CHEM 101A ARMSTRONG SOLUTIONS TO TOPIC C PROBLEMS 1) This problem is a straightforward application of the combined gas law. In this case, the temperature remains the same, so we can eliminate it from the combined gas law equation, giving P 1 V 1 P V (Boyle s law). 1.33 atm 451 ml 1.65 atm V Solving this equation gives V 364 ml. ) This is another application of the combined gas law. Be sure to convert all temperatures to kelvin! 744 torr 3.85 L 591 torr 4.13 L 304 K T Solving this equation gives T 59 K 14ºC. 3) This is an application of the ideal gas law. We are given T and P, and we can calculate n from the given mass of hydrogen. Our only unknown in the ideal gas law is then V. To begin, 5.3 g of hydrogen equals 1.5496 moles (note that elemental hydrogen is diatomic, so the molar mass is.016 g/mol). When we substitute into the ideal gas law, the temperature must be in kelvins, as it is in all gas law equations. Also, we must use the correct value of R. In this case, our pressure is given in atm, so we use R 0.0806 L atm/mol K. 3.88 atm V 1.5496 mol 0.0806 L atm/mol K 304 K Solving this equation gives V 80.7 L. 4) We can use the ideal gas law to calculate the number of moles of methane, then convert moles into grams. However, we must first account for the fact that the methane is collected over water. P methane + (vapor pressure of water) P total Now we can calculate the pressure that the methane exerts: P methane + 19.8 torr 731 torr P methane 711. torr Next, we substitute into the ideal gas law, being sure to use a kelvin temperature. Since our pressure is in torr, we use R 6.36 L torr/mol K. Using this value of R in turn means that our volume must be in liters. (Keeping the units straight in the ideal gas law isn t actually hard at all just be sure that the units in R match the units you use for P, V, n and T!) 711. torr 0.500 L n 6.36 L torr/mol K 95 K Solving this equation gives n 0.00966504 mol. Finally, we convert moles into grams, using the molar mass of CH 4 (16.04 g/mol). Doing so gives us 0.155 g of methane. 5) First, we use stoichiometry to calculate the number of moles of NO that will be formed. Then we use the ideal gas law to determine the volume of the NO. The stoichiometry gives us: 1 mol Cu 4.17 g Cu 63.55 g Cu mol NO 0.1314 mol NO 1 mol Cu Now we use the ideal gas law: 1.0 atm V 0.1314 mol 0.0806 L atm/mol K 304 K
Solving this equation gives V 3.0 L. 6) Each gas behaves as if the other gas was not present. When the valve is opened, the He will expand to fill the entire system, which has a volume of 4 ml + 761 ml 1183 ml: 817 torr 4 ml P 1183 ml P 91.44 torr (final pressure of He) The Ar will also expand to fill the entire system: 685 torr 761 ml P 1183 ml P 440.65 torr (final pressure of Ar) The total pressure in the container is the sum of these two partial pressures: 91.44 torr + 440.65 torr 73 torr 7) a) We can set up an ICE table using partial pressures whenever all of the reactants and products are gases and the volume and temperature do not change. Doing so here gives: C H 6 + 7 Cl CCl 4 + 6 HCl Initial 35.5 torr 318 torr 0 torr 0 torr Change 35.5 torr 48.5 torr + 71.0 torr + 13.0 torr End 0 torr 69.5 torr 71.0 torr 13.0 torr The initial values are given in the problem. To find the changes, use the mole ratios. For instance, to find the amount of Cl that is used up: 7 torr Cl 35.5 torr C H 6 1 torr C H 6 48.5 torr Cl (Note that this calculation assumes that C H 6 is the limiting reactant. Since the amount of Cl used up is less than the amount we start with, the assumption is correct.) To find the amount of CCl 4 that is formed: 35.5 torr C H 6 torr CCl 4 1 torr C H 6 71.0 torr CCl 4 The amount of HCl that is formed is calculated similarly. Finally, we calculate the final amounts doing arithmetic. The final mixture contains 69.5 torr Cl, 71.0 torr CCl 4, and 13.0 torr HCl. b) The total pressure is the sum of the partial pressures, which equals 353.5 torr. 8) Since there is oxygen in the container after the reaction, we can conclude that Cr + is the limiting reactant. This allows us to calculate the amount of oxygen that was consumed in the reaction using stoichiometry. 