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Charles's law

Boyle's law. PV = constant (k). P 1 V 1 = P 2 V 2. Charles's law. RT. RT. V. V. Avogadro’s law. V = k n. Ideal gas equation. PV = nRT. (. (. (. (. P 2 V 2. P 1 V 1. =. T 2. T 1. Combined gas law. P t =(n 1 +n 2 +n 3 …. ). =. n. t. PM. dRT. V 1. V 2. d. M. =. =.

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Charles's law

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  1. Boyle's law PV = constant (k) P1V1 = P2V2 Charles's law RT RT V V Avogadro’s law V = k n Ideal gas equation PV = nRT ( ( ( ( P2V2 P1V1 = T2 T1 Combined gas law Pt=(n1+n2+n3…. ) = n t PM dRT V1 V2 d M = = = RT P T1 T2

  2. Example: Calcium carbonate, CaCO3(s), decomposes upon heating to give CaO(s) and CO2(g). A sample of CaCO3 is decomposed, and the carbon dioxide is collected in a 250-mL flask. After the decomposition is complete, the gas has a pressure of 1.3 atm at a temperature of 31 °C. How many moles of CO2 gas were generated? PV = nRT

  3. Example: The gas pressure in an aerosol can is 1.5 atm at 25 °C. Assuming that the gas inside obeys the ideal-gas equation, what would the pressure be if the can were heated to 450 °C? PV = nRT

  4. Example: A large natural-gas storage tank is arranged so that the pressure is maintained at 2.20 atm. On a cold day in December when the temperature is –15 °C (4 °F), the volume of gas in the tank is 3.25 × 103 m3. What is the volume of the same quantity of gas on a warm July day when the temperature is 31 °C (88 °F)? P1= 2.20 atm V1 = 3.25 x 103 m3 T1= -15oC = -15+273 = 258 K P2 = 2.20 atm V2 = ? T2= -15oC = 31+273 = 304 K P2V2 V2 V1 P1V1 = = Answer: 3.83 × 103 m3 T2 T2 T1 T1

  5. Partial Pressures • When one collects a gas over water, there is water vapor mixed in with the gas. • To find only the pressure of the desired gas, one must subtract the vapor pressure of water from the total pressure.

  6. Collecting gases over water 2KClO3 (s) 2KCl (s)+ 3O2 (g) PT = P + P O2 H2O

  7. Example 2: Oxygen gas generated by the decomposition of potassium chlorate is collected over water. The volume of oxygen collected at 24C and atmospheric pressure of 762 mmHg is 128 mL. Calculate the mass (in grams) of oxygen gas obtained. The pressure of the water vapor at 24 C is 22.4 mmHg. m number of mole, n = = 762 mmHg – 22.4 mmHg M O2 1 atm =739.6 mmHg = 739.6 mmHg  = 0.973 atm 760 mmHg P = 0.973 atm T = 273 + 24 = 297 K V = 128 mL = 0.128 L R = 0.08206 L.atm /K.mol P=PT - P PT=P+P M = molar mass = 32 g / mol m =mass in gram ? H2O H2O O2 PV = nRT = 0.163 g m = RT M PVM 0.973 atm  0.128 L  32 g / mol m = = RT 0.0821 L.atm / K.mol  297 K

  8. 10.7 Kinetic-molecular theory The kinetic-molecular theory is summarized by the following statements: Gases consist of large numbers of molecules that are in continuous, random motion The combined volume of all the molecules of the gas is negligible relative to the total volume in which the gas is contained Attractive and repulsive forces between gas molecules are negligible Energy can be transferred between molecules during collisions, but the average kinetic energy of the molecules does not change with time The average kinetic energy of the molecules is proportional to the absolute temperature

  9. Distribution of molecular speed Average kinetic energy of a gas molecules:  = u, root-mean-square (rms) m, mass of individual molecule rms and average speed : If we have four objects with speeds of 4.0, 6.0, 10.0 and 12.0 m/s The effect of temperature on molecular speeds 4.0 + 6.0 + 10.0 + 12.0 Average speed = = 8 m/s 1 mu2 4 2 ( ( 4.02 + 6.02 + 10.02 + 12.02 rms speed = 4 ( ( 16 + 36 + 100+ 144 = 74 = 8.6 m/s 4 =

