Chapter 4 The Properties of Gases Significant Figure Convention At least one extra significant figure is displayed in all intermediate calculations. The final answer is expressed with the correct number of significant figures. The Nature of Gases (Sections 4.1 4.3) physical states (solid, liquid, gas), bulk matter, compressibility, pressure, gauge pressure, barometer, pascal (Pa), manometer, mmhg (Torr), atmosphere (atm), bar 4.1 Observing Gases Example 4.1 Relate the general properties of gas behavior to the molecular level. The fact that gases are compressed with ease and expand rapidly to fill the available space suggests that molecules of gases are widely separated and in ceaseless chaotic motion. 4. Pressure Example 4.a Suppose the height of the column of mercury in a barometer is exactly 760 mm. Given that the density of mercury at 0 C is 13 595.1 kg m 3, calculate the atmospheric pressure. P = dhg = (13 595.1 kg m 3 )(0.760 m)(9.806 65 m s ) = 101 35 kg m s = 101 35 Pa = 101.35 kpa Note: 1 Pa (pascal) = 1 kg m s SI unit of pressure Example 4.b Example 4.c Example 4.d At the same atmospheric pressure obtained in Example 4.a, calculate the height of liquid in a barometer that employs water at 0 C. [d (H O) = 0.998 g cm 3 ] h = 10.4 m The height of the system-side column in an open-tube mercury manometer was 15 mm above that of the open side when the atmospheric pressure was 76 mmhg at 0 C. Calculate the pressure in the system in units of mmhg and kpa. In this case, the pressure in the system is equal to atmospheric pressure minus the difference in height of the two columns. The system is at lower pressure than the atmosphere (open end). The conversion to kpa requires the density of Hg at 0 C, which is 13 546 kg m s. P = 76 15 = 747 mmhg = 0.747 mhg P = dhg = (13 546 kg m s ) (0.747 m) (9.806 65 m s ) = 9.9 10 4 Pa = 99. kpa Calculate the pressure inside a system when a closed-tube mercury manometer shows a height difference of 1 cm (higher on the closed side) at 0 C. P = 10 mmhg and 15.9 kpa 1
Chapter 4 4.3 Alternative Units of Pressure Example 4.3 Convert an atmospheric pressure of 740 mmhg into each of the following pressure units: Torr, atm, Pa, kpa, and bar. P = 740 mmhg = 740 Torr = 0.974 atm = 9.87 10 4 Pa = 98.7 kpa = 0.987 bar The Gas Laws (Sections 4.4 4.11) Boyle s law, isotherm, hyberbola, Charles s law, extrapolation, Kelvin scale, molar volume, Avogadro s principle, ideal gas law, gas constant, equation of state, ideal gas, limiting law, combined gas law, standard temperature and pressure (STP), standard ambient temperature and pressure (SATP), molar concentration, density, amount-to-volume conversions in reacting gases, gas mixtures, partial pressure, Dalton s law of partial pressures, vapor pressure, mole fraction 4.7 The Ideal Gas Law Example 4.7a A.6-µL ampule of xenon has a pressure of.00 Torr at 15 C. How many xenon atoms are present? Convert the volume units into liters, the pressure units into atmospheres, and the temperature units into kelvins. Solve for the number of moles of xenon, using the ideal gas law, and then use Avogadro s number to determine the number of atoms. V =.6 10 6 L P = (.00/760) atm =.63 10 3 atm T = 15 + 73.15 = 88. K A 14 3 6 PV (.63 10 atm)(.6 10 L) 0 n = = =.9 10 mol RT ( 8.0574 10 L atm K mol )(88. K) n N = 1.7 10 atoms Example 4.7b A 0-L flask at 00 K and 0 Torr contains nitrogen. What mass of nitrogen is present (in grams)? 0.89 g 4.8 Applications of the Ideal Gas Law Example 4.8a A gas sample is compressed to half of its original volume and the absolute temperature is increased by 15%. What is the pressure change? Let indices 1 and represent the initial and final states, respectively. Note the changes, if any, in amount, volume, and temperature of the gas. Use the combined form of the ideal gas law given above to solve for the pressure change.