0.131 mol Cr+ 1 mol O 0.013 L 1 L 4 mol Cr 6.9757 + 10-4 mol O consumed Now, let s calculate the number of moles of O in the container after the reaction, using the ideal gas law. The volume of the oxygen is not equal to the container volume (56 ml), because the solution occupies 1.3 ml, so the volume of the oxygen is only 540.7 ml 0.5407 L. 119 torr 0.5407 L n 6.36 L torr/mol K 94 K Solving this equation gives n 0.0035095 mol O left over The total number of moles of oxygen in the container before the reaction was: 0.0035095 mol + 6.9757 x 10-4 mol 0.004071 mol O
Finally, we can use the ideal gas law to calculate the original pressure of the oxygen. Before the Cr + solution was added, the O occupied the entire container (56 ml 0.56 L). P 0.56 L 0.004071 mol 6.36 L atm/mol K 94 K Solving this equation gives P 137 torr. 9) In this problem, we must consider two processes: each gas expands to fill the entire system, and the gases react with one another. These two processes actually occur at the same time, but we can treat them as if they occur one at a time, with the expansion occurring first. The C 3 H 6 expands from 367 ml to 969 ml (367 ml + 60 ml): 663 torr 367 ml P 969 ml P 51.1 torr The O expands from 60 ml to 969 ml: 16 torr 60 ml P 969 ml P 1376.7 torr Now, we consider the reaction. Since the volume of the entire system and the temperature are constant, we can do our stoichiometry using partial pressures. (Note that we can not use the pressures given in the problem to do stoichiometry!) Here is an ICE table for the reaction. C 3 H 6 + 9 O 6 CO + 6 H O Initial 51.1 torr 1376.7 torr 0 torr 0 torr Change 51.1 torr 1130.0 torr + 753.3 torr + 753.3 torr End 0 torr 46.7 torr 753.3 torr 753.3 torr As in problem 7, I assumed that C 3 H 6 was the limiting reactant and calculated the amount of O that would react: 9 torr O 51.1 torr C 3 H 6 1130.0 torr O consumed torr C 3 H 6 Since this amount is less than the initial O pressure, my assumption was correct. The amounts of CO and H O are calculated similarly. Finally, we calculate the total pressure in the system: 46.7 torr + 753.3 torr + 753.3 torr 1753 torr 10) This problem is similar to #9, except that we do not know the initial pressure of the oxygen. Let s assign it a variable: the initial pressure of oxygen is x torr. Now we proceed just as we did in problem 9. The C 3 H 6 expands from 617 ml to 1550 ml: 317 torr 617 ml P 1550 ml P 16. torr The O expands from 933 ml to 1550 ml: x torr 933 ml P 1550 ml P 0.60194x torr
Note that the final pressure of oxygen contains the variable x. Now we consider the reaction. In this case, we have no obvious way of determining the limiting reactant, so let s just assume that it is C 3 H 6 and see what we get C 3 H 6 + 9 O 6 CO + 6 H O Initial 16. torr 0.60194x torr 0 torr 0 torr Change 16. torr 567.8 torr + 378.6 torr + 378.6 torr End 0 torr (0.60194x 567.8) torr 378.6 torr 378.6 torr The problem tells us that the total pressure after the reaction is 1133 torr. Therefore, the expressions in the End row of our ICE table must add up to 1133 torr. (0.60194x 567.8) + 378.6 + 378.6 1133 Solving this equation gives x 1568 torr. Is this a reasonable value? If so, it should give positive pressures for every gas in the final mixture. The pressures of CO and H O are obviously positive, but is 0.60194x 567.8 a positive number? Let s check: 0.60194(1568) 567.8 376.0 torr This is a positive value, so 1568 torr is a reasonable answer. Now let s see what we get if we assume that O is the limiting reactant. In this case, all of the Change entries are going to contain x, because they will be based on the O pressure. For instance, the amount of C 3 H 6 we consume will be: 0.60194x torr O torr C H 3 6 9 torr O 0.13376x torr C 3 H 6 Here is the ICE table we get when we assume that O is the limiting reactant: C 3 H 6 + 9 O 6 CO + 6 H O Initial 16. torr 0.60194x torr 0 torr 0 torr Change 0.13376x torr 0.60194x torr + 0.4019x torr + 0.4019x torr End (16. 