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Gas Laws

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CHAPTER 5 GASES AND THE KINETICMOLECULAR THEORY
5.1 Plan: Review the behavior of the gas phase vs. the liquid phase. Solution: a) The volume of the liquid remains constant, but the volume of the gas increases to the volume of the larger container. b) The volume of the container holding the gas sample increases when heated, but the volume of the container holding the liquid sample remains essentially constant when heated. c) The volume of the liquid remains essentially constant, but the volume of the gas is reduced. The particles in a gas are further apart than those are in a liquid. a) The greater empty space between gas molecules allows gases to be more compressible than liquids. b) The greater empty space between gas molecules allows gases to flow with less resistance (hindrance) than liquids. c) The large empty space between gas molecules limits their interaction, allowing all mixtures of gases to be solutions. d) The large empty space between gas molecules increases the volume of the gas, therefore decreasing the density. The mercury column in the mercury barometer stays up due to the force exerted by the atmosphere on the mercury in the outer reservoir just balancing the gravitational force on the mercury in the tube. Its height adjusts according to the air pressure on the reservoir. The column of mercury is shorter on a mountaintop as there is less atmosphere to exert a force on the mercury reservoir. On a mountaintop, the air pressure is less, so the height of mercury it balances in the barometer is shorter than at sea level where there is more air pressure. The pressure of mercury is its weight (force) per unit area. The weight, and thus the pressure, of the mercury column is directly proportional to its height. When the mercury level in the arm attached to the flask is higher than the level in the other arm, the pressure in the flask is less than the pressure exerted in the other arm. This is an impossible situation for a closed-end manometer as the flask pressure cannot be less than the vacuum in the other arm. Plan: The ratio of the heights of columns of mercury and water are inversely proportional to the ratio of the densities of the two liquids. Solution: h H2O d Hg = h Hg d H2O h H2O = d Hg d H2O
⎛ 10−3 m ⎞ ⎛ 1 cm ⎞ ⎛ 13.5 g/mL ⎞ xh Hg = ⎜ = 985.5 = 990 cmH2O ( 730 mmHg ) ⎜ ⎟ ⎜ 1 mm ⎟ ⎜ 10−2 m ⎟ ⎟⎜ ⎟ ⎝ 1.00 g/mL ⎠ ⎠ ⎝ ⎠⎝ d Hg d H2O ⎛ 10−3 m ⎞⎛ 1 cm ⎞ ⎛ 13.5 g/mL ⎞ xh Hg = ⎜ = 1019.25 = 1.02 x 103 cmH2O ⎟ ( 755 mmHg ) ⎜ ⎜ 1 mm ⎟⎜ 10−2 m ⎟ ⎟⎜ ⎟ ⎝ 1.00 g/mL ⎠ ⎠ ⎝ ⎠⎝

5.2

5.3

5.4 5.5

5.6

5.7

P (mmH2O) = h H2O =

5-1

5.8

Plan: Use the conversion factors between pressure units: 1 atm = 760 mmHg = 760 torr = 101.325 kPa = 1.01325 bar Solution: ⎛ 760 mmHg ⎞ a) ( 0.745 atm ) ⎜ ⎟ = 566.2 = 566 mmHg ⎝ 1 atm ⎠
⎛ 1.01325 bar ⎞ b) ( 992 torr ) ⎜ ⎟ = 1.32256 = 1.32 bar ⎝ 760 torr ⎠ ⎛ ⎞ 1 atm c) ( 365 kPa ) ⎜ ⎟ = 3.60227 = 3.60 atm 101.325 kPa ⎠ ⎝ ⎛ 101.325 kPa ⎞ d) ( 804 mmHg ) ⎜ ⎟ = 107.191 = 107 kPa ⎝ 760 mmHg ⎠

5.9

⎛ 10−2 m ⎞ ⎛ 1 mm ⎞ ⎛ 1 atm ⎞ = 1.01083 = 1.01 atm a) ( 76.8 cmHg ) ⎜ ⎜ 1 cm ⎟ ⎜ 10−3 m ⎟ ⎜ 760 mmHg ⎟ ⎟⎜ ⎟ ⎠ ⎠⎝ ⎝ ⎠⎝ ⎛ 101.325 kPa ⎞ 3 3 b) ( 27.5 atm ) ⎜ ⎟ = 2.786 x 10 = 2.79 x 10 kPa 1 atm ⎝ ⎠ ⎛ 1.01325 bar ⎞ c) ( 6.50 atm ) ⎜ ⎟ = 6.5861 = 6.59 bar ⎝ 1 atm ⎠ ⎛ 760 torr ⎞ d) ( 0.937 kPa ) ⎜ ⎟ = 7.02808 = 7.03 torr ⎝ 101.325 kPa ⎠

5.10

Plan: Since the height of the mercury column in contact with the gas is higher than the column in contact with the air, the gas is exerting less pressure on the mercury than the air. Therefore the pressure corresponding to the height difference between the two arms is subtracted from the atmospheric pressure. Solution: ⎛ 10−2 m ⎞⎛ 1 mm ⎞ ⎛ 1 torr ⎞ = 23.5 torr ( 2.35 cm ) ⎜ ⎜ 1 cm ⎟⎜ 10−3 m ⎟ ⎜ 1 mmHg ⎟ ⎟⎜ ⎟ ⎠ ⎠⎝ ⎝ ⎠⎝ 738.5 torr - 23.5 torr = 715.0 torr ⎛ 1 atm ⎞ P(atm) = ( 715.0 torr ) ⎜ ⎟ = 0.940789 = 0.9408 atm ⎝ 760 torr ⎠ Since the height of the mercury column in contact with the gas is higher than the column in contact with the air, the gas is exerting less pressure on the mercury than the air. ⎛ 10−2 m ⎞ ⎛ 1 mm ⎞ ⎛ 1 torr ⎞ = 13.0 torr (1.30 cm ) ⎜ ⎜ 1 cm ⎟ ⎜ 10−3 m ⎟ ⎜ 1 mmHg ⎟ ⎟⎜ ⎟ ⎠ ⎠⎝ ⎝ ⎠⎝ 765.2 torr - 13.0 torr = 752.2 torr ⎛ 101.325 kPa ⎞ P(kPa) = ( 752.2 torr ) ⎜ ⎟ = 100.285 = 100.3 kPa ⎝ 760 torr ⎠ The difference in the height of the Hg is directly related to pressure in atmospheres. ⎛ 1 mmHg ⎞ ⎛ 1 atm ⎞ = 0.965789 = 0.966 atm P(atm) = ( 0.734 mHg ) ⎜ −3 ⎜ 10 mHg ⎟ ⎜ 760 mmHg ⎟ ⎟ ⎠ ⎝ ⎠⎝

5.11

5.12

5-2

5.13

⎛ 10−2 mHg ⎞ ⎛ 1 mmHg ⎞ ⎛ 1.01325 x 105 Pa ⎞ = 4746.276 = 4.75 x 103 Pa P(Pa) = ( 3.56 cm ) ⎜ ⎜ 1 cmHg ⎟ ⎜ 10−3 m ⎟ ⎜ 760 mmHg ⎟ ⎟⎜ ⎟⎜ ⎟ ⎠⎝ ⎝ ⎠⎝ ⎠ ⎛ 1 atm ⎞ a) P(atm) = 2.75 x 102 mmHg ⎜ ⎟ = 0.361842 = 0.362 atm ⎝ 760 mmHg ⎠

5.14

(

)

⎛ 1 atm ⎞ b) P(atm) = ( 86 psi ) ⎜ ⎟ = 5.85034 = 5.9 atm ⎝ 14.7 psi ⎠ ⎛ ⎞ 1 atm c) P(atm) = 9.15 x 106 Pa ⎜ = 90.303 = 90.3 atm ⎜ 1.01325 x 105 Pa ⎟ ⎟ ⎝ ⎠

(

)

⎛ 1 atm ⎞ d) P(atm) = 2.54 x 104 torr ⎜ ⎟ = 33.42105 = 33.4 atm ⎝ 760 torr ⎠

(

)

5.15

a) 1 atm = 1.01325 x 105 N/m2 The force on 1 m2 of ocean is 1.01325 x 105 N. F=mxg 1.01325 x 105 N = mg kgim 1.01325 x 105 2 = (mass) (9.81 m/s2) s mass = 1.03287 x 104 = 1.03 x 104 kg kg ⎞ ⎛ 103 g ⎞ ⎛ 10−2 m ⎞ ⎛ b) ⎜1.03287 x 104 2 ⎟ ⎜ = 1.03287 x 103 g/cm2 (unrounded) ⎜ 1 kg ⎟ ⎜ 1 cm ⎟ ⎟⎜ ⎟ m ⎠⎝ ⎝ ⎠⎝ ⎠ g ⎞ ⎛ 1 mL ⎞ ⎛ 1 cm3 ⎞ ⎛ Height = ⎜1.03287 x 103 ⎟ = 45.702 = 45.7 cm Os ⎟⎜ ⎟⎜ ⎟ cm 2 ⎠ ⎝ 22.6 g ⎠ ⎜ 1 mL ⎠ ⎝ ⎝
2

5.16 5.17

The statement is incomplete with respect to temperature and mass of sample. The correct statement is: At constant temperature and moles of gas, the volume of gas is inversely proportional to the pressure. variable a) Volume; Temperature b) Moles; Volume c) Pressure; Temperature fixed Pressure; Moles Temperature; Pressure Volume; Moles

5.18

Plan: Review Boyle’s and Amonton’s Laws. Solution: At constant temperature and volume, the pressure of the gas is directly proportional to the number of moles of the gas. Verify this by examining the ideal gas equation. At constant T and V, the ideal gas equation becomes P = n(RT/V) or P = n x constant. a) n fixed b) P halved c) P fixed d) T doubled

5.19 5.20

Plan: Use the relationship

P1V1 PV VPT = 2 2 or V2 = 1 1 2 T1 T2 P2 T1

Solution: a) As the pressure on a gas increases, the molecules move closer together, decreasing the volume. When the pressure is tripled, the volume decreases to one third of the original volume at constant temperature (Boyle’s VPT (V )(P )(1) V2 = ⅓V1 Law). V2 = 1 1 2 = 1 1 P2 T1 (3P1 )(1)

5-3

b) As the temperature of a gas increases, the gas molecules gain kinetic energy. With higher energy, the gas molecules collide with the walls of the container with greater force, which increases the size (volume) of the container. If the temperature is increased by a factor of 3.0 (at constant pressure) then the volume will increase VPT (V )(1)(3T1 ) by a factor of 3.0 (Charles’s Law). V2 = 1 1 2 = 1 V2 = 3V1 P2 T1 (1)(T1 ) c) As the number of molecules of gas increases, the force they exert on the container increases. This results in an increase in the volume of the container. Adding three moles of gas to one mole increases the number of moles by a factor of four, thus the volume increases by a factor of four (Avogadro’s Law). 5.21 V2 = V1 x (P1/P2) x (T2/T1) a) If P2 is 1/4 of P1, then the volume would be increased by a factor of 4. b) V2 = V1 x (101 kPa/202 kPa) x (155 K/310 K) = 1/4, so the volume would be decreased by a factor of 4. c) V2 = V1 x (202 kPa/101 kPa) x (305 K/305 K) = 2, so the volume would be increased by a factor of 2.
P1V1 PV VPT = 2 2 or V2 = 1 1 2 T1 T2 P2 T1

5.22

Plan: Use the relationship

Solution: a) The temperature is decreased by a factor of 2, so the volume is decreased by a factor of 2 (Charles’s Law). 1 (V1 )(1)( T1 ) V1P1T2 2 V2 = = V2 = ½ V1 P2 T1 (1)(T1 ) b) The temperature increases by a factor of [(500 + 273) / (250 + 273)] = 1.48, so the volume is increased by a VPT (V )(1)(1.48T1 ) factor of 1.48 (Charles’s Law). V2 = 1 1 2 = 1 V2 = 1.48V1 P2 T1 (1)(T1 ) c) The pressure is increased by a factor of 3, so the volume decreases by a factor of 3 (Boyle’s Law). VPT (V )(P )(1) V2 = 1 1 2 = 1 1 V2 = ⅓V1 P2 T1 (3P1 )(1) 5.23 V2 = V1 x (P1/P2) x (T2/T1) x (n2/n1) a) Since n2 = 0.5 n1, the volume would be decreased by a factor of 2. b) This corresponds to both P and T remaining constant, so the volume remains constant. c) Since P2 = 0.25 P1 and T2 = 0.25 T1, these two effects offset one another and the volume remains constant. Plan: This is Charles’s Law. Charles’s Law states that at constant pressure and with a fixed amount of gas, the volume of a gas is directly proportional to the absolute temperature of the gas. The temperature must be lowered to reduce the volume of a gas. Solution: V1 V2 V T2 = T1 2 = at constant n and P T1 T2 V1 V1 = 9.10 L; T1 = 198oC + 273 = 471K; V2 = 2.50 L T2 = ? ⎛ 2.50 L ⎞ T2 = 471K ⎜ ⎟ = 129.396 K – 273 = –143.604 = –144°C ⎝ 9.10 L ⎠
V2 = V1 ⎛ ( 273 − 22 ) K ⎞ T2 = ( 93 L ) ⎜ = 55.844 = 56 L ⎜ ( 273 + 145 ) K ⎟ ⎟ T1 ⎝ ⎠

5.24

5.25

5-4

5.26

Plan: Since the volume, temperature, and pressure of the gas are changing, use the combined gas law. Solution: P1 V1 P2 V2 P1 = 153.3 kPa; V1 = 25.5 L; T1 = 298 K; P2 = 101.325 kPa; T2 = 273 K; V2 = ? = T1 T2
⎛T V2 = V1 ⎜ 2 ⎜T ⎝ 1 ⎞ ⎛ P1 ⎟⎜ ⎟⎜ P ⎠⎝ 2 ⎞ ⎛ 273 K ⎞⎛ 153.3 kPa ⎞ ⎟ = ( 25.5 L ) ⎜ ⎟⎜ ⎟ = 35.3437 = 35.3 L ⎟ ⎝ 298 K ⎠⎝ 101.325 kPa ⎠ ⎠

5.27

P1 V1 P2 V2 = T1 T2 ⎛T ⎞⎛ P V2 = V1 ⎜ 2 ⎟ ⎜ 1 ⎜ T ⎟⎜ P ⎝ 1 ⎠⎝ 2

P1 = 745 torr; V1 = 3.65 L; T1 = 298 K; P2 = 367 torr; T2 = –14oC + 273 = 259 K; V2 = ?
⎞ ⎛ 259 K ⎞ ⎛ 745 torr ⎞ ⎟ = (3.65 L ) ⎜ ⎟⎜ ⎟ = 6.4397 = 6.44 L ⎟ ⎝ 298 K ⎠ ⎝ 367 torr ⎠ ⎠

5.28

Plan: Given the volume, pressure, and temperature of a gas, the number of moles of the gas can be calculated using the ideal gas equation, PV = nRT. The gas constant, R = 0.0821 L•atm/mol•K, gives pressure in atmospheres and temperature in Kelvin. The given pressure in torr must be converted to atmospheres and the temperature converted to Kelvin. Solution: PV = nRT or n = PV/RT ⎛ 1 atm ⎞ V = 5.0 L; T = 37oC + 273 = 310 K P = ( 328 torr ) ⎜ ⎟ = 0.43158 atm; ⎝ 760 torr ⎠ n = PV (0.43158 atm)(5.0 L) = 0.08479 = 0.085 mol chlorine = Liatm RT (0.0821 )(310 K) moliK

5.29

PV = nRT P = nRT / V =

(1.47 x 10

−3

Liatm ⎞ ⎛ mol ⎜ 0.0821 ⎟ ( ( 273 + 26 ) K ) ⎛ 1 mL ⎞ ⎛ 760 torr ⎞ moliK ⎠ ⎝ ⎜ −3 ⎟ ⎜ ⎜ 10 L ⎟ 1 atm ⎟ 75.0 mL ⎠ ⎝ ⎠⎝

)

= 365.6655 = 366 torr 5.30 Plan: Solve the ideal gas equation for moles and convert to mass using the molar mass of ClF3. Volume must be converted to L, pressure to atm and temperature to K. Solution: ⎛ 1 atm ⎞ V = 0.357 L PV = nRT or n = PV/RT P = (699 mmHg ) ⎜ ⎜ 760 mmHg ⎟ = 0.91974 atm; ⎟ ⎝ ⎠ T = 45oC + 273 = 318 K
PV (0.91974 atm)(0.357 L) = 0.01258 mol ClF3 = L i atm RT (0.0821 )(318 K) mol i K ⎛ 92.45 g ClF3 ⎞ mass ClF3 = ( 0.01258 mol ClF3 ) ⎜ ⎟ = 1.163021 = 1.16 g ClF3 ⎝ 1 mol ClF3 ⎠ n =

5-5

5.31

PV = nRT

⎛ 1 mol N 2 O ⎞ n = (75.0 g N 2 O )⎜ ⎜ 44.02 g N O ⎟ = 1.70377 mol N2O ⎟ 2 ⎝ ⎠ Li atm (1.70377 mol )(0.0821 )(388 K) nRT mol i K P= = = 17.5075 =18 atm N2O V (3.1 L)

5.32

PV = nRT

The total pressure of the gas is (85 + 14.7)psi. ( (85 + 14.7 ) psi ) (1.5 L ) ⎛ 1 atm ⎞ = 0.41863 = 0.42 mol SO n = PV / RT = 2 ⎟ ⎜ Liatm ⎞ 14.7 psi ⎠ ⎛ 0.0821 273 + 23) K ) ⎝ (( ⎜ moliK ⎟ ⎝ ⎠