The Properties of Gases Change in the amount of gas: n = n 1 50% decrease in the volume: V = V 1 0.50V 1 = 0.50V 1 15% increase in the temperature: T = T 1 + 0.15T 1 = 1.15T 1 V1 n T V1 n1 1.15T1 P = P 1 = P1 = P1( )( 1)( 1.15) =.30P1 V n1 T1 0.50V1 n1 T1 Because P = P 1 + 1.30P 1, the pressure has increased by 130%. Example 4.8b A gas sample has its pressure quadrupled, absolute temperature doubled, and amount doubled. Calculate the volume change. V = V 1 (no change in volume) 4.9 Gas Density Example 4.9a The density of an unknown hydrocarbon gas (molecules containing only carbon and hydrogen atoms) is found to be 0.875 g L at 570 Torr and 0 C. Calculate the molar mass of the gas and identify it. Convert the pressure units into atmospheres and the temperature units into kelvin. Solve for the unknown molar mass. Compare with molar masses of simple hydrocarbons to identify the gas. P = (570/760) atm = 0.7500 atm and T = 0 + 73.15 = 93. K drt ( 0.875 g L )( 8.0574 10 L atm K mol )(93. K) = = = 8.1 g mol P (0.7500 atm) Example 4.9b d = 0.715 g L If the gas contains moles of carbon atoms (4.0 g mol ) and 4 moles of hydrogen atoms (4.03 g mol ), then = 8.1 g mol. Gas is ethene C H 4. Calculate the density of methane gas at STP. 4.10 The Stoichiometry of Reacting Gases Example 4.10a Calculate the volume of each gas produced when 5 g of NH 4 NO 3 (ammonium nitrate) decomposes completely to N O and H O gases at atmospheric pressure (1.00 atm) and 300 C (573 K). [Note the convention for significant figures described at the beginning of this chapter.] Balance the chemical equation for the decomposition of ammonium nitrate solid. Calculate the number of moles of solid and use mole-to-mole calculations to determine the number of moles of each gas formed. Use the ideal gas law to calculate the volume of each gas formed at 1.00 atm and 573 K. NH 4 NO 3 (s) N O (g) + H O (g) (5 g NH 4 NO 3 )/(80.05 g mol 1 ) =.811 mol 1 mol NH 4 NO 3 1 mol N O and.811 mol NH 4 NO 3.811 mol N O 1 mol NH 4 NO 3 mol H O and.811 mol NH 4 NO 3 5.6 mol H O V N = nrt/p = (.811 mol)(0.08 057 4 L atm K O 1 mol 1 )(573 K)/(1.00 atm) = 13 L V HO = nrt/p = (5.6 mol)(0.08 057 4 L atm K 1 mol 1 )(573 K)/(1.00 atm) = 64 L 3
Chapter 4 Example 4.10b An initial volume of 35 L of butane at 50 C and 3.4 atm is burned with excess oxygen: C 4 H 10 (g) + 13 O (g) 8 CO (g) + 10 H O (g). Calculate the mass of carbon dioxide released to the atmosphere. m = 4.94 10 3 g 4.11 ixtures of Gases Example 4.11a A 1.00-L sample of 100% humid air from the atmosphere is taken at 15 C. An analysis reveals the following composition by mass: 934 mg of nitrogen (N ), 85 mg of oxygen (O ), 1.8 mg of water vapor (H O), 15.9 mg of argon (Ar), and 0.658 mg of carbon dioxide (CO ). Calculate the mole fraction and partial pressure of each component of air in the sample. Convert the mass of each component into amount (n A, n B, ) by using the appropriate molar mass. Sum the amounts of each component to obtain the total amount n. Take the ratio of the amount of each component to the total amount to obtain the mole fraction. The total pressure P (atm) is obtained from the total number of moles, the volume of the sample, and the absolute temperature by the ideal gas law. The partial pressures are then obtained from Dalton s law (P A = x A P, ). 