0.13376x) torr 0 torr 0.4019x torr 0.4019x torr Once again, the expressions in the End row must add up to 1133 torr. (16. 0.13376x) + 0.4019x + 0.4019x 1133 Solving this equation gives x 1505 torr. This is a different value from our previous answer (1568 torr). Are there two possible answers? We must again make sure that our final pressures are all positive. The pressures of CO and H O will clearly be positive, but what about the pressure of C 3 H 6? 16. 0.13376(1505) 75.1 torr The final pressure of C 3 H 6 turns out to be a negative number, which is an absurd result. This is an example of a situation where we get two answers that satisfy the mathematics, but only one of the two is physically reasonable. (The final pressure turns out to be negative because we use more C 3 H 6 than we have: 0.13376(1505) 01.3 torr used up, which is greater than the 16. torr we started with).
Therefore, we can conclude that the initial pressure of O was 1568 torr. 11) If you want to memorize the formula in Zumdahl, you can, but you will not be given this formula on your exam data sheet, because you can solve gas density problems using the ideal gas law. The key is to recognize that the density is the number of grams of gas in one liter, and we can calculate the number of grams of any gas in one liter using PVnRT and then converting moles to grams. Start by calculating the number of moles of CO in 1 L: 855 torr 1 L n 6.36 L torr/mol K 34 K n 0.04317 mol Then convert this number of moles to grams: 0.04317 mol 44.01 g 1.864 g 1 mol Since this is the mass of 1 L of CO, the density is 1.86 g/l. 1) First, we can use the percent composition to find the empirical formula of the compound. You did this kind of problem in topic A, so I will just outline the solution here: 100 g of compound contains 88.8 g of carbon 7.394 mol C 11. g of hydrogen 11.11 mol H Dividing both of these numbers by the smaller (7.394) gives: 1 mol C/mol C 1.50 mol H/mol C Multiplying both of these by mol C converts them to whole numbers: mol C 3 mol H So the empirical formula of the compound is C H 3. Now we use the vapor density to determine the molar mass. The density tells us that 1 L of the compound weighs 1.91 g. We use the ideal gas law to determine the number of moles in 1 L. 675 torr 1 L n 6.36 L torr/mol K 306.35 K n 0.035333 mol So 0.035333 moles of the compound weighs 1.91 grams. The molar mass of the compound is therefore: 1.91 g 54.1 g/mol 0.035333 mol If the compound were C H 3, the molar mass would be about 7 g/mol. The actual molar mass is twice this number, so the molecular formula must be C 4 H 6. 13) There are several ways to approach this problem; here is one of them Let s start by using the ideal gas law to get an expression for the number of moles of oxygen in 1 liter:
n PV 1.00 atm 1 L 1 L atm RT RT RT The temperature is unknown, so we cannot insert a value for T here. We can insert a value for R (we would use 0.0806 L atm/mol K), but there is no need to do so yet. Next, let s convert this number of moles into a mass, using the molar mass of oxygen (3.00 g/mol): 1 L atm 3.00 g 3 g L atm/mol RT 1 mol RT This expression tells us that the mass (in grams) of O in 1 L of the sample is 3 divided by RT. The g L atm/mol in the numerator is required to make the units cancel. Now, the nitrogen and the oxygen have the same density, so the mass of N in a 1 L sample is also 3/RT. Let s convert that to moles, using the molar mass of N (8.0 g/mol): 3 g L atm/mol 1 mol 1.14 L atm RT 8.0 g RT This expression tells us that the number of moles of N in 1 L of the sample is 1.14 divided by RT. Finally, we can substitute this expression into the ideal gas law to calculate a pressure for the N : 1.14 L atm P 1 L RT P 1 L 1.14 L atm R T 1.14 L atm P 1.14 atm 1 L Note that since the two gases are at the same temperature, the value of T cancels out here (as does the value of R). The pressure of N must be higher because N is a lighter gas than O ; in order to have equal masses, we need more molecules of N, which gives us a higher pressure. 