5.33

Plan: Assuming that while rising in the atmosphere the balloon will neither gain nor lose gas molecules, the number of moles of gas calculated at sea level will be the same as the number of moles of gas at the higher altitude. Using the ideal gas equation, (n x R) is a constant equal to (PV/T). Given the sea-level conditions of volume, pressure and temperature, and the temperature and pressure at the higher altitude for the gas in the balloon, we can set up an equation to solve for the volume at the higher altitude. Comparing the calculated volume to the given maximum volume of 835 L will tell us if the balloon has reached its maximum volume at this altitude. Solution: P1V1 PV = 2 2 T1 T2
V2 = P1V1T2 P2 T1

V2 = =

( 745 torr )( 65.0 L ) ( ( 273 − 5 ) K ) ⎛ 1 atm ⎞ ( ( 273 + 25) K ) ( 0.066 atm ) ⎜ 760 torr ⎟ ⎝ ⎠

= 868.2217 = 870 L

The calculated volume of the gas at the higher altitude is more than the maximum volume of the balloon. Yes, the balloon will reach its maximum volume. Check: Should we expect that the volume of the gas in the balloon should increase? At the higher altitude, the pressure decreases; this increases the volume of the gas. At the higher altitude, the temperature decreases, this decreases the volume of the gas. Which of these will dominate? The pressure decreases by a factor of 0.98/0.066 = 15. If we label the initial volume V1, then the resulting volume is 15 V1. The temperature decreases by a factor of 296/268 = 1.1, so the resulting volume is V1/1.1 or 0.91 V1. The increase in volume due to the change in pressure is greater than the decrease in volume due to change in temperature, so the volume of gas at the higher altitude should be greater than the volume at sea level. 5.34 Air is mostly N2 (28.02 g/mol), O2 (32.00 g/mol), and argon (39.95 g/mol). These “heavy” gases dominate the density of dry air. Moist air contains H2O (18.02 g/mol). The relatively light water molecules lower the density of the moist air. The molar mass of H2 is less than the average molar mass of air (mostly N2, O2, and Ar), so air is denser. To collect a beaker of H2(g), invert the beaker so that the air will be replaced by the lighter H2. The molar mass of CO2 is greater than the average molar mass of air, so CO2(g) is more dense. Collect the CO2 holding the beaker upright, so the lighter air will be displaced out the top of the beaker. Gases mix to form a solution and each gas in the solution behaves as if it were the only gas present. PA = XA PT The partial pressure of a gas (PA) in a mixture is directly proportional to its mole fraction (XA). a) XA = 0.25; XB = 0.1875; XC = 0.3125; XD2 = 0.25. Gas C has the highest partial pressure. b) Gas B has the lowest partial pressure.. PD2 = 0.25(0.75 atm) PD2 = 0.1875 = 0.19 atm c) PD2 = XD2PT

5.35

5.36 5.37 5.38

5-6

5.39

Plan: Using the ideal gas equation and the molar mass of xenon, 131.3 g/mol, we can find the density of xenon gas at STP. Standard temperature is 0°C and standard pressure is 1 atm. Do not forget that the pressure at STP is exact and will not affect the significant figures. Solution: (131.3 g/mol )(1 atm ) = 5.8581 = 5.86 g/L d = M P / RT = Liatm ⎞ ⎛ 0.0821 ( 273 K ) ⎜ moliK ⎟ ⎝ ⎠ d = M P / RT = = 6.385807663 = 6.4 g/L Liatm ⎞ ⎛ 0.0821 273 + 120 ) K ) (( ⎜ moliK ⎟ ⎝ ⎠ Plan: Apply the ideal gas equation to determine the number of moles. Convert moles to mass and divide by the volume to obtain density in g/L. Do not forget that the pressure at STP is exact and will not affect the significant figures. Solution: (1 atm )( 0.0400 L ) PV = = 1.78465 x 10–3 = 1.78 x 10–3 mol AsH3 n= Liatm ⎞ RT ⎛ ⎜ 0.0821 moliK ⎟ ( 273 K ) ⎝ ⎠
1.78465 x 10−3 mol ( 77.94 g/mol ) mass d= = = 3.47740 = 3.48 g/L volume ( 0.0400 L )

5.40 5.41

(137.36 g/mol )(1.5 atm )

(

)

5.42

d = M P / RT
M = dRT / P =

( 2.71 g/L ) ⎛ 0.0821 ⎜


Liatm ⎞ ( ( 273 + 0 ) K ) moliK ⎟ ⎠ = 20.24668 = 20.2 g/mol ( 3.00 atm )

Therefore, the gas is Ne. 5.43 Plan: Rearrange the formula PV = (m / M)RT to solve for molar mass: M = mRT / PV. Convert the mass in ng to grams and volume in μL to L. Temperature must be in Kelvin and pressure in torr. Solution: ⎛ 1 atm ⎞ P = (388 torr ) ⎜ T = 45oC + 273 = 318 K ⎟ =0.510526 atm ⎝ 760 torr ⎠
⎛ 10 -9 g ⎞ ⎛ 10 -6 L ⎞ –7 ⎟ ⎟ = 2.06 x 10–7 L mass = (206 ng )⎜ V = (0.206 μL )⎜ ⎜ 1 μL ⎟ ⎜ 1 ng ⎟ = 2.06 x 10 g ⎠ ⎝ ⎝ ⎠ Liatm ⎛ ⎞ -7 ⎜ (2.06 x 10 g)(0.0821 moliK )(318 K) ⎟ mRT = ⎜ M= ⎟ = 51.1390 = 51.1 g/mol PV (0.510526 atm)(2.06 x 10-7 L) ⎜ ⎟ ⎜ ⎟ ⎝ ⎠

5.44

M = mRT / PV =

( 0.103 g ) ⎛ 0.0821 ⎜

Liatm ⎞ ⎟ ( ( 273 + 22 ) K ) ⎛ 1 mL ⎞ ⎛ 760 mmHg ⎞ moliK ⎠ ⎝ ⎜ −3 ⎟ ⎜ ⎟ ⎜ 10 L ⎟ ( 747 mmHg )( 63.8 mL ) ⎝ ⎠ ⎝ 1 atm ⎠

= 39.7809 = 39.8 g/mol The molar masses are N2 = 28 g/mol, Ne = 20 g/mol, and Ar = 40 g/mol Therefore, the gas is Ar.

5-7

5.45

Plan: Use the ideal gas equation to determine the number of moles of Ar and O2. The gases are combined (nTOT = nAr + nO) into a 400 mL flask (V) at 27°C (T). Determine the total pressure from nTOT, V, and T. Pressure must be in units of atm, volume in units of L and temperature in K. Solution: PV n= PV = nRT RT (1.20 atm )( 0.600 L ) PV = = 0.017539585 mol Ar (unrounded) Moles Ar = Liatm ⎞ RT ⎛ 0.0821 ( ( 273 + 227 ) K ) ⎜ moliK ⎟ ⎝ ⎠ ⎛ 1 atm ⎞ ( 501 torr )( 0.200 L ) PV Moles O2 = = ⎟ = 0.004014680 mol O2 (unrounded) ⎜ Liatm ⎞ RT ⎛ 0.0821 ( 273 + 127 ) K ) ⎝ 760 torr ⎠ ( ⎜ moliK ⎟ ⎝ ⎠ nTOT = n Ar + nO = 0.017539585 mol + 0.004014680 mol = 0.021554265 mol Liatm ⎞ ( 0.021554265 mol ) ⎛ 0.0821 ( ( 273 + 27 ) K ) ⎛ 1 mL ⎞ ⎜ nRT moliK ⎟ ⎝ ⎠ = P mixture = ⎜ −3 ⎟ V 400 mL ⎝ 10 L ⎠ = 1.32720 = 1.33 atm

5.46

Total moles = nT = PV/RT =

⎛ 1 atm ⎞ ⎛ 10−3 L ⎞ ⎟ ⎜ ⎟⎜ ⎜ ⎟ Liatm ⎞ ⎛ 0.0821 ( ( 273 + 35) K ) ⎝ 760 mmHg ⎠ ⎝ 1 mL ⎠ ⎜ ⎟ moliK ⎠ ⎝ = 0.011563657 mol (unrounded) Moles Ne = nNe = (0.146 g Ne) (1 mol Ne / 20.18 g Ne) = 0.007234886 mol Ne (unrounded) Moles Ar = nT – nNe = (0.011563657 – 0.007234886) mol = 0.004328771 = 0.0043 mol Ar

( 626 mmHg )( 355 mL )

5.47

d = M P / RT At 17°C d = M P / RT = At 60.0°C d = M P / RT =

⎛ 1 atm ⎞ ⎟ = 1.18416 = 1.18 g/L ⎜ Liatm ⎞ ⎛ ⎝ 760 torr ⎠ ⎜ 0.0821 moliK ⎟ ( ( 273 + 17 ) K ) ⎝ ⎠ ⎛ 1 atm ⎞ ⎟ = 1.03125 = 1.03 g/L ⎜ Liatm ⎞ ⎛ ⎝ 760 torr ⎠ ⎜ 0.0821 moliK ⎟ ( ( 273 + 60.0 ) K ) ⎝ ⎠

( 28.8 g/mol )( 744 torr )

( 28.8 g/mol )( 744 torr )

5.48

PV = nRT n / V = P / RT =

⎛ 1 atm ⎞ ⎟ = 0.042005 = 0.0420 mol/L ⎜ Liatm ⎞ ⎛ ⎝ 760 torr ⎠ ⎜ 0.0821 moliK ⎟ ( ( 273 − 25 ) K ) ⎝ ⎠

( 650. torr )

5.49

Plan: The problem gives the mass, volume, temperature and pressure of a gas, so we can solve for molar mass using M = mRT/PV. The problem also states that the gas is a hydrocarbon, which by, definition, contains only carbon and hydrogen atoms. We are also told that each molecule of the gas contains five carbon atoms so we can use this information and the calculated molar mass to find out how many hydrogen atoms are present and the formula of the compound. Convert pressure to atm and temperature to K.

5-8

Solution:
M=
mRT = PV

( 0.482 g ) ⎛ 0.0821 ⎜

Liatm ⎞ ( ( 273 + 101) K ) ⎛ 760 torr ⎞ moliK ⎟ ⎝ ⎠ ⎜ ⎟ = 71.8869 g/mol (unrounded) ( 767 torr )( 0.204 L ) ⎝ 1 atm ⎠

The carbon accounts for [5 (12 g/mol)] = 60 g/mol, thus, the hydrogen must make up the difference (72 – 60) = 12 g/mol. A value of 12 g/mol corresponds to 12 H atoms. (Since fractional atoms are not possible, rounding is acceptable.) Therefore, the molecular formula is C5H12. 5.50 PV = nRT n = PV / RT Molecules of air = n x Avogadro’s number = (PV / RT) x Avogadro’s number ⎛ 6.022 x 1023 molecules ⎞ (1.00 atm )(1.00 L ) Molecules of air = ⎜ ⎟ ⎜ ⎟ Liatm ⎞ mol ⎛ ⎠ 0.0821 ( ( 273 + 25) K ) ⎝ ⎜ ⎟ moliK ⎠ ⎝ = 2.461395 x 1022 molecules (unrounded) Molecules N2 = (2.461395 x 1022 molecules) (78.08%/100%) = 1.921857 x 1022 = 1.92 x 1022 molecules N2 Molecules O2 = (2.461395 x 1022 molecules) (20.94%/100%) = 5.154161 x 1021 = 5.15 x 1021 molecules O2 Molecules CO2 = (2.461395 x 1022 molecules) (0.05%/100%) = 1.2306975 x 1019 = 1 x 1019 molecules CO2 Molecules Ar = (2.461395 x 1022 molecules) (0.93%/100%) = 2.289097 x 1020 = 2.3 x 1020 molecules Ar Plan: Since you have the pressure, volume and temperature, use the ideal gas equation to solve for the total moles of gas. Solution: a) PV = nRT ⎛ 1 atm ⎞ (850. torr )( 21 L ) n = PV / RT = ⎟ = 0.89961 = 0.90 mol gas ⎜ Liatm ⎞ ⎛ 0.0821 ( ( 273 + 45) K ) ⎝ 760 torr ⎠ ⎜ moliK ⎟ ⎝ ⎠ b) The information given in ppm is a way of expressing the proportion, or fraction, of SO2 present in the mixture. Since n is directly proportional to V, the volume fraction can be used in place of the mole fraction used in equation 5.12. There are 7.95 x 103 parts SO2 in a million parts of mixture, so volume fraction = (7.95 x 103 / 1 x 106) = 7.95 x 10–3. Therefore, PSO = volume fraction x PTOT = (7.95 x 10–3) (850. torr) = 6.7575 = 6.76 torr.
2

5.51

5.52

Plan: We can find the moles of oxygen from the standard molar volume of gases (1 L of gas occupies 22.4 L at STP) and use the stoichiometric ratio from the balanced equation to determine the moles of phosphorus that will react with the oxygen. Solution: P4(s) + 5 O2(g) → P4O10(s) ⎛ 1 mol O 2 ⎞ ⎛ 1 mol P4 ⎞ ⎛ 123.88 g P4 ⎞ Mass P4 = ( 35.5 L O 2 ) ⎜ ⎟⎜ ⎟⎜ ⎟ = 39.2655 = 39.3 g P4 ⎝ 22.4 L O 2 ⎠ ⎝ 5 mol O 2 ⎠ ⎝ 1 mol P4 ⎠ 2 KClO3(s) → 2 KCl(s) + 3 O2(g) Moles O2 = n = PV / RT Moles of O2 x mole ratio x molar mass KClO3 = g KClO3 ⎡ ⎤ ⎢ ⎛ 1 atm ⎞ ⎛ 10−3 L ⎞ ⎥ ⎛ 2 mol KClO3 ⎞ ⎛ 122.55 g KClO3 ⎞ ( 752 torr )( 638 mL ) Mass KClO3 = ⎢ ⎟⎥ ⎜ ⎟ ⎟⎜ ⎜ ⎟⎜ Liatm ⎞ 760 torr ⎠ ⎜ 1 mL ⎟ ⎥ ⎝ 3 mol O 2 ⎠⎝ 1 mol KClO3 ⎠ ⎢⎛ ⎝ ⎠ 0.0821 273 + 128 ) K ) ⎝ (( ⎢⎜ ⎥ moliK ⎟ ⎠ ⎣⎝ ⎦ = 1.56660 = 1.57 g KClO3

5.53

5-9

5.54

Plan: To find the mass of PH3, write the balanced equation and find the number of moles of PH3 produced by each reactant. The smaller number of moles of product indicates the limiting reagent. Solve for moles of H2 using the standard molar volume (or use ideal gas equation). Solution: P4(s) + 6 H2(g) → 4 PH3(g) ⎛ 1 mol ⎞ Moles hydrogen = ( 83.0 L ) ⎜ ⎟ = 3.705357 mol H2 (unrounded) ⎝ 22.4 L ⎠
⎛ 1 mol P4 ⎞ ⎛ 4 mol PH 3 ⎞ PH3 from P4 = ( 37.5 g P4 ) ⎜ ⎟ = 1.21085 mol PH3 ⎟⎜ ⎝ 123.88 g P4 ⎠ ⎝ 1 mol P4 ⎠ ⎛ 4 mol PH 3 ⎞ PH3 from H2 = ( 3.705357 mol H 2 ) ⎜ ⎟ = 2.470238 mol PH3 ⎝ 6 mol H 2 ⎠ P4 is the limiting reactant. ⎛ 1 mol P4 ⎞⎛ 4 mol PH3 ⎞ ⎛ 33.99 g PH3 ⎞ Mass PH3 = ( 37.5 g P4 ) ⎜ ⎟ = 41.15676 = 41.2 g PH3 ⎟⎜ ⎟⎜ ⎝ 123.88 g P4 ⎠⎝ 1 mol P4 ⎠ ⎝ 1 mol PH3 ⎠

5.55

4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(l) The moles are directly proportional to the volumes of the gases at the same temperature and pressure. Thus, the limiting reactant may be found by comparing the volumes of the gases. ⎛ 4 L NO ⎞ Mol NO from NH3 = ( 35.6 L NH3 ) ⎜ ⎟ = 35.6 L NO ⎝ 4 L NH 3 ⎠ ⎛ 4 L NO ⎞ Mol NO from O2 = ( 40.5 L O 2 ) ⎜ ⎟ = 32.4 L NO ⎝ 5 L O2 ⎠ O2 is the limiting reactant. ⎛ 1 mol O 2 ⎞ ⎛ 4 mol NO ⎞ ⎛ 30.01 g NO ⎞ Mass NO = ( 40.5 L O 2 ) ⎜ ⎟⎜ ⎟⎜ ⎟ = 43.4073 = 43.4 g NO ⎝ 22.4 L O 2 ⎠ ⎝ 5 mol O 2 ⎠ ⎝ 1 mol NO ⎠ Plan: First, write the balanced equation. The moles of hydrogen produced can be calculated from the ideal gas equation and then the stoichiometric ratio from the balanced equation is used to determine the moles of aluminum that reacted. The problem specifies “hydrogen gas collected over water,” so the partial pressure of water must first be subtracted. Table 5.3 reports pressure at 26°C (25.2 torr) and 28°C (28.3 torr), so take the average of the two values to obtain the partial pressure of water at 27oC. Solution: 2 Al(s) + 6 HCl(aq) → 2 AlCl3(aq) + 3 H2(g) Hydrogen pressure = total pressure – pressure of water vapor = (751 mmHg) – [(28.3 + 25.2) torr / 2] = 724.25 torr (unrounded) Moles of hydrogen: ⎡ ⎤ ⎢ ( 724.25 torr )( 35.8 mL ) ⎥ −3 PV ⎛ 1 atm ⎞ ⎛ 10 L ⎞ ⎥ n= = ⎢ PV =nRT ⎜ 760 torr ⎟ ⎜ 1 mL ⎟ = 0.00138514 mol H2 Liatm ⎞ RT ⎢⎛ ⎝ ⎠⎝ ⎠⎥ ⎢ ⎜ 0.0821 moliK ⎟ ( 273 + 27 ) K ⎥ ⎝ ⎠ ⎣ ⎦
⎛ 2 mol Al ⎞ ⎛ 26.98 g Al ⎞ Mass of Al = ( 0.00138514 mol H 2 ) ⎜ ⎟⎜ ⎟ = 0.024914 = 0.0249 g Al ⎝ 3 mol H 2 ⎠ ⎝ 1 mol Al ⎠