  10. 10.8 Molecular effusion and diffusion M, molar mass T, temperature R, gas constant The effect of molecular mass on molecular speeds Example : Calculate the rms speed, u, of an N2 molecule at 25C ( ( 3RT u = T = 25 +273 =298 K ; M = 28.0 g/mol = 28.0  10-3 kg/mol R= 8.314 J/mol-K = 8.314 kg-m2/s2-mol-K M 3(8.314 kg-m2/s2-mol-K) 298 K u = 28.0  10-3kg/mol = 5.15  102 m/s

  11. The dependence of molecular speeds on mass has two consequences: Effusion : escape of gas molecule through a tiny hole into an evacuated space Diffusion : spread of one substance through a space or throughout a second substance Gas molecule effuse through a pinhole in the partitioning wall only when they happen to hit the hole Each short segment of line represents travel between collisions. The blue arrow indicates the net distance traveled by the molecule

  12. Graham’s law of effusion At same temperature and pressure, and in containers with identical pinholes, if rates of effusion of two substance are r1 and r2 and their respective molar mass are M1 and M2, Graham’s law states r1 = r2 M2 M1 At identical pressure and temperature, the lighter gas effuses more rapidly

  13. Example : Calculate the ratio of the effusion rates of N2 and O2, M = 28 g/mol N2 M = 32 g/mol O2 O2 r r r = O2 O2 O2 r r r N2 N2 N2 = M 32 28 M N2 = 1.07

  14. 10.9 Real gases : Deviations from ideal behavior For a gas to exhibit idea behavior two assumptions are considered: • Intermolecular forces: The attractions and repulsions between the particles are negligible (i.e., small or non existent intermolecular forces) Negligible volume: A gas is made up of a large number of particles (atoms or molecules) whose size is negligible compared to the size of the container. These particles are also small compared to the distances between particles. Real molecules, however, do have finite volumes and they do attract one another

  15. For ideal gas: PV = nRT PV PV =1 = n RT RT for one mole of gas For real gas is true only at moderately low pressure ( 5 atm) PV =1 Intermolecular attractions, increase with molecular weight, cause the PV product to decrease as higher pressures bring the molecules closer together, attractive forces; molecules begin to intrude on each others' territory, repulsive forces always eventually win out. RT

  16. For ideal gas: PV = nRT PV PV = 1 = n RT RT for one mole of gas The effect of temperature and pressure on the behavior of nitrogen gas For real gas nearly true only at very high temperature PV =1 RT At high pressure and low temperature gases will not behave ideally

  17. The Van der Waals equation The two factor leading to deviations from ideal behavior 1. molecular attractions 2. molecular volume Pideal = Preal + Van der Waal’s suggested correction for real gas : an2 “a” is a correction for intermolecular forces Pressure correction : V 2 “nb” represents volume occupied by n moles of the gas Videal = (Vreal – nb) Volume correction : an2 Taking into account the corrections for pressure and volume: V 2 Van der Waals equation = nRT V – nb a and b are van der Waals constants anddifferent for different gases P+

  18. Example : If a sample of 1.00 mol of CO2(g) is confined to a volume of 3.00 L at 0.0 C, calculate the pressure of the gas using (a) the ideal-gas equation and (b) the van der Waals equation. [For CO2(g) a= 3.59 atm.L2/mol2 and b = 0.0427 L/mol] V = 3.00 L R = 0.0821 L.atm/K.mol n = 1.00 mol T = 0 +273 = 273 K (a) For ideal gas : = 7.471 atm 1 mol  0.0821 L.atm/K.mol  273 K P = = 3.00 L nRT nRT (b) For real gas : V -nb V = 0.399 atm 3.59 atm.L2/mol2(1 mol)2 (3.00 L)2 an2 an2 = V 2 V – nb = 3.00 L – (1 mol  0.0427 L/mol) = 2.957 L V 2 P + 0.399 atm P + = 2.957 L 1.0 mol  0.0821 L.atm/K.mol  273 K P + 0.399 atm = 7.580 atm P = 7.181 atm =

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