14) First, let s calculate the number of moles of gaseous H S in the original container, using the ideal gas law: 169.1 torr.50 L n 6.36 L torr/mol K 86.15 K n 0.03691 mol Next, let s calculate the number of moles of gaseous H S in the container after the system has reached equilibrium. The gas volume is now.50 L 0.000 L.30 L, and the partial pressure of the H S is 155.9 torr 11. torr 144.7 torr. 144.7 torr.30 L n 6.36 L torr/mol K 86.15 K n 0.018651 mol Now we can calculate the number of moles of H S that dissolved; it is the difference between the initial and final moles of gaseous H S. 0.03691 mol 0.018651 mol 0.005040 mol
The molar concentration of H S in the water is therefore: 0.005040 mol 0.000 L 0.05 M 15) a) For a moving object, KE 1 / mv. In the metric system, the unit of energy is the joule, which equals a kg m /sec. Therefore, the mass must be in kg and the velocity must be in m/sec. We must start by converting Janice s mass from pounds to kilograms. 454 g 131 pounds 1 pound 1 kg 59.474 kg 1000 g Now we can calculate Janice s speed. In order to make the units cancel, we must make the substitution J kg m /sec KE 1 mv 1.00 J 1 (59.474 kg)v 1.00 kg m /sec 1 (59.474 kg)v 0.03368 m /sec v v 0.03368 m /sec 0.183 m/sec (This is a very slow walking speed. At this speed, it would take Janice about ten minutes to walk the length of a football field, including the end zones.) b) Again, we use KE 1 / mv. KE 1 59.474 kg ( ) 1.34 m/sec 53.4 kg m /sec 53.4 J ( ) 16) a) If one mole (6.0 x 10 3 ) atoms has an energy of 615 joules, then one atom must have 615 J divided by Avogadro s number. Writing this as a formal unit conversion looks like this: 1 mol 1 atom 6.0 10 3 atoms 615 J 1.03 10-0 J 1 mol b) Kinetic energy equals 1 / mv, but we cannot calculate the velocity unless we know the mass. We have two options here Option 1: do everything on a per mole basis. The kinetic energy of 1 mole is 615 J, and the mass of 1 mole of neon is 0.18 g 0.0018 kg. 615 J 1 (0.0018 kg)v 615 kg m /sec 1 (0.0018 kg)v 6.160 10 5 m /sec v v 6.160 10 5 m /sec 784.8 m/sec
Option : do everything based on a single atom of neon. We already know that one neon atom has a kinetic energy of 1.03 x 10-0 J. A single neon atom weighs 1 mol 1 atom 6.0 10 3 atoms 0.18 g 1 kg 1 mol 1000 g 3.351 10-6 kg Now we can use KE 1 / mv. 1.03 10-0 J 1 (3.361 10-6 kg)v 1.03 10-0 kg m /sec 1 (3.361 10-6 kg)v 6.160 10 5 m /sec v v 6.160 10 5 m /sec We end up with the same velocity, as we must. 784.8 m/sec c) To get the total kinetic energy, we multiply the average KE (in J/mol) by the number of moles of neon. 1.50 g of Ne equals 0.061943 mol, so we have: 0.061943 mol 615 J 1 mol 385.0 J Strictly, this only works if we have a very large number of atoms. In this case, 0.061943 moles equals about 37,300,000,000,000,000,000,000 atoms, which is certainly a very large number! d) For any gas, the average KE equals 3 / RT, where R 8.314 J/mol K: 615 J/mol 3 T 498.4 K (5. C) ( 8.314 J/mol K) T e) The most probable KE for a gas equals 1 / RT. Now that we know the temperature, we can calculate KE mp : KE mp 1 ( 8.314 J/mol K) 498.4 K 07 J/mol f) For the root-mean-square speed, we have:
v rms 3RT M ( )( 498.4 K) 3 8.314 J/mol K 0.0018 kg/mol 6.160 10 5 J/kg Before we take the final square root, let s make sense of the unit. A joule is a kg m /sec, so a J/kg equals a m /sec (because the kg units cancel out). v rms 6.160 10 5 m /sec 784.9 m/sec g) For the average speed, we have: v ave 8RT πm ( )( 498.4 K) 3.14159( 0.0018 kg/mol) 8 8.314 J/mol K 5.9 10 5 J/kg 5.9 10 5 m /sec 73.1 m/sec h) For the most probable speed, we have: v mp RT M ( )( 498.4 K) 8.314 J/mol K 0.0018 kg/mol 4.107 10 5 J/kg 4.107 10 5 m /sec 640.8 m/sec 17) a) First, we need to calculate the mass of a single water molecule. One mole of H O weighs 18.016 g, so one molecule weighs 18.016 g divided by Avogadro s number, which equals.9917 x 10-3 g.9917 x 10-6 kg. Now we can calculate the kinetic energy of the molecule.