5.56

5-10

5.57

First, write the balanced equation. The problem specifies “hydrogen gas collected over water,” so the partial pressure of water must first be subtracted. Table 5.3 reports the vapor pressure of water at 18°C (15.5 torr). 2 Li(s) + 2 H2O(l) → 2 LiOH(aq) + H2(g) Pressure H2 = total pressure – vapor pressure of H2O = 725 mmHg – 15.5 torr (1mmHg/1torr) = 709.5 mmHg
⎛ 1 mol Li ⎞ ⎛ 1 mol H 2 Moles H2 = ( 0.84 g Li ) ⎜ ⎟⎜ ⎝ 6.941 g Li ⎠ ⎝ 2 mol Li ⎞ ⎟ = 0.0605100 mol H2 ⎠

⎡ ⎤ Liatm ⎞ ⎢ ( 0.0605100 mol ) ⎛ 0.0821 ( 273+ 18) ⎥ ⎜ nRT moliK ⎟ ⎢ ⎥ ⎝ ⎠ Volume H2 = = ⎢ ⎥ = 1.5485 = 1.5 L H2 P ⎛ ⎞ ⎛ 1 atm ⎞ ⎢ ⎥ ⎜ 709.5 mmHg ⎜ ⎟⎟ ⎜ ⎟ ⎢ ⎥ ⎝ 760 mmHg ⎠ ⎠ ⎝ ⎣ ⎦

5.58

Plan: To find mL of SO2, write the balanced equation, convert the given mass of P4S3 to moles, use the molar ratio to find moles of SO2, and use the ideal gas equation to find volume. Solution: P4S3(s) + 8 O2(g) → P4O10(s) + 3 SO2(g) ⎛ 1 mol P4S3 ⎞ ⎛ 3 mol SO 2 ⎞ Moles SO2 = ( 0.800 g P4S3 ) ⎜ ⎟⎜ ⎟ = 0.010905 mol SO2 ⎝ 220.09 g P4S3 ⎠ ⎝ 1 mol P4S3 ⎠ Liatm ⎞ ( 0.010905 mol SO2 ) ⎛ 0.0821 ( ( 273 + 32) K ) ⎛ 760 torr ⎞ ⎛ 1 mL ⎞ ⎜ nRT moliK ⎟ ⎝ ⎠ Volume SO2 = = ⎜ ⎟ ⎜ −3 ⎟ ⎜ ⎟ 725 torr P ⎝ 1 atm ⎠ ⎝ 10 L ⎠ = 286.249 = 286 mL SO2

5.59

CCl4(g) + 2 HF(g) → CF2Cl2(g) + 2 HCl(g) Moles Freon = n = PV / RT Mass CCl4 = mol Freon x mole ratio x molar mass of CCl4 = (PV / RT) x mole ratio x molar mass of CCl4 ⎡ ⎤ ⎢ (1.20 atm ) 16.0 dm3 ⎛ 1 L ⎞ ⎥ ⎛ 1 mol CCl4 ⎞ ⎛ 153.81 g CCl4 ⎞ ⎥ = ⎢ ⎜ 3 ⎟ ⎜ ⎜ ⎟ ⎥ 1 mol CF Cl ⎟ ⎜ 1 mol CCl ⎟ Liatm ⎞ ⎢⎛ 2 2 ⎠⎝ 4 ⎠ 0.0821 273 + 27 ) K ) ⎝ 1 dm ⎠ ⎥ ⎝ (( ⎢⎜ moliK ⎟ ⎠ ⎣⎝ ⎦ = 119.9006 = 1.20 x 102 g CCl4

(

)

5.60

Plan: First, write the balanced equation. Given the amount of xenon hexafluoride that reacts, we can find the number of moles of silicon tetrafluoride gas formed. Then, using the ideal gas equation with the moles of gas, the temperature and the volume, we can calculate the pressure of the silicon tetrafluoride gas. Solution: 2 XeF6(s) + SiO2(s) → 2 XeOF4(l) + SiF4(g) ⎛ 1 mol XeF6 ⎞⎛ 1 mol SiF4 ⎞ Mole SiF4 = n = ( 2.00 g XeF6 ) ⎜ ⎟ = 0.0040766 mol SiF4 ⎟⎜ ⎝ 245.3 g XeF6 ⎠⎝ 2 mol XeF6 ⎠ Liatm ⎞ ( 0.0040766 mol SiF4 ) ⎛ 0.0821 (( 273 + 25) K ) ⎜ nRT moliK ⎟ ⎝ ⎠ Pressure SiF4 = P = = 1.00 L V = 0.099737 = 0.0997 atm SiF4
PV = nRT. At constant T and P, V α n. Since the volume of the products has been decreased to ½ the original volume, the moles (and molecules) must have been decreased by a factor of ½ as well. Cylinder A best represents the products as there are 2 product molecules (there were 4 reactant molecules).

5.61

5-11

5.62

2 PbS(s) + 3 O2(g) → 2 PbO(g) + 2 SO2(g) ⎛ 103 g ⎞ ⎛ 1 mol PbS ⎞ ⎛ 2 mol SO 2 ⎞ Moles SO2 from PbS = ( 3.75 kg PbS ) ⎜ = 15.6707 mol SO2 (unrounded) ⎜ 1 kg ⎟ ⎜ 239.3 g PbS ⎟ ⎜ 2 mol PbS ⎟ ⎟ ⎠ ⎠⎝ ⎝ ⎠⎝

( 2.0 atm )( 228 L ) PV = = 11.266 mol O2 Liatm ⎞ RT ⎛ ⎜ 0.0821 moliK ⎟ ( ( 273 + 220 ) K ) ⎝ ⎠ ⎛ 2 mol SO 2 ⎞ Moles SO2 from O2 = (11.266 mol O 2 ) ⎜ ⎟ = 7.5107mol SO2 ⎝ 3 mol O 2 ⎠ O2 is the limiting reagent. ⎛ 2 mol SO 2 ⎞ ⎛ 22.4 L ⎞ 2 Volume SO2 = (11.266 mol O 2 ) ⎜ ⎟⎜ ⎟ = 168.23 = 1.7 x 10 L SO2 3 mol O 2 ⎠ ⎝ 1 mol SO 2 ⎠ ⎝
Moles of O2 = 5.63 2 HgO(s) → 2 Hg(l) + O2(g) Moles O2 = n = amount of HgO x 1/molar mass HgO x mole ratio Pressure O2 = P = nRT / V = (amount of HgO x 1/molar mass HgO x mole ratio)RT / V ⎡ Liatm ⎞ ⎛ 20.0% ⎞ ⎛ 1 mol HgO ⎞⎛ 1 mol O2 ⎞⎤ ⎛ ⎢( 40.0 g HgO ) ⎜ ⎟ ⎜ 216.6 g HgO ⎟⎜ 2 mol HgO ⎟⎥ ⎜ 0.0821 moliK ⎟ ( ( 273 + 25.0 ) K ) ⎛ 1 mL ⎞ ⎝ 100% ⎠ ⎝ ⎠ ⎢ ⎠⎝ ⎠⎥ ⎝ ⎦ P= ⎣ ⎜ −3 ⎟ ⎜ 10 L ⎟ 502 mL ⎝ ⎠ = 0.9000305 = 0.900 atm O2 5.64 As the temperature of the gas sample increases, the most probable speed increases. This will increase both the number of collisions per unit time and the force of each collision with the sample walls. Thus, the gas pressure increases. At STP (or any identical temperature and pressure), the volume occupied by a mole of any gas will be identical. This is because at the same temperature, all gases have the same average kinetic energy, resulting in the same pressure. The rate of effusion is much higher for a gas than its rate of diffusion. Effusion occurs into an evacuated space, whereas diffusion occurs into another gas. It is reasonable to expect that a gas will escape faster into a vacuum than it will into a space already occupied by another gas. The ratio of the rates of effusion and diffusion for two gases will be the same since both are inversely proportional to the square root of their molar masses. a) Mass O2 > mass H2 b) d O > d H c) Identical d) Identical e) Average speed H2 > average speed O2 f) Effusion time H2 < effusion time O2
2 2

5.65

5.66

5.67

5.68

Plan: The molar masses of the three gases are 2.016 for H2 (Flask A), 4.003 for He (Flask B), and 16.04 for CH4 (Flask C). Since hydrogen has the smallest molar mass of the three gases, 4 g of H2 will contain more gas molecules (about 2 mole’s worth) than 4 g of He or 4 g of CH4. Since helium has a smaller molar mass than methane, 4 g of He will contain more gas molecules (about 1 mole’s worth) than 4 g of CH4 (about 0.25 mole’s worth). Solution: a) PA > PB > PC The pressure of a gas is proportional to the number of gas molecules. So, the gas sample with more gas molecules will have a greater pressure. b) EA = EB = EC Average kinetic energy depends only on temperature. The temperature of each gas sample is 273 K, so they all have the same average kinetic energy.

5-12

c) rateA > rateB > rateC When comparing the speed of two gas molecules, the one with the lower mass travels faster. d) total EA > total EB > total EC Since the average kinetic energy for each gas is the same (part b of this problem) then the total kinetic energy would equal the average times the number of molecules. Since the hydrogen flask contains the most molecules, its total kinetic energy will be the greatest. e) dA = dB = dC Under the conditions stated in this problem, each sample has the same volume, 5 L, and the same mass, 4 g. Thus, the density of each is 4 g/5 L = 0.8 g/L. f) Collision frequency (A) > collision frequency (B) > collision frequency (C) The number of collisions depends on both the speed and the distance between gas molecules. Since hydrogen is the lightest molecule it has the greatest speed and the 5 L flask of hydrogen also contains the most molecules, so collisions will occur more frequently between hydrogen molecules than between helium molecules. By the same reasoning, collisions will occur more frequently between helium molecules than between methane molecules. 5.69 Plan: To find the ratio of effusion rates, calculate the inverse of the ratio of the square roots of the molar masses (Graham’s Law). Solution: Molar Mass UF6 Rate H 2 352.0 g/mol = = 13.2137 = 13.21 = Molar Mass H 2 2.016 g/mol Rate UF6 To find the ratio of effusion rates, calculate the inverse of the ratio of the square roots of the molar masses . Rate O 2 Molar Mass Kr 83.80 g/mol = = = 1.618255 = 1.618 Molar Mass O 2 32.00 g/mol Rate Kr Plan: Recall that the heavier the gas, the slower the molecular speed. The molar mass of Ar is 39.95 g/mol while the molar mass of He is 4.003 g/mol. Solution: a) The gases have the same average kinetic energy because they are at the same temperature. The heavier Ar atoms are moving slower than the lighter He atoms to maintain the same average kinetic energy. Therefore, Curve 1 better represents the behavior of Ar. b) A gas that has a slower molecular speed would effuse more slowly, so Curve 1 is the better choice. c) Fluorine gas exists as a diatomic molecule, F2, with M = 38.00 g/mol. Therefore, F2 is much closer in mass to Ar (39.95 g/mol) than He (4.003 g/mol), so Curve 1 more closely represents F2’s behavior. a) Curve 1 b) Curve 2 c) Curve 2 Plan: To find the ratio of effusion rates, calculate the inverse of the ratio of the square roots of the molar masses (Graham’s Law). Then use the ratio of effusion rates to find the time for the F2 effusion. Solution: Molar Mass F2 38.00 g/mol Rate He = = 3.08105 (unrounded) = Molar Mass He 4.003 g/mol Rate F2
Time F2 Rate He = Time He Rate F2

5.70

5.71

5.72

5.73

Time F2 3.08105 = 1.00 4.85 min He

Time F2 = 14.94309 = 14.9 min

5-13

5.74

Rate H 2 Time Unk = = Rate Unknown Time H 2 11.1 min = 2.42 min Molar Mass Unk 2.016 g/mol Molar Mass Unk 2.016 g/mol

Molar Mass Unk Molar Mass H 2

4.586777 =

Molar Mass unknown = 42.41366 = 42.4 g/mol 5.75 Plan: White phosphorus is a molecular form of the element phosphorus consisting of some number, x, of phosphorus atoms. Use the relative rates of effusion of white phosphorus and neon to determine the molar mass of white phosphorous. From the molar mass of white phosphorus, determine the number of phosphorus atoms, x, in one molecule of white phosphorus. Solution: Rate Px Molar Mass Ne = 0.404 = Molar Mass Px Rate Ne 0.404 =
20.18 g/mol Molar Mass Px 20.18 g/mol Molar Mass Px

(0.404)2 = 0.163216 =

20.18 g/mol Molar Mass Px Molar Mass Px = 123.6398 g/mol ⎛ 123.6398 g ⎞ ⎛ 1 mol P ⎞ ⎜ ⎟⎜ ⎟ = 3.992244 = 4 mol P/mol Px or 4 atoms P/molecule Px ⎝ mol Px ⎠ ⎝ 30.97 g P ⎠

Thus, 4 atoms per molecule, so Px = P4. 5.76 a) 0°C = 273 K urms He (at 0°C) =

30°C = 303 K

(4.003 g He/mol) (1 kg / 103 g) = 0.004003 kg/mol
⎞ 3 3 ⎟ = 1.3042 x 10 = 1.30 x 10 m /s ⎟ ⎠

J ⎞ ⎛ 3 ⎜ 8.314 ( 273 K ) ⎛ kgim 2 /s2 moliK ⎟ ⎝ ⎠ ⎜ ⎜ 0.004003 kg/mol J ⎝

urms He (at 30°C) =

J ⎞ ⎛ 3 ⎜ 8.314 ( 303 K ) ⎛ kgim 2 /s2 moliK ⎟ ⎝ ⎠ ⎜ ⎜ 0.004003 kg/mol J ⎝ J ⎞ ⎛ 3 ⎜ 8.314 ( 303 K ) ⎛ kgim 2 /s2 moliK ⎟ ⎝ ⎠ ⎜ ⎜ 0.1313 kg/mol J ⎝

⎞ 3 3 ⎟ = 1.3740 x 10 = 1.37 x 10 m /s ⎟ ⎠

b) (131.3 g Xe/mol) (1 kg / 103 g) = 0.1313 kg/mol
⎞ ⎟ = 239.913 m/s (unrounded) ⎟ ⎠ Rate He / Rate Xe = (1.3740 x 103 m/s) / (239.913 m/s) = 5.727076 = 5.73 He molecules travel at almost 6 times the speed of Xe molecules. c) EHe = 1/2 mv2 = (1/2) (0.004003 kg/mol) (1.3740 x 103 m/s)2(1J/(kg•m2/s2) = 3778.58 = 3.78 x 103 J/mol EXe = 1/2 mv2 = (1/2) (0.1313 kg/mol) (239.913 m/s)2(1J/(kg•m2/s2) = 3778.69 = 3.78 x 103 J/mol d) (3778.58 J/mol) (1 mol He / 6.022 x 1023 atoms He) = 6.2746 x 10–21 = 6.27 x 10–21 J/He atom

urms Xe (at 30°C) =

5-14

5.77

a) S2F2 = 102.14 g/mol; N2F4 = 104.02 g/mol; SF4 = 108.07 g/mol. rateSF4 < rate N2 F4 < rateS 2F2 When comparing the speed of gas molecules, the one with the lowest mass travels the fastest. rateS2 F2 Molar Mass N 2 F4 104.02 g/mol = = 1.009161 = 1.0092:1 b) = rate N 2 F4 Molar Mass S2 F2 102.14 g/mol c)
Rate X = 0.935 = Rate SF4 108.07 g/mol Molar Mass X 108.07 g/mol Molar Mass X Molar Mass SF4 Molar Mass X

0.935 =

(0.935)2 = 0.874225 =

108.07 g/mol Molar Mass X Molar Mass X = 123.61806 = 124 g/mol

5.78

Intermolecular attractions cause the real pressure to be less than ideal pressure, so it causes a negative deviation. The size of the intermolecular attraction is related to the constant a. According to Table 5.5, a N = 1.39, a Kr = 2.32
2

and a CO = 3.59. Therefore, CO2 experiences a greater negative deviation in pressure than the other two gases: N2
2

< Kr < CO2.

5.79

Molecular size causes a positive deviation from ideal behavior. Thus, VReal Gases > VIdeal Gases. The molecular volume is related to the constant b. According to Table 5.5, bH = 0.0266, bO = 0.0318and
2 2

bCl = 0.0562. Therefore, the order is H2 < O2 < Cl2.
2

5.80

Nitrogen gas behaves more ideally at 1 atm than at 500 atm because at lower pressures the gas molecules are farther apart. An ideal gas is defined as consisting of gas molecules that act independently of the other gas molecules. When gas molecules are far apart, they act ideally, because intermolecular attractions are less important and the volume of the molecules is a smaller fraction of the container volume.
At 150°C. At higher temperatures, intermolecular attractions become less important and the volume occupied by the molecules becomes less important.

