature and Pressure

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Effusion And Diffusion Effusion: Flow of gas particles through tiny pores or pinholes due to pressure differences. Describes the passage of gas into an evacuated chamber. Diffusion: Movement of gas particles through space, from a region of high concentration to one of low concentration. Describes the mixing of gases. The rate of diffusion is the rate of gas mixing.

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Grahams law of effusion: The rate of effusion of a gas is inversely proportional to the square root of the mass of its particle. Relative rates of effusion of two gases at the same temperature and pressure are given by the inverse ratio of the square roots of the masses of the gas particle: Rate of effusion for gas 1 Rate of effusion for gas 2 M 1 and M 2 represents the molar masses of the gases. Diffusion: Distance traveled by gas 1 Distance traveled by gas 2

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The Effusion of a Gas into an Evacuated Chamber

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HCI(g) and NH 3 (g) Meet in a Tube

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Real Gases An ideal gas is a hypothetical concept. No gas exactly follows the ideal gas law. Real gas typically exhibit behavior that is closest to ideal behavior at low pressure and high temperature. Must correct ideal gas behavior when at high pressure (smaller volume) and low temperature (attractive forces become important) Rearrangement gives vander Waals equation:

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Plots of PV/nRT Versus P for Several Gases (200 K)

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Plots of PV/nRT Versus P for Nitrogen Gas at Three Temperatures

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Chemistry in the Atmosphere The principal components are N 2 and O 2. Other important gases are H 2 O, CO 2, Ar, Ne, He, CH 4, Kr, H 2, NO, Xe etc. Because of gravitational effects, the composition of the earths atmosphere is not constant. Heavier molecules tend to be near the earths surface and light molecules tend to migrate to higher altitudes. Troposphere is the lowest layer of the atmosphere.

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…Chemistry in the Atmosphere continued… Ozone in the upper atmosphere helps prevent high energy ultraviolet radiation from penetrating to the earth. Two main sources of pollution are– transportation and the production of electricity. The combustion of petroleum produces CO, CO 2, NO, and NO 2, along with unburned molecules from petroleum.

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Reactions Occurring in the Atmosphere NO 2(g) NO (g) + O (g) O (g) + O 2(g) O 3(g) O 3 Light O 2 * + O* O* + H 2 O2OH OH + NO 2 HNO 3 S(in coal) + O 2(g) SO 2(g) 2SO 2(g ) + O 2(g) 2SO 3(g ) SO 3(g ) + H 2 O (l) H 2 SO 4(aq) [corrosive] acid rain

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Summary Gas properties depends on four variables – pressure, volume, temperature, and amount of gas. Boyles law: PV = constant Charles law: V/T = constant Avogadros law: V/n = constant Ideal gas law: PV = nRT Molar mass = dRT/P Daltons law of partial pressures: P total = P 1 + P 2 + P 3 + ---

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…Summary continued… Mole fraction: 1 = n 1 /n total = P 1 /P total Kinetic molecular theory of gasses: Postulates Grahams law of effusion: Rate of effusion for gas 1 Rate of effusion for gas 2 Chemistry in atmosphere

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