( ) 46 m/sec KE 1.9917 10-6 kg.71 10-1 kg m /sec.71 10-1 J ( ) b) One mole equals 6.0 x 10 3 molecules, so the kinetic energy expressed in J/mol is:.7146 10-1 J 6.0 10 3 mol -1 1.63 10 3 J/mol c) Calculating the average speed of water molecules at 5 C gives us v ave 8RT πm ( )( 98.15 K) 3.14159( 0.018016 kg/mol) 8 8.314 J/mol K 3.5037 10 5 J/kg 3.5037 10 5 m /sec 591.9 m / sec Our original water molecule was moving at 46 m/sec, so it is moving slower than the average speed of water molecules at 5 C. 18) a) The average kinetic energy of a gas depends only on the temperature, so the krypton would have to be at 5ºC (the same temperature as the argon). b) The molecular speed depends on both the temperature and the molar mass. The most straightforward way (although not the shortest way) to answer this question is to calculate the average molecular speed of the argon, and then use that speed to calculate the temperature of the krypton. The average speed of the argon at 5ºC is: v ave 8RT πm 8(8.314 J/mol K)(98 K) 3.14159(0.03995 kg/mol) 397.4 m/sec Using this speed, we can calculate the necessary temperature for the krypton: 397.4 m/sec ( 397.4 m/sec) T 8(8.314 J/mol K)T 3.14159(0.08380 kg/mol) 8(8.314 J/mol K)T 3.14159(0.08380 kg/mol) ( 397.4 m/sec) (3.14159)(0.08380 kg/mol) 8(8.314 J/mol K) 65 K (35ºC)
The quick way to answer this question is to recognize that if the molecular speeds are equal, we can say: 8RT Kr 8RT Ar πm Kr πm Ar Since the factor 8R/π is the same for both expressions, this reduces to T Kr T Ar M Kr M Ar Substituting in the molar masses of Kr and Ar and the temperature of the Ar allows us to calculate the temperature of the Kr. 19) a) The two gases have the same average kinetic energy, because kinetic energy depends only on temperature, and both gases are at the same temperature. b) Average speed is directly related to temperature and inversely related to molar mass, but since the gases are at the same temperature, we need only consider the molar mass. The CO has the higher average speed, because the molar mass of CO is lower than the molar mass of CO. c) The two gases have the same fraction of molecules with KE > 5 kj/mol. As in part a, the kinetic energy distribution depends only on temperature. d) The CO has the higher fraction of molecules with speeds greater than 500 m/sec, for the same reason that it has the higher average speed. If two gases are at the same temperature, the lighter gas will always have the greater fraction of fast-moving molecules, regardless of what cutoff we use to define fast. e) Since the gases are have the same temperature, pressure, and volume, they must contain the same number of moles. However, a mole of CO weighs more than a mole of CO, so the CO weighs more. 0) a) The O has the higher average kinetic energy, because it is at a higher temperature. b) For this question, we need to calculate the average speeds, because the two relevant factors (molar mass and temperature) are opposing each other; the O is hotter (which would give it a higher average speed), but it also has a higher molar mass (which would give it a lower average speed). Calculating