5.81 5.82

Do not forget to multiply the area of each side by two. Surface area of can = 2 (40.0 cm) (15.0 cm) + 2 (40.0 cm) (12.5 cm) + 2 (15.0 cm) (12.5 cm) = 2.575 x 103 cm2 (unrounded) Total force = (2.575 x 103 cm2) (1 in/2.54 cm)2(14.7 lb/in2) = 5.8671 x 103 = 5.87 x 103 lbs Plan: Use the Ideal Gas Equation to find the number of moles of O2. Moles of O2 combine with Hb in a 4:1 ratio. Divide the mass of Hb by the number of moles to obtain molar mass, g/mol. Solution: PV = nRT ⎛ 1 atm ⎞ ⎛ 10−3 L ⎞ ( 743 torr )(1.53 mL ) PV = Moles O2 = n = ⎟ ⎜ ⎟⎜ ⎜ ⎟ Liatm ⎞ RT ⎛ 0.0821 ( ( 273 + 37 ) K ) ⎝ 760 torr ⎠ ⎝ 1 mL ⎠ ⎜ ⎟ moliK ⎠ ⎝ = 5.87708 x 10–5 mol O2 (unrounded) ⎛ 1 mol Hb ⎞ –5 Moles Hb = 5.87708 x 10-5 mol O 2 ⎜ ⎟ = 1.46927 x 10 mol Hb (unrounded) ⎝ 4 mol O 2 ⎠ Molar mass hemoglobin = (1.00 g Hb) / (1.46927 x 10–5 mol Hb) = 6.806098 x 104 = 6.81 x 104 g/mol

5.83

(

)

5-15

5.84

Reaction 1: 2 NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g) Moles CO2 = (1.00 g NaHCO3) (1 mol NaHCO3 / 84.01 g NaHCO3) (1 mol CO2 /2 mol NaHCO3) = 5.95167 x 10–3 mol CO2 (unrounded) Liatm ⎞ ⎛ 5.95167 x10−3 mol ⎜ 0.0821 ( ( 273 + 200.) K ) ⎛ 1 mL ⎞ moliK ⎟ ⎝ ⎠ V = nRT / P = ⎜ −3 ⎟ ⎜ 10 L ⎟ 0.975 atm ⎝ ⎠

(

)

= 237.049 = 237 mL CO2 Reaction 1 Reaction 2: NaHCO3(s) + H+(aq) → H2O(l) + CO2(g) + Na+(aq) Moles CO2 = (1.00 g NaHCO3) (1 mol NaHCO3 / 84.01 g NaHCO3) (1 mol CO2 /1 mol NaHCO3) = 1.1903 x 10–2 mol CO2 (unrounded) Liatm ⎞ ⎛ 1.1903 x 10−2 mol ⎜ 0.0821 ( ( 273 + 200.) K ) ⎛ 1 mL ⎞ moliK ⎟ ⎝ ⎠ V = nRT / P = ⎜ −3 ⎟ ⎜ 10 L ⎟ 0.975 atm ⎝ ⎠ = 474.0986 = 474 mL CO2 Reaction 2

(

)

5.85

Rearrange PV = nRT to R = PV / nT This gives: PiVi / niTi = PfVf / nfTf Pi = 1.01 atm Pf = 1.01 atm (Thus, P has no effect, and does not need to be included.) Ti = 305 K Tf = 250 K n i = ni nf = 0.75 ni Vi = 600. L Vf = ? ( 0.75 ni ( 250 K )( 600. L ) ) = 368.852 = 369 L Vf = (nfTfVi) / (niTi) = ( n i )( 305 K ) Plan: Convert the mass of Cl2 to moles and use the ideal gas law and van der Waals equation to find the pressure of the gas. Solution: ⎛ 103 g ⎞ ⎛ 1 mol Cl2 ⎞ a) Moles Cl2: ( 0.5950 kg Cl2 ) ⎜ ⎟ = 8.3921016 mol ⎟⎜ ⎝ 1 kg ⎠ ⎝ 70.90 g Cl 2 ⎠ Ideal gas equation: PV = nRT Liatm ⎞ ⎛ 8.3921016 mol ⎜ 0.0821 ( ( 273 + 225) K ) moliK ⎟ ⎝ ⎠ PIGL = nRT/V = 15.50 L = 22.1366 = 22.1 atm ⎛ n 2a ⎞ b) ⎜ P + 2 ⎟ ( V − nb ) = nRT ⎜ V ⎟ ⎝ ⎠ PVDW = nRT n 2a atmiL2 From Table 5.5: a = 6.49 ; − 2 V − nb V mol2 n = 8.3921016 mol from Part (a)

5.86

b = 0.0562

L mol
2

atmiL 2⎛ Liatm ⎞ ⎛ mol Cl2 ) ⎜ 0.08206 ( ( 273 + 225) K ) (8.3921016 mol Cl2 ) ⎜ 6.49 mol2 ⎟ ⎜ moliK ⎠ ⎝ ⎝ − PVDW = 2 L ⎞ ⎛ (15.50 L ) 15.50 L − ( 8.3921016 mol Cl2 ) ⎜ 0.0562 mol ⎟ ⎝ ⎠ = 20.91773 = 20.9 atm

(8.3921016

⎞ ⎟ ⎟ ⎠

5-16

5.87

a) Molar Mass I = mRT / PV =

( 0.1000 g ) ⎛ 0.08206 ⎜

Liatm ⎞ ( ( 273.15 + 70.00 ) K ) ⎛ 1 mL ⎞ moliK ⎟ ⎝ ⎠ ⎜ −3 ⎟ ⎜ 10 L ⎟ ( 0.05951 atm )( 750.0 mL ) ⎝ ⎠

= 63.0905 = 63.09 g I / mol Molar Mass II = mRT / PV =

( 0.1000 g ) ⎛ 0.08206 ⎜


Liatm ⎞ ( ( 273.15 + 70.00 ) K ) ⎛ 1 mL ⎞ moliK ⎟ ⎠ ⎜ −3 ⎟ ⎜ 10 L ⎟ ( 0.07045 atm )( 750.0 mL ) ⎝ ⎠

= 53.293 = 53.29 g II / mol Molar Mass III = mRT / PV =

( 0.1000 g ) ⎛ 0.08206 ⎜

Liatm ⎞ ( ( 273.15 + 70.00 ) K ) ⎛ 1 mL ⎞ moliK ⎟ ⎝ ⎠ ⎜ −3 ⎟ ⎜ 10 L ⎟ ( 0.05767 atm )( 750.0 mL ) ⎝ ⎠

= 65.10349 = 65.10 g III / mol b) % H in I = 100% – 85.63% = 14.37% H % H in II = 100% – 81.10% = 18.90% H % H in III = 100% – 82.98% = 17.02% H Assume 100 grams of each so the mass percentages are also the grams of the element. I (85.63 g B) (1 mol B / 10.81 g B) = 7.921369 mol B (unrounded) (14.37 g H) (1 mol H / 1.008 g H) = 14.25595 mol H (unrounded) Dividing by the smaller value (7.921369) gives B = 1 and H = 1.7997. The hydrogen value is not close enough to a whole number to round. Thus, both amounts need to be multiplied by the smallest value to get near whole numbers. This value is 5. Multiplying by 5 gives B = 5 and H = 9. The empirical formula is B5H9, which has a formula mass of 63.12 g/mol. The empirical formula mass is near the molecular mass from part a. Therefore, the empirical and molecular formulas are both B5H9. II (81.10 g B) (1 mol B / 10.81 g B) = 7.50231 mol B (unrounded) (18.90 g H) (1 mol H / 1.008 g H) = 18.750 mol H (unrounded) Dividing by the smaller value (7.50231) gives B = 1 and H = 2.4992. The hydrogen value is not close enough to a whole number to round. Thus, both amounts need to be multiplied by the smallest value to get near whole numbers. This value is 2. Multiplying by 2 gives B = 2 and H = 5. The empirical formula is B2H5, which has a formula mass of 26.66 g/mol. Dividing the molecular formula mass by the empirical formula mass gives the relationship between the formulas: (53.29) / (26.66) = 2. The molecular formula is two times the empirical formula, or B4H10. III (82.98 g B) (1 mol B / 10.81 g B) = 7.6762 mol B (unrounded) (17.02 g H) (1 mol H / 1.008 g H) = 16.8849 mol H (unrounded) Dividing by the smaller value (7.6762) gives B = 1 and H = 2.2. The hydrogen value is not close enough to a whole number to round. Thus, both amounts need to be multiplied by the smallest value to get near whole numbers. This value is 5. Multiplying by 5 gives B = 5 and H = 11. The empirical formula is B5H11, which has a formula mass of 65.14 g/mol. The empirical formula mass is near the molecular mass from part a. Therefore, the empirical and molecular formulas are both B5H11. Rate SO 2 Molar Mass IV c) = Molar Mass SO 2 Rate IV
⎛ 250.0 mL ⎞ ⎜ ⎟ ⎝ 13.04 min ⎠
Molar Mass Unk 64.07 g/mol

⎛ 350.0 mL ⎞ ⎜ ⎟ ⎝ 12.00 min ⎠ Molar Mass IV = 27.6825 = 27.68 g/mol

= 0.657318 =

5-17

IV % H in IV = 100% – 78.14% = 21.86% H Assume 100 grams of each so the mass percentages are also the grams of the element. (78.14 g B) (1 mol B / 10.81 g B) = 7.22849 mol B (unrounded) (21.86 g H) (1 mol H / 1.008 g H) = 21.6865 mol H (unrounded) Dividing by the smaller value (7.22849) gives B = 1 and H = 3.000. The empirical formula is BH3, which has a formula mass of 13.83 g/mol. Dividing the molecular formula mass by the empirical formula mass gives the relationship between the formulas: (27.68) / (13.83) = 2. The molecular formula is two times the empirical formula, or B2H6. 5.88 a) I. XA =
3 A particles 4 A particles 5 A particles = 0.33; II. XA = = 0.33; III. XA = = 0.33 9 total particles 12 total particles 15 total particles The partial pressure of A is the same in all 3 samples since the mole fraction of A is the same in all samples. 3 B particles 3 B particles 3 B particles b) I. XB = = 0.33; II. XB = = 0.25; III. XB = = 0.20 9 total particles 12 total particles 15 total particles The partial pressure of B is the same is lowest in Sample III since the mole fraction of B is the smallest in that Sample. c) All samples are at the same temperature, T, so all have the same average kinetic energy.

5.89

a) Vf = Vi(Tf / Ti) (Pi / Pf) Vf = Vi((273+100K) / (273+200K) (2 atm / 1 atm) Vf = Vi((373K) / (473K) (2 atm / 1 atm) = 1.577 Vi (unrounded) b) Vf = Vi(Tf / Ti) (Pi / Pf) Vf = Vi((573K) / (373K) (1 atm / 3 atm) = 0.51206 Vi (unrounded) c) Vf = Vi(Tf / Ti) (Pi / Pf) Vf = Vi((400K) / (200K) (3 atm / 6 atm) = Vi d) Vf = Vi(Tf / Ti) (Pi / Pf) Vf = Vi((423K) / (573K) (0.2 atm / 0.4 atm) = 0.3691 Vi (unrounded)

Increase Decrease Unchanged Decrease

5.90

Plan: Partial pressures and mole fractions are calculated from Dalton’s Law of Partial Pressures: PA = XA(Ptotal) Remember that 1 atm = 760 torr. Solution: a) Convert each mole percent to a mole fraction by dividing by 100%. ⎛ 760 torr ⎞ PNitrogen = XNitrogen PTotal = 0.786 (1.00 atm ) ⎜ ⎟ = 597.36 = 597 torr N2 ⎝ 1 atm ⎠
⎛ 760 torr ⎞ POxygen = XOxygen PTotal = 0.209 (1.00 atm ) ⎜ ⎟ = 158.84 = 159 torr O2 ⎝ 1 atm ⎠

⎛ 760 torr ⎞ PCarbon Dioxide = XCarbon Dioxide PTotal = 4.00 x 10-4 (1.00 atm ) ⎜ ⎟ = 0.304 = 0.3 torr CO2 ⎝ 1 atm ⎠
⎛ 760 torr ⎞ PWater = XWater PTotal = 0.0046 (1.00 atm ) ⎜ ⎟ = 3.496 = 3.5 torr O2 ⎝ 1 atm ⎠ b) Mole fractions can be calculated by rearranging Dalton’s Law of Partial Pressures: XA = PA/Ptotal and multiply by 100 to express mole fraction as percent. PTotal = (569 + 104 + 40 + 47) torr = 760 torr N2: [(569 torr) / (760 torr)] x 100% = 74.8684 = 74.9 mol% N2 O2: [(104 torr) / (760 torr)] x 100% = 13.6842 = 13.7 mol% O2 CO2: [(40 torr) / (760 torr)] x 100% = 5.263 = 5.3 mol% CO2 H2O: [(47 torr) / (760 torr)] x 100% = 6.1842 = 6.2 mol% CO2

5-18

c) Number of molecules of O2 can be calculated using the Ideal Gas Equation and Avogadro’s number. PV = nRT ⎛ 1 atm ⎞ (104 torr )( 0.50 L ) PV = Moles O2 = ⎜ ⎟ = 0.0026883 mol O2 Liatm ⎞ 760 torr ⎠ RT ⎛ 0.0821 273 + 37 ) K ) ⎝ (( ⎜ moliK ⎟ ⎝ ⎠ ⎛ 6.022 x 1023 molecules O 2 ⎞ Molecules O2 = ( 0.0026883 mol O 2 ) ⎜ ⎟ 1 mol O 2 ⎝ ⎠ = 1.6189 x 1021 = 1.6 x 1021 molecules O2
⎛ ⎛ 1.373 x 104 Rn atoms ⎞ ⎞ ⎜⎜ ⎟⎟ ⎛ 1 mol Ra ⎞ ⎛ 6.022 x 1023 Ra atoms ⎞ ⎜ ⎜ 1.0 x 1015 Ra atoms ⎟ ⎟ ⎛ 3600 s ⎞ ⎛ 24 h ⎞ ⎝ ⎠ Atoms Rn = (1.0 g Ra ) ⎜ ⎟⎜ ⎟⎜ ⎟ ⎟⎜ ⎟⎜ ⎜ ⎟ 1 mol Ra s ⎝ 226 g Ra ⎠ ⎝ ⎠⎜ ⎟ ⎝ h ⎠ ⎝ day ⎠ ⎜ ⎟ ⎝ ⎠ = 3.16094 x 1015 Rn atoms / day (unrounded) ⎡ ⎛ ⎞⎤ ⎛ 1 mol Rn Liatm ⎞ 15 ⎢ 3.16094 x 10 Rn atoms ⎜ ( 273 K ) ⎜ ⎟ ⎥ ⎜ 0.0821 23 ⎟ ⎝ moliK ⎟ ⎠ ⎢ ⎝ 6.022 x 10 Rn atoms ⎠ ⎥ ⎣ ⎦ V = nRT / P = (1.00 atm )

5.91

(

)

= 1.17647 x 10–7 = 1.2 x 10–7 L Rn 5.92 Plan: For part a, since the volume, temperature, and pressure of the gas are changing, use the combined gas law. For part b, use the ideal gas equation to solve for moles of air and then moles of N2. Solution: PV P V ⎛ 1 atm ⎞ a) 1 1 = 2 2 P1 = (1450. mmHg ) ⎜ ⎟ = 1.9079 atm; V1 = 208 mL; T1 T2 ⎝ 760 mmHg ⎠ T2 = 298 K; V2 = ? T1 = 286 K; P2 = 1 atm; ⎛T ⎞⎛ P ⎞ ⎛ 298 K ⎞ ⎛ 1.9079 atm ⎞ 2 V2 = V1 ⎜ 2 ⎟ ⎜ 1 ⎟ = ( 208 mL ) ⎜ ⎟⎜ ⎟ = 4.13494 mL = 4 x 10 mL ⎜ T ⎟⎜ P ⎟ ⎝ 286 K ⎠ ⎝ 1 atm ⎠ ⎝ 1 ⎠⎝ 2 ⎠ b) Mole air = n =

(1.9079 atm )( 0.208 L ) PV = = 0.016901 mol air Liatm ⎞ RT ⎛ 0.0821 ( 286 K ) ⎜ moliK ⎟ ⎝ ⎠ ⎛ 77% N 2 ⎞ Mole N2 = ( 0.016901 mol ) ⎜ ⎟ = 0.01301 = 0.013 mol N2 ⎝ 100% ⎠

5.93

a) Moles = (5.14 x 1015 t) (1000 kg / t) (103 g / kg) (1 mol / 28.8 g) = 1.78472 x 1020 = 1.78 x 1020 mol gas Liatm ⎞ ⎛ 1.78472 x 1020 mol ⎜ 0.0821 ( ( 273 + 25) K ) moliK ⎟ ⎝ ⎠ = 4.36646 x 1021 = 4 x 1021 L b) V = nRT / P = 1 atm ) (

(

)

( 4.36646 x 10 L ) ⎛ 10 m c) Height = ⎜ ⎝ (5.100 x 10 km ) ⎜ 1 L
21
−3

3

8

2

⎞ ⎛ 1 km ⎞ ⎟ ⎜ 3 ⎟ = 8.561686 = 9 km ⎟ ⎜ 10 m ⎟ ⎠ ⎠⎝

3

The pressure exerted by the atmosphere is 1 atm only at the surface of the earth. Above the earth’s surface, the pressure is lower than this value, so the air expands and extends much further up than 9 km.