1 g 1 mol n N = 934 mg = 3.333 10 mol 1000 mg 8.0 g n 1 g 1 mol 3 O = 85 mg 8.906 10 mol 1000 mg 3.00 g = n HO 1 g 1 mol 4 = 1.8 mg = 7.105 10 mol 1000 mg 18.016 g n 1 g 1 mol 4 Ar = 15.9 mg 3.980 10 mol 1000 mg 39.95 g = n 1 g 1 mol 5 CO = 0.658 mg 1.495 10 mol 1000 mg 44.01 g = n = n N + no + nh O + nar + nco = 4.336 10 mol xn = n N/ n = 0.7687; xo = no/ n = 0.054; xho= nho/ n = 0.016 39; xar = n Ar/ n = 0.009 179; xco = nco / n = 0.000 3448 P = nrt/v = (0.043 33 mol)(0.08 057 4 L atm K 1 mol 1 )(88. K)/(1.00 L) = 1.05 atm PN = x N P = 0.788 atm; PO = xop = 0.11 atm; PHO= xhop = 0.0168 atm; PAr = x ArP = 0.00941 atm; PCO = xco P = 0.000 353 atm Example 4.11b A sample of damp air in a 1.00-L container exerts a pressure of 76.0 Torr at 0 C; but when it is cooled to 0.0 C, the pressure falls to 607.1 Torr as the water condenses. What mass of water was present? Assume the vapor pressure of ice at 0 C is negligible. m = 8.44 10 g 4
The Properties of Gases olecular otion (Section 4.1 4.14) diffusion, effusion, Graham s law of effusion, kinetic model of gases, pressure-volume product proportional to square of speed, momentum, root mean square speed, axwell distribution of speeds 4.1 Diffusion and Effusion Example 4.1a It takes a certain amount of hydrogen gas 45 s to effuse through a porous barrier. How long, in minutes, does it take the same amount of uranium hexafluoride (UF 6 ) to effuse under the same conditions? First determine the molar mass of H and UF 6. The time for effusion is proportional to the square root of the molar mass. Because UF 6 has a higher molar mass, it takes longer to effuse. H 6 UF = (1.0079) =.0158 g mol ; = 38.03+ 6(19.00) = 35.03 g mol 6 (time for UF to effuse) = (time for H to effuse) = (45 s) 6 ( ) = (45 s) 174.635 = (45 s) 13.15 = 5.9 10 s = 9.9 min UF H 35.03.0158 Example 4.1b It takes 50.0 ml of nitrogen 55.0 s to effuse through a porous barrier. An unknown compound takes 164 s to effuse through the same barrier under the same conditions. What is the molar mass of this compound? of unknown compound = 48 g mol 1 Example 4.1c How much does the rate of effusion (and average speed) in H change when the temperature is decreased from 5 C to 0 C? Convert temperature to the absolute scale before using the above relationship. T 1 = 5 + 73.15 = 98. K and T = 0 + 73.15 = 153. K Rate of effusion (average speed) at 153 K 153. K = = 0.5137 = 0.717 Rate of effusion (average speed) at 98 K 98. K Example 4.1d The average speed of molecules in a gas increases by 50% when the temperature is increased from an initial value of 30 K. What is the final temperature of the gas? T = 70 K Example 4.1e Which has the higher average speed, N O molecules at 50 C or HCN molecules at 5 C? Calculate the molar mass of N O and HCN. Convert Celsius temperature to the absolute scale. Determine which gas has the larger value of the square root of the temperature over the molar mass. NO = (14.01) + 16.00 = 44.0 g mol HCN = 1.007 9 + 1.01+ 14.01 = 7.07 9 g mol TNO= 50 + 73. = 33. K and THCN= 5+ 73. = 98. K Average speed of N O (33./44.0) 1/ =.71 Average speed of HCN (98./7.07 9) 1/ = 3.3 Therefore, HCN has the greater average speed under these conditions. 5