the average speeds gives the following values: For O : 480 m/sec For NO: 459 m/sec So the O has the higher average speed. c) The NO has a higher fraction of molecules with KE < 5 kj/mol. Since the NO is at the lower temperature, it has more low-energy molecules, regardless of the cutoff we use to define low. d) The NO has a higher fraction of molecules with speeds below 500 m/sec. In part b, we found that O has the higher average speed, so O must also have the greater fraction of fastmoving molecules. Therefore, NO must have the higher fraction of slow-moving molecules. e) We need to use PVnRT to calculate the moles of each gas, then convert to grams. The volume is not given, but it is the same for both gases, so we can just leave it as a variable. We can also leave R as a variable (although if you want to plug it in, you can). For the O : n PV RT 1 atm V R 348 K 0.0087 V R
mass 0.0087 V R 3.00 g 1 mol 0.090 V R For the NO: n PV RT 1 atm V R 98 K 0.00336 V R mass 0.00336 V R 30.01 g 1 mol 0.101 V R Since 0.101V/R is larger than 0.090V/R, the NO weighs more. 1) a) The most probable kinetic energy is around 700 J/mol. This is the kinetic energy that corresponds to the highest point on the curve. (This is a rough estimate anything within 100 J/mol of this value is a reasonable answer.) b) Since KE mp 1 / RT, we can calculate a rough value for the temperature from the most probable KE we estimated in part a. You should get somewhere around 170 K. (Using 700 J/mol gives T 168 K.) c) When x 000 J/mol, y 0.0003. This is the fraction of molecules that have kinetic energies between 1999.5 J/mol and 000.5 J/mol. (A fraction of 0.0003 equals 0.03%.) d) This value tells us that the fraction of molecules that have kinetic energies between 000 J/mol and 4000 J/mol is 0.59. Another way to say this is that 5.9% of the molecules have kinetic energies in this range. ) a) Curve B represents the KE distribution for Ar(g) at 300 K. Neon at 300 K and argon at 300 K have exactly the same KE distribution, because they are at the same temperature. b) Curve C could represent the KE distribution for Ne(g) at 600 K. As the temperature increases, the curve shifts toward higher kinetic energies. 3) a) Curve A could represent the speed distribution for H (g) at 15ºC. As the temperature decreases, the distribution shifts toward lower speeds. b) Curve A could represent the speed distribution for N (g) at 5ºC. Comparing the distributions for two different gases at the same temperature (H and N in this case), we expect the heavier gas to have slower speeds. c) Curve B could represent the speed distribution for He(g) at 319ºC. In this case, we need to compare the average speeds for the two gases. When we calculate the average speed for each gas, we get: H at 5ºC: v ave 1769.5 m/sec He at 319ºC: v ave 1769.7 m/sec The two gases have essentially the same average speed. Therefore, they must have the same distribution of speeds. 4) a) The most probable speed is around 40 m/sec. This is the speed that corresponds to the highest point on the curve. (This is a rough estimate anything within 10 m/sec of this value is a reasonable answer.)