5-19

5.94

Plan: The balanced equation and reactant amounts are given, so the first step is to identify the limiting reactant. Then use the limiting reactant and molar ratios to find the moles of NO2 produced. The volume of NO2 is found from the Ideal Gas Equation. Solution: Cu(s) + 4 HNO3(aq) → Cu(NO3)2(aq) + 2 NO2(g) + 2 H2O(l)
⎛ 8.95 g Cu ⎞ ⎛ 1 mol Cu ⎞ ⎛ 2 mol NO 2 ⎞ Moles NO2 from Cu = 4.95 cm3 ⎜ ⎟⎜ ⎟ = 1.394256 mol NO2 (unrounded) ⎟⎜ ⎝ cm3 ⎠ ⎝ 63.55 g Cu ⎠ ⎝ 1 mol Cu ⎠

(

)

⎛ 68.0% HNO3 ⎞ ⎛ 1 cm3 ⎞ ⎛ 1.42 g ⎞ ⎛ 1 mol HNO3 ⎞ ⎛ 2 mol NO 2 ⎞ Moles NO2 from HNO3 = ( 230.0 mL ) ⎜ ⎜ ⎟ ⎟ ⎟⎜ ⎟ ⎜ 1 mL ⎟ ⎜ 3 ⎟⎜ 100% ⎝ ⎠⎝ ⎠ ⎝ cm ⎠ ⎝ 63.02 g ⎠⎝ 4 mol HNO3 ⎠ = 1.7620 mol NO2 (unrounded) Since less product can be made from the copper, it is the limiting reactant and excess nitric acid will be left after the reaction goes to completion. Use the calculated number of moles of NO2 and the given temperature and pressure in the ideal gas equation to find the volume of nitrogen dioxide produced. Note that nitrogen dioxide is the only gas involved in the reaction. Liatm ⎞ (1.394256 mol NO2 ) ⎛ 0.0821 ( ( 273.2 + 28.2 ) K ) ⎛ 760 torr ⎞ ⎜ nRT moliK ⎟ ⎝ ⎠ = V= ⎜ ⎟ P ( 735 torr ) ⎝ 1 atm ⎠

= 35.67428 = 35.7 L NO2 5.95 a) n = PV / RT =
⎛ 10−3 L ⎞ = 0.047149 = 0.047 mol air ⎜ ⎜ 1 mL ⎟ ⎟ Liatm ⎞ ⎛ ⎝ ⎠ ⎜ 0.0821 moliK ⎟ ( ( 273 + 37 ) K ) ⎝ ⎠ b) Molecules = (0.047149 mol air) (6.022 x 1023 molecules air / mol air) = 2.83931 x 1022 = 2.8 x 1022 molecules air

(1.0 atm )(1200 mL )

5.96

5 NaBr(aq) + NaBrO3(aq) + 3 H2SO4(aq) → 3 Br2(g) + 3 Na2SO4(aq) + 3 H2O(g) ⎛ 1 mol NaBr ⎞⎛ 3 mol Br2 ⎞ Moles Br2 from NaBr = ( 275 g NaBr ) ⎜ ⎟⎜ ⎟ = 1.60365 mol Br2 (unrounded) ⎝ 102.89 g NaBr ⎠⎝ 5 mol NaBr ⎠
⎛ 1 mol NaBrO3 ⎞ ⎛ 3 mol Br2 ⎞ Moles Br2 from NaBrO3 = (175.6 g NaBrO3 ) ⎜ ⎟ ⎟⎜ ⎝ 150.89 g NaBrO3 ⎠ ⎝ 1 mol NaBrO3 ⎠ = 3.491285 mol Br2 (unrounded) The NaBr is limiting. Liatm ⎞ (1.60365 mol ) ⎛ 0.0821 ( ( 273 + 300 ) K ) ⎜ moliK ⎟ ⎝ ⎠ Volume = nRT / P = = 88.235 = 88.2 L Br2 ( 0.855 atm )

5.97

The reaction is: 2 NaN3(s) → 2 Na(s) + 3 N2(g) The vapor pressure of water (Table 5.3) at 26°C is 25.2 torr (= 25.2 mmHg). Volume = nRT/ P = ⎡ ⎛ 1 mol NaN3 ⎞⎛ 3 mol N 2 ⎞ ⎤ ⎛ Liatm ⎞ ( ( 273 + 26 ) K ) ⎢50.0 g NaN3 ⎜ ⎟⎜ ⎟ ⎥ ⎜ 0.0821 moliK ⎟ ⎠ ⎛ 760 mmHg ⎞ ⎢ ⎝ 65.02 g NaN3 ⎠⎝ 2 mol NaN3 ⎠ ⎥ ⎝ ⎣ ⎦ ⎜ ⎟ ( ( 745.5 − 25.2) mmHg ) ⎝ 1 atm ⎠ = 29.8764 = 29.9 L N2

5-20

5.98

It is necessary to determine both the empirical formula and the molar mass of the gas. Empirical formula: Assume 100.0 grams of sample to make the percentages of each element equal to the grams of that element. Thus, 64.81% C = 64.81 g C, 13.60% H = 13.60 g H, and 21.59% O = 21.59 g O. Moles C = (64.81 g C) (1 mol C / 12.01 g C) = 5.39634 mol C (unrounded) Moles H = (13.60 g H) (1 mol H / 1.008 g H) = 13.49206 mol H (unrounded) Moles O = (21.59 g O) (1 mol O / 16.00 g O) = 1.349375 mol O (unrounded) Divide by the smallest number of moles (O): C = (5.39634 mol) / (1.349375 mol) = 4 H = (13.49206 mol) / (1.349375 mol) = 10 O = (1.349375 mol) / (1.349375 mol) = 1 Empirical formula = C4H10O (empirical formula mass = 74.12 g/mol) Molar Mass: Liatm ⎞ ( 2.57 g ) ⎛ 0.0821 ( ( 273 + 25) K ) ⎜ moliK ⎟ ⎝ ⎠ M = mRT / PV = = 74.85 g/mol (unrounded) ( 0.420 atm )( 2.00 L ) Molecular formula: Since the molar mass and the empirical formula mass are similar, the empirical and molecular formulas must both be: C4H10O

5.99

Plan: The empirical formula for aluminum chloride is AlCl3 (Al3+ and Cl–). The empirical formula mass is (133.33 g/mol). Calculate the molar mass of the gaseous species from the ratio of effusion rates. This molar mass, divided by the empirical weight, should give a whole number multiple that will yield the molecular formula. Solution: Molar Mass He Rate Unk = 0.122 = Molar Mass Unk Rate He 0.122 =
4.003 g/mol Molar Mass Unk

Molar mass Unknown = 268.9465 g/mol The whole number multiple is 268.9465/133.33, which is about 2. Therefore, the molecular formula of the gaseous species is 2 x (AlCl3) = Al2Cl6. 5.100 Four moles of gas (NH3, CO, N2 and HCNO are formed from the decomposition of 1 mole of azodicarbonamide. Two of those moles of gas, NH3 and HCNO, further react to form solid nonvolatile polymers. So the decomposition of azodicarbonamide leads to the overall formation of two moles of gas. ⎛ 1 mol C2 H 4 N 4 O 2 ⎞ ⎛ 2 moles of gas ⎞ Moles of gas formed = (1.00 g C2 H 4 N 4 O 2 ) ⎜ ⎟ = 0.017228 mol ⎟⎜ ⎝ 116.09 g C2 H 4 N 4 O 2 ⎠ ⎝ 1 mol C2 H 4 N 4 O 2 ⎠ Liatm ⎞ ( 0.017228 mol ) ⎛ 0.0821 ( 273 K ) 1 mL ⎜ moliK ⎟ ⎞ ⎛ ⎝ ⎠ Volume = nRT / P = ⎜ -3 ⎟ = 386.136 = 386 mL gas (1.00 atm ) ⎝ 10 L ⎠ a) 2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O (g) ⎛ 1 mol C8 H18 ⎞ ⎛ 34 mol gas ⎞ Moles products = (100. g C8 H18 ) ⎜ ⎟ = 14.883558 mol gas (unrounded) ⎟⎜ ⎝ 114.22 g C8 H18 ⎠ ⎝ 2 mol C8 H18 ⎠ Liatm ⎞ (14.883558 mol gas ) ⎛ 0.0821 ( ( 273 + 350 ) K ) ⎛ 760 torr ⎞ ⎜ moliK ⎟ ⎝ ⎠ V = nRT / P = ⎜ ⎟ ( 735 torr ) ⎝ 1 atm ⎠ = 787.162 = 787 L gas

5.101

5-21

⎛ 1 mol C8 H18 ⎞ ⎛ 25 mol O 2 ⎞ b) Moles O2 = (100. g C8 H18 ) ⎜ ⎟ = 10.94379 mol O2 (unrounded) ⎟⎜ ⎝ 114.22 g C8 H18 ⎠ ⎝ 2 mol C8 H18 ⎠ Moles other gases = (10.94379 mol O2) [(78 + 1.0)%] / (21%) = 41.1695 mol gas (unrounded) Liatm ⎞ ( 41.1695 mol gas ) ⎛ 0.0821 ( ( 273 + 350 ) K ) ⎛ 760 torr ⎞ ⎜ moliK ⎟ ⎝ ⎠ V = nRT / P = ⎜ ⎟ ( 735 torr ) ⎝ 1 atm ⎠

= 2177.37 L residual air (unrounded) Total volume of gaseous exhaust = 787.162 L + 2177.37 L = 2964.53 = 2.96 x 103 L 5.102 Plan: First, write the balanced equation for the reaction: 2 SO2 + O2 → 2 SO3. The total number of moles of gas will change as the reaction occurs since 3 moles of reactant gas forms 2 moles of product gas. From the volume, temperature, and pressures given, we can calculate the number of moles of gas before and after the reaction using ideal gas equation. For each mole of SO3 formed, the total number of moles of gas decreases by 1/2 mole. Thus, twice the decrease in moles of gas equals the moles of SO3 formed. Solution: Moles of gas before and after reaction: (1.90 atm )( 2.00 L ) PV = = 0.05785627 mol (unrounded) Initial moles = Liatm ⎞ RT ⎛ 0.0821 (800. K ) ⎜ moliK ⎟ ⎝ ⎠ (1.65 atm )( 2.00 L ) PV = = 0.050243605 mol (unrounded) Final moles = Liatm ⎞ RT ⎛ 0.0821 (800. K ) ⎜ moliK ⎟ ⎝ ⎠ Moles of SO3 produced = 2 x decrease in the total number of moles = 2 x (0.05785627 mol – 0.050243605 mol) = 0.01522533 = 1.52 x 10–2 mol Check: If the starting amount is 0.0578 total moles of SO2 and O2, then x + y = 0.0578 mol, where x = mol of SO2 and y = mol of O2. After the reaction: (x – z) + (y – 0.5z) + z = 0.0502 mol Where z = mol of SO3 formed = mol of SO2 reacted = 2(mol of O2 reacted). Subtracting the two equations gives: x – (x – z) + y – (y – 0.5z) – z = 0.0578 – 0.0502 z = 0.0152 mol SO3 The approach of setting up two equations and solving them gives the same result as above. The reaction is: 2 NCl3(l) ⎯Δ N2(g) + 3 Cl2(g) ⎯→ The decomposition of all the NCl3 means that the final pressure must be due to the N2 and the Cl2. The product gases are present at a 1:3 ratio, and the total moles are 4. a) Partial pressure N2 = Pnitrogen = Xnitrogen Ptotal = (1 mol N2 / 4 mol total) (754 mmHg) = 188.5 = 188 mmHg N2 Partial pressure Cl2 = Pnitrogen = Xnitrogen Ptotal = (3 mol Cl2 / 4 mol total) (754 mmHg) = 565.5 = 566 mmHg Cl2 b) The mass of NCl3 may be determined several ways. Using the partial pressure of Cl2 gives: ( 565.5 mmHg )( 2.50L ) ⎛ 1 atm ⎞ PV = Moles Cl2 = n = ⎜ ⎟ = 0.0615698 mol Cl2 Liatm ⎞ RT ⎛ 0.0821 ( 273 + 95 ) K ) ⎝ 760 mmHg ⎠ ( ⎜ moliK ⎟ ⎝ ⎠ ⎛ 2 mol NCl3 ⎞ ⎛ 120.36 g NCl3 ⎞ Mass NCl3 = ( 0.0615698 mol Cl2 ) ⎜ ⎟ = 4.94036 = 4.94 g NCl3 ⎟⎜ ⎝ 3 mol Cl2 ⎠⎝ 1 mol NCl3 ⎠

5.103

5-22

5.104

a) Use the density to find the volume of one mole of gas: ⎛ 28.05 g C 2 H 4 ⎞ ⎛ 1 mL ⎞ ⎛ 10-3 L ⎞ (1 mole C2 H 4 ) ⎜ ⎟ = 0.130465 L = 0.130 L ⎟⎜ ⎟⎜ ⎜ ⎟ ⎝ 1 mole C2 H 4 ⎠ ⎝ 0.215 g ⎠ ⎝ 1 mL ⎠ Use van der Waals equation with 1.00 mol (the mole ratio is 1:1, so the number of moles of gas remains the same). ⎛ n 2a ⎞ ⎜ P + 2 ⎟ ( V − nb ) = nRT ⎜ V ⎟ ⎝ ⎠ PVDW = nRT n 2a − 2 V − nb V

From Table 5.5: a = 2.25

atmi L2 ; mol2

b = 0.0428
⎞ ⎟ ⎟ ⎠

L mol

⎛ atmiL2 Liatm ⎞ ⎛ (1.00 mol )2 ⎜ 2.25 1.00 mol ⎜ 0.0821 (1233 K ) ⎜ mol2 moliK ⎟ ⎝ ⎠ ⎝ PVDW = − 0.130 L − 1.00 mol ( 0.0428 L/mol ) ( 0.130 L )2

PVDW = 1027.7504 = 1028 atm (1028 atm )( 01.30 L ) PV = 1.32 = b) Liatm ⎞ RT ⎛ 0.0821 (1233 K ) ⎜ moliK ⎟ ⎝ ⎠ This value is smaller than that shown in Figure 5.21 for CH4. The temperature in this situation is very high (1233 K). At high temperatures, the gas particles have high kinetic energy. Thus the gas particles have the energy to overcome the effects of intermolecular attraction and the gas behaves more ideally. 5.105 The reaction is: 2 NH4NO3(s) → 2 N2(g) + O2(g) + 4 H2O(g) ⎛ 103 g ⎞ ⎛ 1 mol NH 4 NO3 ⎞ ⎛ 7 mol Gas ⎞ Moles gas = (15.0 kg NH 4 NO3 ) ⎜ ⎜ 1 kg ⎟ ⎜ 80.05 g NH NO ⎟ ⎜ 2 mol NH NO ⎟ ⎟ 4 3 ⎠⎝ 4 3 ⎠ ⎝ ⎠⎝ = 655.840 mol gas (unrounded) Liatm ⎞ ( 655.840 mol Gas ) ⎛ 0.0821 ( ( 273 + 307 ) K ) ⎜ moliK ⎟ ⎝ ⎠ = 3.1229789 x 104 = 3.12 x 104 L V = nRT / P = (1.00 atm ) The balanced equation is: I2(aq) + 2 S2O32−(aq) → 2 I−(aq) + S4O62−(aq) ⎛ 10−3 L ⎞ ⎛ 0.01017 mol I 2 ⎞ –4 Initial moles of I2 = ( 20.00 mL ) ⎜ ⎟ = 2.034 x 10 mol I2 initial ⎜ 1 mL ⎟ ⎜ ⎟⎝ L ⎠ ⎝ ⎠
2 ⎛ 10−3 L ⎞ ⎛ 0.0105 mol S2 O3 − ⎞⎛ 1 mol I 2 ⎞ Moles I2 reacting with S2O32– = (11.37 mL ) ⎜ ⎜ ⎟⎜ ⎟⎜ ⎜ 1 mL ⎟ ⎟⎜ 2 mol S O 2 − ⎟ ⎟ L ⎝ ⎠⎝ 2 3 ⎠ ⎠⎝ 2– –5 = 5.96925 x 10 mol I2 reacting with S2O3 (not reacting with SO2) Moles I2 reacting with SO2 = 2.034 x 10–4 mol - 5.96925 x 10–5 mol = 1.437075 x 10–4 mol I2 reacted with SO2 (unrounded) The balanced equation is: SO2(aq) + I2(aq) + 2 H2O(l) → HSO4−(aq) + 2 I−(aq) + 3 H+(aq). Moles SO2 = (1.437075 x 10–4 mol I2) (1 mol SO2 / 1 mol I2) = 1.437075 x 10–4 mol SO2 (unrounded) ⎛ 1 atm ⎞ ⎛ 10−3 L ⎞ ( 700. torr )( 500. mL ) Moles air = ⎟ = 0.018036 mol air (unrounded) ⎜ ⎟⎜ ⎜ ⎟ Liatm ⎞ ⎛ 0.0821 ( ( 273 + 38) K ) ⎝ 760 torr ⎠ ⎝ 1 mL ⎠ ⎜ ⎟ moliK ⎠ ⎝ Volume % SO2 = mol % SO2 = [(1.437075 x 10–4 mol SO2) / (0.018036 mol air)] x 100% = 0.796781 = 0.797%

5.106

5-23

5.107

Plan: The balanced equation for the reaction is Ni(s) + 4 CO(g) → Ni(CO)4(g). Although the problem states that Ni is impure, you can assume the impurity is unreactive and thus not include it in the reaction. The mass of Ni can be calculated from the stoichiometric relationship between moles of CO (using the ideal gas equation) and Ni. Mass Ni = mol Ni x molar mass Solution:
⎛ ⎞ ⎛ 1 L ⎞ ⎛ 1 mol Ni ⎞⎛ 58.69 g Ni ⎞ 1 atm ⎟ ⎜ ⎟ ⎜ −3 3 ⎟ ⎜ ⎟⎜ ⎜ ⎟ Liatm ⎞ ⎛ 0.0821 ( ( 273 + 50 ) K ) ⎝ 101.325 kPa ⎠ ⎝ 10 m ⎠ ⎝ 4 mol CO ⎠⎝ 1mol Ni ⎠ ⎜ moliK ⎟ ⎝ ⎠ = 1952.089 = 1.95 x 103 g Ni b) Assume the volume is 1 m3. Use the ideal gas equation to solve for moles of Ni(CO)4, which equals the moles of Ni, and convert moles to grams using the molar mass.

a) Mass Ni =

(100.7 kPa ) ( 3.55 m3 )

⎛ 1 L ⎞ ⎛ 1 mol Ni ⎞ ⎛ 58.69 g Ni ⎞ ⎜ −3 3 ⎟ ⎜ ⎜ 10 m ⎟ 1 mol Ni(CO) ⎟ ⎜ 1 mol Ni ⎟ Liatm ⎞ ⎛ ⎠ 4 ⎠⎝ ⎝ ⎠⎝ ⎜ 0.0821 moliK ⎟ ( ( 273 + 155 ) K ) ⎝ ⎠ = 3.50749 x 104 = 3.5 x 104 g Ni The pressure limits the significant figures. c) The amount of CO needed to form Ni(CO)4 is the same amount that is released on decomposition. The vapor pressure of water at 35°C (42.2 torr) must be accounted for (see Table 5.3). Use the ideal gas equation to calculate the volume of CO, and compare this to the volume of Ni(CO)4. The mass of Ni from a m3 (part b) can be used to calculate the amount of CO released. PCO = Ptotal - Pwater = 769 torr - 42.2 torr = 726.8 torr (unrounded) V = nRT / P = ⎡ ⎛ 1 mol Ni ⎞⎛ 4 mol CO ⎞⎤ ⎛ Liatm ⎞ 4 ⎢3.50749 x 10 g Ni ⎜ ⎟⎜ ⎟⎥ ⎜ 0.0821 ⎟ ( ( 273 + 35) K) moliK ⎠ ⎛ 760 torr ⎞ ⎛ 10−3 m3 ⎞ ⎢ ⎝ 58.69 g Ni ⎠⎝ 1 mol Ni ⎠⎥ ⎝ ⎣ ⎦ ⎟ ⎜ ⎟⎜ ⎜ ⎟ 726.8 torr ⎝ 1 atm ⎠ ⎝ 1 L ⎠ = 63.209866 = 63 m3 CO The answer is limited to two significant figures because the mass of Ni comes from part b.