Chapter 4 Example 4.1f What ratio of temperatures is required for methane (CH 4 ) and helium (He) to have the same average speed? TCH 4 CH 16.0416 4 = = = 4.01 THe He 4.00 4.13 The Kinetic odel of Gases Example 4.13a Calculate the root mean square speed of helium and radon gases at 300 K and 100 K. Look up the molar masses of helium and radon and convert them to kg mol. Use the above expression for root mean square speed to calculate c at the two temperatures. Be careful to use the gas constant in the proper SI units. He = (4.00 g mol ) (10 3 kg g ) = 4.00 10 3 kg mol = ( g mol ) (10 3 kg g ) =. 10 kg mol Rn 3RT 3 (8.314 47 J K mol )(300 K) 6 3 vrms = = = 1.871 10 m s = 1.37 10 m s 3 4.00 10 kg mol The ratio of temperatures is (100/300) = 4. Because v rms T, the speed at 100 K will be 4 = times that at 300 K The final speed for helium is therefore v r m s =.74 10 3 m s at 100 K. For radon, the same procedure yields v r m s = 184 m s at T = 300 K and v r m s = 368 m s at T = 100 K Example 4.13b Calculate the root mean square speed for ethene (ethylene, C H 4 ) gas at 98 K. At what temperature will ethyne (acetylene, C H ) gas have the same root mean square speed? v r m s = 515 m s 1 and T = 77 K for ethyne gas The Impact on aterials: Real Gases (Sections 4.15 4.17) real gases, compression factor ( Z ), intermolecular forces, gas liquefaction, Joule-Thomson effect, equation of state, virial equation, virial coefficients, van der Waals equation, van der Waals parameters 6
The Properties of Gases 4.17 Equations of State of Real Gases Example 4.17a Use both the ideal gas law and van der Waals equation to calculate the pressure of CO gas at 98 K and 00 K (5 K above the sublimation temperature). Assume the molar volume is exactly 1 L mol. Note: The van der Waals parameters are a = 3.640 L atm mol and b = 4.67 10 L mol. Use the ideal gas law and the rearranged form of the van der Waals equation. T = 98 K nrt ( 0.08057 L atm K mol ) (98 K) P = = = 4.45 atm V (1 L mol ) nrt n (0.08 057 L atm K mol )(98 K) (3.640 L atm mol ) P = a = V nb V (1 4.67 10 )(L mol ) (1 L mol ) = ( 5.543 3.640 ) atm = 1.90 atm Change from ideal = (4.45 1.90)/(4.45) = (.55/4.45) = 0.104 or 10.4% T = 00 K nrt (0.08 057 L atm K mol )(00 K) P = = = 16.41 atm V (1 L mol ) nrt n (0.08 057 L atm K mol )(00 K) (3.640 L atm mol ) P = a = V nb V (1 4.67 10 )(L mol ) (1 L mol ) = ( 17.143 3.640 ) atm = 13.50 atm Change from ideal = (16.41 13.50)/(16.41) = (.91/16.41) = 0.177 or 17.7% Example 4.17b Use both the ideal gas law and van der Waals equation to calculate the pressure of H O gas at 98 K and 373 K (boiling temperature of water). Use the molar volumes of 78.3 L mol at 98 K and 30.61 L mol at 373 K, which correspond to the vapor pressure of liquid water. Note: The van der Waals parameters are a = 5.536 L atm mol and b = 3.049 10 L mol. P = 0.031 6 atm (ideal gas) and P = 0.031 5 atm (van der Waals) at 98 K P = 0.999 9 atm (ideal gas) and P = 0.995 0 atm (van der Waals) at 373 K Surprisingly, water vapor behaves as an ideal gas up to its boiling temperature. 7