RT b) Since v mp, we can calculate a rough value for the molar mass from the most M probable speed we estimated in part a. Using v mp 40 m/sec and T 473 K gives M 0.137 kg/mol 137 g/mol. (Did you remember that M is in kg/mol in this kind of formula???) This value is closest to the molar mass of xenon (131.3 g/mol). c) When x 00 m/sec, y 0.0033. This is the fraction of molecules that have speeds between 199.5 J/mol and 00.5 m/sec. (A fraction of 0.0033 equals 0.33%.) d) This value tells us that the fraction of molecules that have speeds above 400 m/sec is 0.111. Another way to say this is that 11.1% of the molecules have speeds in this range. e) The entire area under the curve is 1 ( 100%), so the area of the region that isn t shaded must be 1 0.111 0.889. This value tells us that the fraction of molecules having speeds below 400 m/sec is 0.889. 5) We can use Graham s law of effusion to calculate the molar mass of the gas: rate 1 rate M M 1 Let s let gas 1 be argon and gas be the unknown substance. (You can do it the other way, too.) To calculate the rates, we need to express the times in a single unit, either minutes or seconds. I ll express the times in seconds. For the argon: time of effusion (5 min x 60 sec/min) + 13 sec 313 sec rate 1 5.00 ml/313 sec 0.01597 ml/sec M 1 39.97 g/mol (it s okay to use grams per mole in Graham s law) For the unknown: time of effusion (6 min x 60 sec/min) + 10 sec 370 sec rate 5.00 ml/370 sec 0.01351 ml/sec M is unknown Squaring both sides gives 0.01597 ml/sec 0.01351 ml/sec M 39.97 g/mol 0.01597 ml/sec 0.01351 ml/sec M 39.97 g/mol 0.01597 ml/sec 0.01351 ml/sec 39.97 g/mol M M 55.9 g/mol
The empirical formula is CH, which has a molar mass of about 14 g/mol. 55.9/14 4 So the molecular formula is C 4 H 8. 6) a) Water has a higher a value because the attraction between H O molecules is stronger than the attraction between N molecules. This agrees with the fact that H O is a liquid at room temperature while N is a gas; the attraction between H O molecules is strong enough to keep them in contact with one another at room temperature. b) Nitrogen has a higher b value because N molecules have a larger volume than H O molecules. This is not surprising; it is reasonable that a nitrogen atom should be larger than two hydrogen atoms. (N and O should be similar sizes.) 7) a) Using the ideal gas law gives us: 150 atm 53 L n 0.0806 L atm/mol K 53 K n 884.6 mol 884 moles of O (to three sig figs) b) The van der Waals equation for gases is P + a n V V - nb ( ) nrt For oxygen, a 1.36 atm L /mol and b 0.0318 L/mol. We already have n 884.6 mol, V 53 L, and T 53 K. Substituting these values into the equation gives us: P + 1.36 atm L /mol 884.6 mol 53 L - 884.6 mol 0.0318 L/mol 53 L ( ) 884.6 mol 0.0806 L atm/mol K 53 K Doing all of the arithmetic and canceling units gives us:
( P + 16.6134 atm) ( 4.881 L) 37950.1 L atm P + 16.6134 atm 37950.1 L atm 4.881 L P + 16.6134 atm 168.756 atm P 15.143 atm 15 atm (to 3 sig figs) This is slightly above the safety limit for the reactor (150 atm), so the engineer should decrease the maximum number of moles of O accordingly. 8) a) The actual pressure is lower than the ideal gas prediction when the volume is greater than about 0.1 L. Over this range, P real /P ideal is less than 1, which means that P real is less than P ideal. In this range, the attraction between O molecules reduces the force with which the molecules collide with the container walls, giving a lower pressure than we expect based on the ideal gas model. b) The actual pressure is higher than the ideal gas prediction when the volume is less than about 0.1 L. In this range, the O molecules are so tightly packed together that there is almost no empty space between them. Reducing the volume requires extremely high pressures, because you must compress the molecules themselves. 9) a) The kinetic energy of the molecules in a gas is proportional to the temperature (by the kinetic theory); raising the temperature increases the kinetic energy. Since the kinetic energy is related to the speed of the molecules by KE 1 / mv, increasing the kinetic energy is equivalent to increasing the molecular speeds. The molecules therefore collide with the container walls more often and at higher speeds. Since pressure is simply the collective result of molecules colliding with the walls, the pressure increases. b) Gases are mostly empty space (by the kinetic theory), so compressing a gas simply involves moving the molecules closer to one another. c) The kinetic energy of a mole of a gas is related only to its temperature (by the kinetic theory), and is not a function of the molar mass of the gas. Therefore, the amount of energy you must put into a mole of a gas in order to raise the temperature is likewise a function only of the temperature; it is the difference between the energy of the gas when it s cool and the energy of the gas when it s hot. Note: strictly, this is only valid for monatomic gases. For gases that are molecules, the kinetic energy of the gas depends on the structure of the gas; the more complex the molecules, the higher the energy.