Mass Ni =

( 21 atm ) ( m3 )

5.108

Assume 100 grams of sample, thus 33.01% Si is equivalent to 33.01 grams Si, and 66.99% F is equivalent to 66.99 grams of F. Moles Si = (33.01 g Si) (1 mol Si / 28.09 g Si) = 1.17515 mol Si (unrounded) Moles F = (66.99 g F) (1 mol F / 19.00 g F) = 3.525789 mol F (unrounded) Dividing by the smaller value (1.17515) gives an empirical formula of SiF3 (empirical formula mass = 85 g/mol). Liatm ⎞ ⎛ ( 2.60 g ) ⎜ 0.0821 ( ( 273 + 27 ) K ) moliK ⎟ ⎝ ⎠ M = mRT / PV = = 170.768 g/mol (unrounded) (1.50 atm )( 0.250 L ) The molar mass is twice the empirical formula mass, so the molecular formula must be twice the empirical formula, or 2 x SiF3 = Si2F6.

5.109

a) A preliminary equation for this reaction is __CxHyNz + n O2 → 4 CO2 + 2 N2 + 10 H2O. Since the organic compound does not contain oxygen, the only source of oxygen as a reactant is oxygen gas. To form 4 volumes of CO2 would require 4 volumes of O2 and to form 10 volumes of H2O would require 5 volumes of O2. Thus, 9 volumes of O2 was required. b) Since the volume of a gas is proportional to the number of moles of the gas we can equate volume and moles. From a volume ratio of 4 CO2:2 N2:10 H2O we deduce a mole ratio of 4 C:4 N:20 H or 1 C:1 N:5 H for an empirical formula of CH5N.

5-24

5.110

The final pressure will be the sum of the air pressure (0.980 atm) and the pressure generated by the carbon dioxide. ⎡ ⎛ 1 mol CO 2 ⎞ ⎤ ⎛ Liatm ⎞ ( ( 273.2 + 550.0 ) K ) ⎢10.0 g CO 2 ⎜ ⎟ ⎥ 0.0821 44.01 g CO 2 ⎠ ⎥ ⎜ moliK ⎟ ⎝ ⎠ ⎢ ⎝ ⎣ ⎦ Carbon dioxide pressure = nRT / V = ( 0.800 L ) = 19.1958 atm (unrounded) Ptotal = (0.980 atm) + (19.1958 atm) = 20.1758 = 20.2 atm

5.111

a) 2 x 106 blue particles and 2 x 106 black particles b) 2 x 106 blue particles and 2 x 106 black particles c) Final pressure in C = (2/3) (750 torr) = 500 torr d) Final pressure in B = (2/3) (750 torr) = 500 torr At this altitude, the atmosphere is very thin. The collision frequency is very low and thus there is very little transfer of kinetic energy. Satellites and astronauts would not become “hot” in the usual sense of the word. a) C6H14(l) + 19 O2(g) →12 CO2(g) + 14 H2O(g) Assume that you have 1.00 L sample of air at STP. ⎛ 20.9 L O 2 Volume of C6H14 vapor needed = (1.00 L air ) ⎜ ⎝ 100 L air

5.112 5.113

⎞ ⎛ 2 L C6 H14 ⎞ ⎟ ⎜ 19 L O ⎟ = 0.0220 L C6H14 ⎠⎝ 2 ⎠ C H volume 0.0220 L C6 H14 Volume % of C6H14 = 6 14 (100 ) = (100 ) = 2.2 %C6H14 air volume 1.00 L air

LFL = 0.5(2.2%) = 1.1% C6H14 ⎛ 1 L ⎞ ⎛ 1.1% C6 H14 ⎞ b) Volume of C6H14 vapor = 1.000 m3 air ⎜ −3 3 ⎟ ⎜ ⎟ = 11.0 L C6H14 ⎝ 10 m ⎠ ⎝ 100% air ⎠ (1 atm ) (11.0 L C6 H14 ) PV = = 0.490780 mol C6H14 Moles of C6H14 = Liatm ⎞ RT ⎛ 0.0821 ( 273 K ) ⎜ mol•K ⎟ ⎝ ⎠ ⎞ ⎛ 86.17 g C6 H14 ⎞ ⎛ 1 mL Volume of C6H14 liquid = ( 0.490780 mol C6 H14 ) ⎜ ⎟ = 64.0765 = 64 mL C6H14 ⎟⎜ ⎝ 1mol C6 H14 ⎠ ⎝ 0.660 g C6 H14 ⎠ 5.114 Plan: To find the factor by which a diver’s lungs would expand, find the factor by which P changes from 125 ft to the surface, and apply Boyle’s Law. To find that factor, calculate Pseawater at 125 ft by converting the given depth from ft-seawater to mmHg to atm and adding the surface pressure (1.00 atm). Solution: −2 ⎛ 12 in ⎞ ⎛ 2.54 cm ⎞ ⎛ 10 m ⎞ ⎛ 1 mm ⎞ 4 P (H2O) = (125 ft ) ⎜ ⎟ ⎜ 1 in ⎟ ⎜ 1 cm ⎟ ⎜ -3 ⎟ = 3.81 x 10 mmH2O 1 ft ⎠ ⎝ ⎝ ⎠⎝ ⎠ ⎝ 10 m ⎠ P (Hg): hH2O hHg
=

(

)

d Hg d H2O

3.81 x 104 mmH 2 O 13.5 g/mL = hHg 1.04 g/mL

hHg = 2935.1111 mmHg

⎛ 1 atm ⎞ P (Hg) = ( 2935.11111 mmHg ) ⎜ ⎟ = 3.861988 atm (unrounded) ⎝ 760 mm Hg ⎠ Ptotal = (1.00 atm) + (3.861988 atm) = 4.861988 atm (unrounded) Use Boyle’s Law to find the volume change of the diver’s lungs: P1V1 = P2V2 V2 P V2 4.861988 atm = 4.86 = 1 = V1 P2 V1 1 atm

5-25

To find the depth to which the diver could ascend safely, use the given safe expansion factor (1.5) and the pressure at 125 ft, P125, to find the safest ascended pressure, Psafe. P125 / Psafe = 1.5 Psafe = P125 / 1.5 = (4.861988 atm) / 1.5 = 3.241325 atm (unrounded) Convert the pressure in atm to pressure in ft of seawater using the conversion factors above. Subtract this distance from the initial depth to find how far the diver could ascend. ⎛ 760 mmHg ⎞ h (Hg): ( 4.861988 - 3.241325 atm ) ⎜ ⎟ = 1231.7039 mmHg ⎝ 1 atm ⎠ hH2O hH2 O d Hg 13.5 g/mL = = h H2 O = 15988.464 mm 1231.7039 mmHg 1.04 g/mL hHg d H2 O

(15988.464 mmH 2 O ) ⎜

⎛ 10−3 m ⎞ ⎛ 1.094 yd ⎞ ⎛ 3 ft ⎞ ⎟⎜ ⎟ = 52.4741 ft ⎟⎜ ⎝ 1 mm ⎠ ⎝ 1 m ⎠ ⎝ 1 yd ⎠

Therefore, the diver can safely ascend 52.5 ft to a depth of (125 – 52.4741) = 72.5259 = 73 ft. 5.115 The balanced equation is: CaF2(s) + H2SO4(aq) → 2 HF(g) + CaSO4(s) ⎛ 1 atm ⎞ (875 torr )(8.63 L ) T = PV / nR = ⎜ ⎟ ⎡ ⎤⎛ ⎛ 1 mol CaF2 ⎞ ⎛ 2 mol HF ⎞ Liatm ⎞ ⎝ 760 torr ⎠ 15.0 g CaF2 ⎜ 0.0821 ⎢ ⎟⎜ ⎟⎥ ⎜ moliK ⎟ ⎠ ⎢ ⎝ 78.08 g CaF2 ⎠ ⎝ 1 mol CaF2 ⎠ ⎥ ⎝ ⎣ ⎦ = 314.9783 K The gas must be heated to 315 K. 5.116 First, write the balanced equation: 2 H2O2(aq) → 2 H2O(l) + O2(g) According to the description in the problem, a given volume of peroxide solution (0.100 L) will release a certain number of “volumes of oxygen gas” (20). Assume that 20 is exact. 0.100 L solution will produce (20 x 0.100 L) = 2.00 L O2 gas You can convert volume of O2(g) to moles of gas using the ideal gas equation. (1.00 atm )( 2.00 L ) = 8.92327 x 10–2 mol O2 (unrounded) Moles O2 = PV / RT = Liatm ⎞ ⎛ ⎜ 0.0821 moliK ⎟ ( 273 K ) ⎝ ⎠ The grams of hydrogen peroxide can now be solved from the stoichiometry using the balanced chemical equation. Mass H2O2 = (8.92327 x 10–2 mol O2) (2 mol H2O2 / 1 mol O2) (34.02 g H2O2 / 1 mol H2O2) = 6.071395 = 6.07 g H2O2 The moles may be found using the ideal gas equation. Moles times Avogadro’s number gives the molecules. Molecules =
⎛ 1 atm ⎞ ⎛ 10−3 L ⎞ ⎛ 6.022 x 1023 molecules ⎞ ⎟⎜ ⎟ = ⎜ ⎟⎜ ⎜ ⎟⎜ ⎟ Liatm ⎞ mol ⎛ ⎠ 0.0821 ( 500 K ) ⎝ 760 mmHg ⎠ ⎝ 1 mL ⎠ ⎝ ⎜ ⎟ moliK ⎠ ⎝ = 1.93025 x 108 = 108 molecules (The 10–8 mmHg limits the significant figures.)

5.117

(10

−8

mmHg (1 mL )

)

5.118

3 ( 8.314 J/mol • K )( 273 K ) ⎛ 103 g ⎞ ⎛ kg • m 2 /s 2 ⎞ ⎟ = 461.2878 = 461 m/s ⎜ ⎟ ⎜ 1 kg ⎟ ⎜ ⎟⎜ J ( 32.00 g/mol ) ⎠ ⎝ ⎠⎝ b) Collision frequency = (urms / mean free path) = (461.2878 m/s) / (6.33 x 10–8 m) = 7.2873 x 109 = 7.29 x 109 s–1

urms =

5-26

5.119

3 ⎞ ⎛ ⎞ ⎜ 10−2 m ⎟ ⎛ 1 L ⎞ ⎟⎜ = 2831.685 L ⎟ ⎜ (1 cm )3 ⎟ ⎜ 10-3 m3 ⎟ ⎠ ⎟⎝ ⎠⎝ ⎠ ( 2.5 atm )( 2831.685 L ) PV = Moles of gas = n = = 142.996 mol gas L•atm ⎞ RT ⎛ 0.0821 (330 + 273) K ⎜ mol•K ⎟ ⎝ ⎠ moles of propylene Xpropylene = moles of mixture x moles of propylene 0.07 = 142.996 moles of mixture Moles of propylene = 10.00972 moles ⎛ 42.08 g C3 H 6 ⎞ ⎛ 1 lb C3 H 6 ⎞ ⎛ 1 ⎞ ⎛ 3600 s ⎞ Pounds of propylene = (10.00972 moles C3 H 6 ) ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⎠ ⎝ 1 mole C3 H 6 ⎠ ⎝ 453.6 g C3 H 6 ⎠ ⎝ 1.8 s ⎠ ⎝

⎛ (12 in )3 Volume of tubes = 100 ft ⎜ ⎜ (1 ft )3 ⎝

(

3

)

⎞ ⎛ ( 2.54 cm )3 ⎟⎜ ⎟ ⎜ (1 in )3 ⎠⎝

(

)

(

)

= 1857.183 = 1900 lb C3H6 5.120 Molar volume uses exactly one mole of gas along with the temperature and pressure given in the problem. Liatm ⎞ ⎛ ⎜ 0.0821 moliK ⎟ ( 730. K ) ⎝ ⎠ V / n = RT / P = = 0.66592 = 0.67 L/mol 90 atm ) ( Plan: The diagram below describes the two Hg height levels within the barometer. To find the mass of N2, find P and V (T given) and use the ideal gas equation. The PN is directly related to the change in column height of Hg.
2

5.121

The volume of the space occupied by the N2(g) is calculated from the dimensions of the barometer.

vacuum (74.0 cm)

with N2 (64.0 cm)

Solution:
⎛ 10−2 m ⎞ ⎛ 1 mm ⎞ ⎛ 1 atm ⎞ Pressure of the nitrogen = ( 74.0 cm - 64.0 cm ) ⎜ = 0.1315789 atm ⎜ 1 cm ⎟ ⎜ 10-3 m ⎟ ⎜ 760 atm ⎟ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠
−3 ⎞ ⎛ 10 L ⎞ ⎜ ⎟ ⎜ 1 mL ⎟ = 0.0432 L ⎟ ⎠⎝ ⎠ The moles of nitrogen may be found using the ideal gas equations, and the moles times the molar mass of nitrogen will give the mass of nitrogen. ( 0.1315789 atm )( 0.0432 L ) ⎛ 28.02 g N 2 ⎞ Mass of nitrogen = ⎜ ⎟ Liatm ⎞ ⎛ ⎝ 1 mol ⎠ ⎜ 0.0821 moliK ⎟ ( ( 273 + 24 ) K ) ⎝ ⎠ = 6.5319 x 10–3 = 6.53 x 10–3 g N2

⎛ 1 mL Volume of the nitrogen = 1.00 x 102 cm - 64.0 cm 1.20 cm 2 ⎜ ⎝ 1 cm3

(

)(

)

5-27

5.122

Molar concentration is moles per liter. Use the original volume (10.0 L) to find the moles of gas, and the final volume (0.750 L) to get the molar concentration. ⎛ 1 atm ⎞ ( 735 torr )(10.0 L ) Moles NH3 = PV / RT = ⎜ ⎟ Liatm ⎞ 760 torr ⎠ ⎛ 0.0821 273 + 33) K ) ⎝ (( ⎜ moliK ⎟ ⎝ ⎠ = 0.384954 mol (unrounded) Molar concentration = mol / volume, in liters, of solution. M = (0. 384954 mol) / (0.750 L) = 0.513272 = 0.513 M Determine the total moles of gas produced. The total moles times the fraction of each gas gives the moles of that gas which may be converted to metric tons. m3 /d ⎛ 1 L ⎞ ⎛ 365.25 d ⎞ Moles total = n = PV / RT = ⎜ −3 3 ⎟ ⎜ ⎟ ⎜ ⎟ Liatm ⎞ ⎛ 0.0821 ( 298 K ) ⎝ 10 m ⎠ ⎝ 1 y ⎠ ⎜ ⎟ moliK ⎠ ⎝ 7 = 2.23935 x 10 mol/year (unrounded) Mass CO2 = (0.4896) (2.23935 x 107 mol/year) (44.01 g CO2/mol) (1 kg/103 g) (1 t/103 kg) = 482.519 = 4.8 x 102 t CO2/y Mass CO = (0.0146) (2.23935 x 107 mol/year) (28.01 g CO/mol) (1 kg/103 g) (1 t/103 kg) = 9.15773 = 9.16 t CO/y Mass H2O = (0.3710) (2.23935 x 107 mol/year) (18.02 g H2O/mol) (1 kg/103 g) (1 t/103 kg) = 149.70995 = 1.50 x 102 t H2O/y Mass SO2 = (0.1185) (2.23935 x 107 mol/year) (64.07 g SO2/mol) (1 kg/103 g) (1 t/103 kg) = 170.018 = 1.70 x 102 t SO2/y Mass S2 = (0.0003) (2.23935 x 107 mol/year) (64.14 g S2/mol) (1 kg/103 g) (1 t/103 kg) = 0.4308957 = 4 x 10–1 t S2/y Mass H2 = (0.0047) (2.23935 x 107 mol/year) (2.016 g H2/mol) (1 kg/103 g) (1 t/103 kg) = 0.21218 = 2.1 x 10–1 t H2/y Mass HCl = (0.0008) (2.23935 x 107 mol/year) (36.46 g HCl/mol) (1 kg/103 g) (1 t/103 kg) = 0.6531736 = 6 x 10–1 t CO2/y Mass H2S = (0.0003) (2.23935 x 107 mol/year) (34.09 g H2S/mol) (1 kg/103 g) (1 T/103 kg) = 0.229018 = 2 x 10–1 T CO2/y

5.123

(1.00 atm ) (1.5 x 103

)

5.124

The balanced chemical equation is: 2 H2(g) + O2(g) → 2 H2O(l) Moles H2 = (28.0 mol H2O) (2 mol H2/2 mol H2O) = 28.0 mol H2 Moles O2 = (28.0 mol H2O) (1 mol O2/2 mol H2O) = 14.0 mol O2 a) P = nRT / V Liatm ⎞ ( 28.0 mol H 2 ) ⎛ 0.0821 ( ( 273.2 + 23.8) K ) ⎜ moliK ⎟ ⎝ ⎠ = 34.137 = 34.1 atm H2 P H2 = ( 20.0 L ) P O2 =

(14.0 mol O2 ) ⎛ 0.0821 ⎜


Liatm ⎞ ( ( 273.2 + 23.8) K ) moliK ⎟ ⎠ = 17.06859 = 17.1 atm O2 ( 20.0 L )

⎛ n 2a ⎞ b) ⎜ P + 2 ⎟ ( V − nb ) = nRT ⎜ V ⎟ ⎝ ⎠ Use Table 5.5 to find the values of the van der Waals constants. b = 0.0266 L/mol H2: a = 0.244 atm•L2/mol2 O2: a = 1.36 atm•L2/mol2 b = 0.0318 L/mol nRT n 2a − PVDW = V − nb V 2

5-28

atmiL2 ⎞ 2⎛ Liatm ⎞ ⎟ ( ( 273.2 + 23.8) K ) ( 28.0 mol H2 ) ⎜ 0.244 mol2 ⎟ ⎟ ⎜ moliK ⎠ ⎝ ⎝ ⎠ − PVDW H2 = L ⎞ ⎛ ( 20.0 L )2 20.0 L − ( 28.0 mol H2 ) ⎜ 0.0266 ⎟ mol ⎠ ⎝ = 34.9631 = 35.0 atm H2 atmiL2 ⎞ 2⎛ Liatm ⎞ ⎛ ⎟ .8) K ) (14.0 mol O2 ) ⎜1.36 (14.0 mol O2 ) ⎜ 0.08206 ( ( 273.2 + 23 ⎜ mol2 ⎟ moliK ⎟ ⎝ ⎠ ⎝ ⎠ − PVDW O2 = L ⎞ ⎛ ( 20.0 L )2 20.0 L − (14.0 mol O2 ) ⎜ 0.0318 ⎟ mol ⎠ ⎝ = 16.78228 = 16.8 atm O2 c) The van der Waals value for hydrogen is slightly higher than the value from the ideal gas equation. The van der Waals value for oxygen is slightly lower than the value from the ideal gas equation.

( 28.0 mol H2 ) ⎛ 0.08206 ⎜

5.125

Plan: Deviations from ideal gas behavior are due to attractive forces between particles which reduce the pressure of the real gas and due to the size of the particle which affects the volume. Compare the size and/or attractive forces between the substances. Solution: a) Xenon would show greater deviation from ideal behavior than argon since xenon is a larger atom than argon. The electron cloud of Xe is more easily distorted so intermolecular attractions are greater. Xe’s larger size also means that the volume the gas occupies becomes a greater proportion of the container’s volume at high pressures. b) Water vapor would show greater deviation from ideal behavior than neon gas since the attractive forces between water molecules are greater than the attractive forces between neon atoms. We know the attractive forces are greater for water molecules because it remains a liquid at a higher temperature than neon (water is a liquid at room temperature while neon is a gas at room temperature). c) Mercury vapor would show greater deviation from ideal behavior than radon gas since the attractive forces between mercury atoms is greater than that between radon atoms. We know that the attractive forces for mercury are greater because it is a liquid at room temperature while radon is a gas. d) Water is a liquid at room temperature; methane is a gas at room temperature (think about where you have heard of methane gas before - Bunsen burners in lab, cows’ digestive system). Therefore, water molecules have stronger attractive forces than methane molecules and should deviate from ideal behavior to a greater extent than methane molecules. Moles of HBr in the Hydrobromic acid: (1.20 mol HBr / L) (3.50 L) = 4.20 mol HBr Liatm ⎞ ( 4.20 mol HBr ) ⎛ 0.0821 ⎜ ⎟ ( ( 273 + 29 ) K ) moliK ⎠ ⎝ = 107.9126 = 108 L HBr V HBr = nRT / P = ( 0.965 atm ) The mole fraction of CO must be found from the moles of the gases. The mole fraction plus the total pressure can be used to find the partial pressure. Moles CO = (7.0 g CO) (1 mol CO / 28.01 g CO) = 0.24991 mol CO (unrounded) Moles SO2 = (10.0 g CO) (1 mol SO2 / 64.07 g SO2) = 0.156079 mol SO2 (unrounded) P CO = XCO Ptotal = [(0.24991 mol CO) /(0.24991 + 0.156079)mol](0.33 atm) = 0.20313 = 0.20 atm CO First, balance the equation: 4 FeS2(s) + 11 O2(g) → 8 SO2(g) + 2 Fe2O3(s) Pressure of N2 = 1.05 atm – 0.64 atm O2 = 0.41 atm N2 Pressure of unreacted O2 = (0.64 atm O2) [(100 – 85)% / 100%] = 0.096 atm O2 Pressure of SO2 produced = (0.64 atm O2) (8 atm SO2 / 11 atm O2) = 0.46545 = 0.47 atm SO2 Total Pressure = (0.41 atm) + (0.096 atm) + (0.46545 atm) = 0.97145 = 0.97 atm total

5.126

5.127

5.128

5-29

5.129

Plan: V and T are not given, so the ideal gas equation cannot be used. The total pressure of the mixture is given. Use PA = XA x Ptotal to find the mole fraction of each gas and then the mass fraction. The total mass of the two gases is 35.0 g. Solution: Ptotal = Pkrypton + Pcarbon dioxide = 0.708 atm The NaOH absorbed the CO2 leaving the Kr, thus Pkrypton = 0.250 atm. Pcarbon dioxide = Ptotal – Pkrypton = 0.708 atm – 0.250 atm = 0.458 atm Determining mole fractions: PA = XA x Ptotal PCO2 0.458 atm Carbon dioxide: X = = = 0.64689 (unrounded) 0.708 atm Ptotal Krypton: X =
PKr 0.250 atm = = 0.353107 (unrounded) 0.708 atm Ptotal

⎡ ⎛ 83.80 g Kr ⎞ ⎤ ⎢ ( 0.353107 ) ⎜ ⎟⎥ mol ⎝ ⎠ ⎥ = 1.039366 (unrounded) Relative mass fraction = ⎢ ⎢ ⎛ 44.01 g CO 2 ⎞ ⎥ ⎢ ( 0.64689 ) ⎜ ⎟⎥ mol ⎝ ⎠⎦ ⎣

35.0 g = x g CO2 + (1.039366 x) g Kr 35.0 g = 2.039366 x Grams CO2 = x = (35.0 g) / (2.039366) = 17.16219581 = 17.2 g CO2 Grams Kr = 35.0 g – 17.162 g CO2 = 17.83780419 = 17.8 g Kr 5.130 5.131 As the car accelerates, the air in the car is pressed toward the back. The helium balloon floats on “top” of this denser air, which pushes it toward the front of the car. P = nRT / V = ⎡⎛ 2.5 x 1012 molecules ⎞ ⎡ ⎤⎤ ⎛ ⎛ −3 3 ⎞ 1 mol Liatm ⎞ ⎢⎜ ⎜ ⎟⎢ ( ( 273 + 22 ) K ) ⎜ 101 Lm ⎟ ⎥ ⎥ ⎜ 0.0821 ⎟ ⎢ 6.022 x 1023 molecules ⎥ ⎥ ⎝ ⎜ ⎟ moliK ⎟ m3 ⎠ ⎢⎝ ⎦⎦ ⎠⎣ ⎝ ⎠ ⎣ = 1.005459 x 10–13 = 1.0 x 10–13 atm •OH The mole percent is the same as the pressure percent as long as the other conditions are the same. Mole percent = [(1.005459 x 10–13 atm) / (1.00 atm)] 100% = 1.005459 x 10–11 = 1.0 x 10–11% The balanced chemical equations are: SO2(g) + H2O(l) → H2SO3(aq) H2SO3(aq) + 2 NaOH(aq) → Na2SO3(aq) + 2 H2O(l) Combining these equations gives: SO2(g) + 2 NaOH(aq) → Na2SO3(aq) + H2O(l) M NaOH = [(mol SO2) (2 mol NaOH / 1 mol SO2) / (L NaOH solution) Moles SO2 = PV / RT M NaOH = [(PV / RT) (2 mol NaOH / 1 mol SO2) / (L NaOH solution) M NaOH = = 1.635598 = 1.64 M NaOH

5.132

5-30

5.133

Write the balanced chemical equations: LiH(s) + H2O(l) → LiOH(aq) + H2(g) MgH2(s) + 2 H2O(l) → Mg(OH)2(s) + 2 H2(g) LiOH is water soluble, however, Mg(OH)2 is not water soluble. V = nRT / P LiH V= ⎡ ⎛ 1 kg ⎞ ⎛ 103 g ⎞ ⎛ 1 mol LiH ⎞⎛ 1 mol H2 ⎞⎤ ⎛ Liatm ⎞ ⎢1.00 lb LiH ⎜ ( ( 273 + 27) K ) ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎥ 0.0821 2.205 lb ⎠ ⎜ 1 kg ⎟ ⎝ 7.946 g LiH ⎠⎝ 1 mol LiH ⎠⎦ ⎜ moliK ⎟ ⎠ ⎛ 760 torr ⎞ ⎢ ⎥⎝ ⎝ ⎝ ⎠ ⎣ ⎜ ⎟ ( 750. torr ) ⎝ 1 atm ⎠

MgH2

= 1424.490595 = 1420 L H2 from LiH V= ⎡ ⎛ 1 kg ⎞ ⎛ 103 g ⎞ ⎛ 1 mol MgH2 ⎞⎛ 2 mol H2 ⎞⎤ ⎛ Liatm ⎞ ⎢1.00 lb LiH ⎜ ( ( 273 + 27) K) ⎟⎜ ⎟⎜ ⎟⎥ 0.0821 ⎟⎜ 2.205 lb ⎠ ⎜ 1 kg ⎟ ⎝ 26.33 g MgH2 ⎠⎝ 1 mol MgH2 ⎠⎦ ⎜ moliK ⎟ ⎝ ⎠ ⎛ 760 torr ⎞ ⎢ ⎥ ⎝ ⎝ ⎠ ⎣ ⎜ ⎟ 750.torr ) ( ⎝ 1 atm ⎠ = 859.7798912 = 8.60 x 102 L H2 from MgH2

5.134

The root mean speed is related to temperature and molar mass as shown in equation: 3 RT urms = M urms Ne =
3 ( 8.314 J/moliK )( 370 K ) ⎛ 103 g ⎞ ⎛ kgim 2 /s 2 ⎜ ⎜ kg ⎟ ⎜ ⎟⎜ J ( 20.18 g/mol ) ⎝ ⎠⎝ ⎞ ⎟ = 676.24788 = 676 m/s Ne ⎟ ⎠

urms Ar =

3 ( 8.314 J/moliK )( 370 K ) ⎛ 103 g ⎞⎛ kg im 2 /s 2 ⎜ ⎜ kg ⎟⎜ ⎟⎜ J ( 39.95 g/mol ) ⎝ ⎠⎝

⎞ ⎟ = 480.6269 = 481 m/s Ar ⎟ ⎠ ⎞ 3 ⎟ = 1518.356 = 1.52 x 10 m/s He ⎟ ⎠

urms He = 5.135

3 ( 8.314 J/moliK )( 370 K ) ⎛ 103 g ⎞⎛ kg im 2 /s 2 ⎜ ⎜ kg ⎟⎜ ⎟⎜ J ( 4.003 g/mol ) ⎝ ⎠⎝

a) Moles = n = PV / RT ⎛ 1 atm ⎞ ( 30.0 torr )( 300 L ) Moles = ⎜ ⎟ = 0.464990 mol (unrounded) Liatm ⎞ ⎛ ⎝ 760 torr ⎠ ⎜ 0.0821 moliK ⎟ ( ( 273.2 + 37.0 ) K ) ⎝ ⎠ Mass CO2 = (0.464990 mol) (44.01 g/mol) = 20.4642 = 20.5 g CO2 Mass H2O = (0.464990 mol) (18.02 g/mol) = 8.3791 = 8.38 g H2O b) C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(g) Mass C6H12O6 = (0.464990 mol CO2/h) (1 mol C6H12O6/6 mol CO2) (180.16 g C6H12O6 / 1 mol C6H12O6) (8 h) = 111.6968 = 1 x 102 g C6H12O6 (= body mass lost) (This assumes the significant figures are limited by the 8 h.) a) The number of moles of water gas can be found using the ideal gas equation, and moles times molar mass gives the mass of water. ⎛ 10−3 L ⎞ ⎛ 75% ⎞ ( 9.0 atm )( 0.25 mL ) PV Moles H2O = = ⎜ ⎟⎜ ⎟ ⎜ ⎟ Liatm ⎞ nT ⎛ 0.0821 ( ( 273 + 170 ) K ) ⎝ 1 mL ⎠ ⎝ 100% ⎠ ⎜ moliK ⎟ ⎝ ⎠

5.136

5-31

⎛ 18.02 g H 2 O ⎞ ⎛ 100% ⎞ Mass = 4.639775 x 10−5 mol H 2 O ⎜ ⎟⎜ ⎟ = 0.052255 = 0.052 g ⎝ 1 mol H 2 O ⎠ ⎝ 1.6% ⎠ The last percent adjustment compensates for the water content being 1.6% of the mass. Liatm ⎞ ⎛ 4.639775 x 10−5 mol H 2 O ⎜ 0.0821 ( ( 273 + 25) K ) ⎛ 1 mL ⎞ moliK ⎟ ⎝ ⎠ b) V = nRT / P = ⎜ −3 ⎟ 1.00 atm ⎝ 10 L ⎠ = 1.13516 = 1.1 mL

(

)

(

)

5.137

The balanced chemical equations are:

CaCO3(s) + SO2(g) → CaSO3(s) + CO2(g) 2 CaSO3(s) + O2(g) → 2 CaSO4(s) Assuming the pressure and temperature from part b applies to part a: Mole fraction SO2 = 1000 (2 x 10–10) = 2 x 10–7 ⎛ 109 L ⎞ (1.00 atm)( 4 GL) −7 ⎛ 95% ⎞ ⎛ 1 mol CaSO4 ⎞⎛ 136.15 g CaSO4 ⎞ ⎛ 1 kg ⎞ a) ⎜ ⎟ 2 x 10 ⎜ ⎟ ⎟⎜ ⎟⎜ ⎟⎜ ⎜ 1 GL ⎟ Liatm ⎞ ⎛ ⎝ 100% ⎠ ⎝ 1 mol SO2 ⎠⎝ 1 mol CaSO4 ⎠ ⎜ 103 g ⎟ ⎝ ⎠ ⎝ ⎠ ⎜ 0.0821 moliK ⎟ ( ( 273 + 25) K ) ⎝ ⎠ = 4.2293 = 4 kg CaSO4 b) ⎛ 1 mol CaSO4 ⎞⎛ 1 mol O2 ⎞ ⎛ ⎛ ⎞ 1 Liatm ⎞ ( 4.2293 kg CaSO4 ) ⎜ ( ( 273 + 25) K) ⎜ 1.001atm ⎟ ⎛ 0.209 ⎞ ⎟⎜ ⎟ 0.0821 ⎜ ⎟ 136.15 g CaSO4 ⎠⎝ 2 mol CaSO4 ⎠ ⎜ moliK ⎟ ⎝ ⎠ ⎠ ⎝ ⎠⎝ ⎝

(

)

= 1.8183 = 2 L air 5.138 a) n = PV / RT =
⎛ 80.0% ⎞ 3 3 ⎜ 100% ⎟ = 2.36248 x 10 = 2.36 x 10 mol Cl2 Liatm ⎞ ⎛ ⎝ ⎠ ⎜ 0.0821 moliK ⎟ ( 298 K ) ⎝ ⎠

(85.0 atm )(850. L )

⎛ n 2a ⎞ b) ⎜ P + 2 ⎟ ( V − nb ) = nRT ⎜ V ⎟ ⎠ ⎝ nRT n 2a − 2 V − nb V Use Table 5.5 to find the values of the van der Waals constants. Cl2: a = 6.49 atm•L2/mol2 b = 0.0562 L/mol

PVDW =

2⎛ atmiL2 ⎞ 3 Liatm ⎞ ⎛ ⎟ 2.36248 x 103 mol Cl2 ⎜ 0.08206 298 K ) 2.36248 x 10 mol Cl2 ⎜ 6.49 ( ⎜ mol2 ⎟ moliK ⎟ ⎝ ⎠ ⎝ ⎠ − L ⎞ ⎛ (850. L )2 850. L − 2.36248 x 103 molCl2 ⎜ 0.0562 ⎟ mol ⎠ ⎝ = 30.4134 = 30.4 atm c) The engineer did not completely fill the tank. She should have filled it to (80.0%/100%) (85.0 atm) = 68 atm, but only filled it to 30.4 atm.

(

)

(

)

(

)

5.139

The height of the column (measured parallel to the original vertical position of the tube) would remain constant, so the length of liquid column (measured along the no-longer-vertical tube) would increase. Part of the mass of the liquid column is now being supported by the glass of the tube, so its mass (and hence length) would be greater. a) cylinder B b) cylinder B c) none d) cylinder C e) cylinder D

5.140

5-32

5.141

Ideal Gas Equation: PV = nRT ⎡ ⎛ 1 mol NH3 ⎞ ⎤ ⎛ Liatm ⎞ ⎢51.1 g NH3 ⎜ ⎟ ⎥ ⎜ 0.0821 ⎟ ( ( 273 + 0 ) K ) 17.03 g NH3 ⎠ ⎥ ⎝ moliK ⎠ ⎢ ⎝ ⎣ ⎦ P0° = nRT / V = ( 3.000 L ) = 22.417687 = 22.4 atm at 0°C ⎡ ⎛ 1 mol NH3 ⎞ ⎤ ⎛ Liatm ⎞ ( ( 273 + 400 ) K ) ⎢51.1 g NH 3 ⎜ ⎟ ⎥ ⎜ 0.0821 moliK ⎟ ⎠ ⎢ ⎝ 17.03 g NH 3 ⎠ ⎥ ⎝ ⎦ ⎣ P400° = nRT / V = ( 3.000 L ) = 55.2641 = 55.3 atm at 400°C ⎛ n 2a ⎞ van der Waals Equation: ⎜ P + 2 ⎟ ( V − nb ) = nRT ⎜ V ⎟ ⎝ ⎠ NH3: a = 4.17 atm•L2 / mol2 b = 0.0371 L/mol 2 nRT n a − PVDW = V − nb V 2 Moles NH3 = (51.1 g NH3) (1 mol NH3 / 17.03 g NH3) = 3.000587 mol NH3 (unrounded) P0° NH3 = ⎛ atmiL2 ⎞ Liatm ⎞ ⎛ ( 3.000587 mol NH3 )2 ⎜ 4.17 ⎟ ( 3.000587 mol NH3 ) ⎜ 0.08206 moliK ⎟ ( ( 273 + 0) K ) ⎜ mol2 ⎟ ⎠ ⎝ ⎠ ⎝ − 2 L ⎞ ⎛ ( 3.000 L ) 3.000 L − ( 3.000587 mol H2 ) ⎜ 0.0371 mol ⎟ ⎝ ⎠ = 19.0986 = 19.1 atm NH3 P400° NH3 = atmiL2 2⎛ Liatm ( 3.000587 mol NH3 ) ⎛ 0.08206 moliK ⎞ ( ( 273 + 400) K ) ( 3.000587 mol NH3 ) ⎜ 4.17 mol2 ⎜ ⎟ ⎜ ⎝ ⎠ ⎝ − L ⎞ ⎛ ( 3.000 L )2 3.000 L − ( 3.000587 mol H2 ) ⎜ 0.0371 ⎟ mol ⎠ ⎝ = 53.2243 = 53.2 atm NH3

⎞ ⎟ ⎟ ⎠

5.142

Xmethane = 1.00 – Xargon – Xhelium = 1.00 – 0.35 – 0.25 = 0.40 Pmethane = Xmethane Ptotal = (0.40) (1.75 atm) = 0.70 atm CH4 Moles x Avogadro’s number = molecules Moles = PV / RT ⎛ 6.022 x 1023 molecules CH 4 ⎞ ( 0.70 atm )( 6.0 L ) Molecules = ⎜ ⎟ ⎜ ⎟ Liatm ⎞ 1 mol ⎛ ⎝ ⎠ ⎜ 0.0821 moliK ⎟ ( ( 273 + 45 ) K ) ⎝ ⎠ = 9.6876795 x 1022 = 9.7 x 1022 molecules CH4 Plan: Find the number of moles of carbon dioxide produced by converting the mass of glucose in grams to moles and using the stoichiometric ratio from the balanced equation. Then use the T and P given to calculate volume from the ideal gas equation. Solution: a) C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(g) ⎛ 1 mol C6 H12 O6 ⎞⎛ 6 mol CO 2 ⎞ Moles CO2: ( 20.0 g C6 H12 O6 ) ⎜ ⎟⎜ ⎟ = 0.666075 mol CO2 ⎝ 180.16 g C6 H12 O6 ⎠⎝ 1 mol C6 H12 O6 ⎠

5.143

5-33

V=

nRT = P

( 0.666075

Liatm ⎞ ⎛ mol ) ⎜ 0.0821 ( ( 273 + 37 ) K ) ⎛ 760 torr ⎞ moliK ⎟ ⎝ ⎠ ⎜ ⎟ ( 780. torr ) ⎝ 1 atm ⎠

= 16.5176 = 16.5 Liters CO2 This solution assumes that partial pressure of O2 does not interfere with the reaction conditions. b) Plan: From the stoichiometric ratios in the balanced equation, calculate the moles of each gas and then use Dalton’s law of partial pressures to determine the pressure of each gas. Solution: ⎛ 1 mol C6 H12 O6 ⎞⎛ ⎞ 6 mol Moles CO2 = Moles O2 = (10.0 g C6 H12 O6 ) ⎜ ⎟⎜ ⎟ ⎝ 180.16 g C6 H12 O6 ⎠⎝ 1 mol C6 H12 O 6 ⎠ = 0.333037 mol CO2 = mol O2 At 37°C, the vapor pressure of water is 48.8 torr. No matter how much water is produced, the partial pressure of H2O will still be 48.8 torr. The remaining pressure, 780 torr – 48.8 torr = 731.2 torr is the sum of partial pressures for O2 and CO2. Since the mole fractions of O2 and CO2 are equal, their pressures must be equal, and be one–half of sum of the partial pressures just found. Pwater = 48.8 torr (731.2 torr) / 2 = 365.6 = 3.7 x 102 torr Poxygen = Pcarbon dioxide 5.144 The average kinetic energies are the same for any gases at the same temperature: Average kinetic energy = (3/2)RT = (3/2) (8.314 J/mol•K) (273 K) = 3.40458 x 103 = 3.40 x 103 J for both gases rms speed: 3RT urms = M urmsN2 =
3 ( 8.314 J/moliK )( 273 K ) ⎛ 103 g ⎞⎛ kg im 2 /s 2 ⎜ ⎜ kg ⎟⎜ ⎟⎜ J ( 28.02 g/mol ) ⎝ ⎠⎝ 3 ( 8.314 J/moliK )( 273K ) ⎛ 103 g ⎞⎛ kg im 2 /s 2 ⎜ ⎜ kg ⎟⎜ ⎟⎜ J ( 2.016 g/mol ) ⎝ ⎠⎝ ⎞ 2 2 ⎟ = 4.92961x 10 = 4.93 x 10 m/s N2 ⎟ ⎠ ⎞ 3 3 ⎟ = 1.83781 x 10 = 1.84 x 10 m/s H2 ⎟ ⎠

urms H2 = 5.145

a) Assuming the total pressure is 760 torr. (0.2 torr Br2 / 760 torr) (106) = 263.15789 = 300 ppm Br2 Unsafe b) Assuming the total pressure is 760 torr. (0.2 torr CO2 / 760 torr) (106) = 263.15789 = 300 ppm CO2 Safe c) Moles bromine = (0.0004 g Br2) (1 mol Br2 / 159.80 g Br2) = 2.5031 x 10–6 mol Br2 (unrounded) (1.00 atm )(1000 L ) Moles air = PV / RT = = 44.616 mol air (unrounded) Liatm ⎞ ⎛ ⎜ 0.0821 moliK ⎟ ( 273 K ) ⎝ ⎠ Concentration of Br2 = [(2.5031 x 10–6 mol) / (44.616 mol)] (106) = 0.056103 = 0.06 ppm Br2 Safe d) Moles CO2 = (2.8 x 1022 molecules CO2) (1 mol CO2 / 6.022 x 1023 molecules CO2) = 0.046496 mol CO2 (unrounded) Concentration of CO2 = [(0.046496 mol) / (44.616 mol)] (106) = 1042.1 = 1.0 x 103 ppm CO2 Safe 4 NH3(g) + 4 NO(g) + O2(g) → 4 N2(g) + 6 H2O(g) a)
⎛ 4 L NH3 ⎞ ⎜ ⎟ Liatm ⎞ ⎛ ⎝ 4 L NO ⎠ ⎜ 0.0821 moliK ⎟ ( ( 273 + 365 ) K ) ⎝ ⎠ = 8.5911x 10–7 = 8.6 x 10–7 L NH3 for every liter of polluted air

5.146

( 4.5 x 10

−5

atm (1.00 L )

)

5-34

b)

⎛ 103 L ⎞ ⎛ 4 mol NH3 ⎞ ⎛ 17.03 g NH3 ⎞ ⎜ ⎜ 1 kL ⎟ ⎜ 4 mol NO ⎟ ⎜ 1 mol NH ⎟ ⎟ Liatm ⎞ ⎛ ⎠⎝ 3 ⎠ ⎝ ⎠⎝ ⎜ 0.0821 moliK ⎟ ( ( 273 + 365 ) K ) ⎝ ⎠ = 0.014631 = 0.015 g NH3 Molar Mass Xe = Molar Mass Ne 131.3 g/mol = 2.55077 enrichment factor (unrounded) 20.18 g/mol

( 4.5 x 10

−5

atm (1 kL )

)

5.147

Rate Ne = Rate Xe

Thus XNe = (2.55077) / (2.55077 + 1) = 0.7183709 = 0.7184 5.148 Plan: To find the number of steps through the membrane, calculate the molar masses to find the ratio of effusion rates. This ratio is the enrichment factor for each step. Solution: Rate 235 Molar Mass 238 UF6 352.04 g/mol UF6 = = 235 Rate 238 349.03 g/mol Molar Mass UF6 UF
6

= 1.004302694 enrichment factor (unrounded) Therefore, the abundance of 235UF6 after one membrane is 0.72% x 1.004302694; Abundance of 235UF6 after “N” membranes = 0.72% * (1.004302694)N Desired abundance of 235UF6 = 3.0% = 0.72% * (1.004302694)N Solving for N: 3.0% = 0.72% * (1.004302694)N 4.16667 = (1.004302694)N ln 4.16667 = ln (1.004302694)N ln 4.16667 = N * ln (1.004302694) N = (ln 4.16667) / (ln 1.004302694) = 332.392957 = 332 steps 5.149
⎛ n 2a ⎞ ⎜ P + 2 ⎟ ( V − nb ) = nRT ⎜ V ⎟ ⎝ ⎠ nRT n 2a − 2 V − nb V Use Table 5.5 to find the values of the van der Waals constants. b = 0.0395 L/mol CO: a = 1.45 atm•L2/mol2

PVDW =

⎛ atmiL2 ⎞ Liatm ⎞ ⎛ (1.000 mol CO)2 ⎜1.45 ⎟ (1.000 mol CO) ⎜ 0.08206 ⎟ ( 273.15 K ) ⎜ ⎟ mol2 ⎠ moliK ⎠ ⎝ ⎝ − L ⎞ ⎛ ( 22.414 L)2 22.414 L − (1.000 mol CO ) ⎜ 0.0395 ⎟ mol ⎠ ⎝ = 0.998909977 = 0.9989 atm 5.150 Plan: The amount of each gas that leaks from the balloon is proportional to its effusion rate. Using 35% as the rate for H2, the rate for O2 can be determined from Graham’s Law. Solution: Rate O 2 Molar Mass H 2 2.016 g/mol Rate O2 = = = Molar Mass O 2 32.00 g/mol 35 Rate H 2 Rate O2 = 0.250998(35) = 8.78493 (unrounded) Amount of H2 that leaks = 35%; 100–35 = 65% H2 remains Amount of O2 that leaks = 8.78493%; 100–8.78493 = % O2 remains O2 91.21507 = = 1.40331 = 1.4 H2 65

5-35

5.151

a) Options for PCl3: All values are g/mol P First Cl Second Cl Third Cl Total 31 35 35 35 136 31 37 35 35 138 31 37 37 35 140 31 37 37 37 142 b) The fraction abundances are 35Cl = 75% / 100% = 0.75, and 37Cl = 25% / 100% = 0.25. The relative amount of each mass comes from the product of the relative abundances of each Cl isotope. Mass 136 = (0.75) (0.75) (0.75) = 0.421875 = 0.42 (most abundant) Mass 138 = (0.25) (0.75) (0.75) = 0.140625 = 0.14 Mass 140 = (0.25) (0.25) (0.75) = 0.046875 = 0.047 Mass 142 = (0.25) (0.25) (0.25) = 0.015625 = 0.016 c) Rate P37Cl3 / Rate P35Cl3 =
= 0.978645 = 0.979 Molar Mass P 35 Cl 3 = Molar Mass P 37 Cl 3
136 g/mol 142 g/mol

5.152

a) Pf = PiTf / Ti = (35.0 psi) (318 K) / (295 K) = 37.7288 = 37.7 psi b) New tire volume = Vi (102% /100%) = 1.02 Vi = Vf Pf = PiViTf / TiVf = [(35.0 psi) (Vi) (318 K)] / [(295 K) (1.02 Vi)] = 36.9890 = 37.0 psi ( ( 36.9890 − 35.0 ) psi ) ( 218 L ) ⎛ 1 atm ⎞ = 1.217892 mol lost c) Δn = ΔPV / RT = ⎜ ⎟ Liatm ⎞ 14.7 psi ⎠ ⎛ 295 K ) ⎝ 0.0821 ( ⎜ moliK ⎟ ⎠ ⎝ Time = (1.217892 mol) (28.8 g/mol) (1 min / 1.5 g) = 23.384 = 23 min a) d = M P / RT O2: d =

5.153

⎛ 1 atm ⎞ ⎜ ⎟ = 1.42772 = 1.43 g O2 / L Liatm ⎞ 760 torr ⎠ ⎛ 0.0821 273 + 0 ) K ) ⎝ (( ⎜ moliK ⎟ ⎝ ⎠ ( 48.00 g/mol )( 760 torr ) ⎛ 1 atm ⎞ O3: d = ⎜ ⎟ = 2.141585576 = 2.14 g O3 / L Liatm ⎞ ⎛ ⎝ 760 torr ⎠ ⎜ 0.0821 moliK ⎟ ( ( 273 + 0 ) K ) ⎝ ⎠ b) dozone / doxygen = (2.141585576) / (1.42772) = 1.5 The ratio of the density is the same as the ratio of the molar masses.

( 32.00 g/mol )( 760 torr )

5.154

The balanced chemical equation is: 7 F2(g) + I2(s) → 2 IF7(g). It will be necessary to determine the limiting reactant. ( 350. torr )( 2.50 L ) ⎛ 1 atm ⎞ ⎛ 2 mol IF7 ⎞ Moles IF7 from F2 = ⎟ ⎜ ⎟⎜ Liatm ⎞ ⎛ ⎝ 760 torr ⎠ ⎝ 7 mol F2 ⎠ ⎜ 0.0821 moliK ⎟ ( 250. K ) ⎝ ⎠ = 0.016026668 mol IF7 (unrounded) ⎛ 1 mol I 2 ⎞ ⎛ 2 mol IF7 ⎞ Moles IF7 from I2 = ( 2.50 g I 2 ) ⎜ ⎟⎜ ⎟ = 0.019700551 mol IF7 (unrounded) ⎝ 253.8 g I 2 ⎠ ⎝ 1 mol I2 ⎠ F2 is limiting. Mole I2 remaining = original amount of moles of I2 – number of I2 moles reacting with F2 ⎛ 1 mol I 2 ⎞ ( 350. torr )( 2.50 L ) ⎛ 1 atm ⎞ ⎛ 1 mol I2 ⎞ Mole I2 remaining = ( 2.50 g I 2 ) ⎜ ⎟ ⎟ – ⎜ ⎟⎜ Liatm ⎞ ⎛ ⎝ 760 torr ⎠ ⎝ 7 mol F2 ⎠ ⎝ 253.8 g I 2 ⎠ ⎜ 0.0821 moliK ⎟ ( 250. K ) ⎝ ⎠ = 1.83694 x 10–3 mol I2 (unrounded)

5-36

Total moles of gas = (0 mol F2) + (0.016026668 mol IF7) + (1.83694 x 10–3 mol I2) = 0.0178636 mol gas (unrounded) Liatm ⎞ ⎛ ( 0.0178636 mol ) ⎜ 0.0821 ( 550. K ) moliK ⎟ ⎝ ⎠ = 0.322652 = 0.323 atm Ptotal = nRT / V = ( 2.50 L ) Piodine = Xiodine Ptotal = [(1.83694 x 10–3 mol I2) / (0.0178636 mol)] (0.322652 atm) = 0.03317877 = 0.0332 atm

5-37

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    Gas Laws Name Institution The 3 Gas Laws Introduction The three gas laws include: Gay-Lussac’s law, Boyle’s law and Charles’ law. When combined with Avogadro’s law the three laws can be generalized by the ideal gas law. Gases possess observable properties which include, mass, pressure (P), thermodynamic temperature (T) and volume (V). These properties are related to each other and the state of a gas is determined by their values. The three laws are derived from these properties. Discussion…

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