CHEM 101A 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 P1V1 = P2V2 (Boyle’s law).

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1.33 atm × 451 mL = 1.65 atm × V2 Solving this equation gives V2 = 364 mL. 2) This is another application the combined gas law. Be sure to convert all temperatures to kelvin!

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744 torr × 3.85 L304 K

= 591 torr × 4.13 LT2

Solving this equation gives T2 = 259 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 25.3 g of hydrogen equals 12.5496 moles (note that elemental hydrogen is diatomic, so the molar mass is 2.016 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 = 0.08206 L·atm/mol·K.

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3.88 atm × V = 12.5496 mol × 0.08206 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. Pmethane + (vapor pressure of water) = Ptotal To reconcile units, we convert 2645 Pa into torr:

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2645 Pa × 1 atm101,325 Pa

× 760 torr1 atm

= 19.84 torr

Now we can calculate the pressure that the methane exerts: Pmethane + 19.84 torr = 731 torr Pmethane = 711.16 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 = 62.36 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!)

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711.16 torr × 0.2500 L = n × 62.36 L ⋅ torr/mol ⋅K × 295 K Solving this equation gives n = 0.0096645 mol. Finally, we convert moles into grams, using the molar mass of CH4 (16.042 g/mol). Doing so gives us 0.155 g of methane. 5) First, we use stoichiometry to calculate the number of moles of NO2 that will be formed. Then we use the ideal gas law to determine the volume of the NO2. The stoichiometry gives us:

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4.17 g Cu × 1 mol Cu63.55 g Cu

× 2 mol NO2

1 mol Cu = 0.13124 mol NO2

Now we use the ideal gas law:

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1.022 atm × V = 0.13124 mol × 0.08206 L ⋅ atm/mol ⋅K × 304 K Solving this equation gives V = 3.20 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 422 mL + 761 mL = 1183 mL:

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817 torr × 422 mL = P2 × 1183 mL P2 = 291.44 torr (final pressure of He) The Ar will also expand to fill the entire system:

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685 torr × 761 mL = P2 × 1183 mL P2 = 440.65 torr (final pressure of Ar) The total pressure in the container is the sum of these two partial pressures: 291.44 torr + 440.65 torr = 732 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: C2H6 + 7 Cl2 → 2 CCl4 + 6 HCl Initial 35.5 torr 318 torr 0 torr 0 torr Change –35.5 torr –248.5 torr + 71.0 torr + 213.0 torr End 0 torr 69.5 torr 71.0 torr 213.0 torr The initial values are given in the problem. To find the changes, use the mole ratios. For instance, to find the amount of Cl2 that is used up:

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35.5 torr C2H6 × 7 torr Cl2

1 torr C2H6

= 248.5 torr Cl2

(Note that this calculation assumes that C2H6 is the limiting reactant. Since the amount of Cl2 used up is less than the amount we start with, the assumption is correct.) To find the amount of CCl4 that is formed:

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35.5 torr C2H6 × 2 torr CCl4

1 torr C2H6

= 71.0 torr CCl4

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 Cl2, 71.0 torr CCl4, and 213.0 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 Cr2+ is the limiting reactant. This allows us to calculate the amount of oxygen that was consumed in the reaction using stoichiometry.

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0.0213 L × 0.131 mol Cr2+

1 L × 1 mol O2

4 mol Cr2+ = 6.9757 × 10-4 mol O2 consumed

Now, let’s calculate the number of moles of O2 in the container after the reaction, using the ideal gas law. The volume of the oxygen is not equal to the container volume (562 mL), because the solution occupies 21.3 mL, so the volume of the oxygen is only 540.7 mL = 0.5407 L.

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119 torr × 0.5407 L = n × 62.36 L ⋅ torr/mol ⋅K × 294 K

Solving this equation gives n = 0.0035095 mol O2 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 = 0.0042071 mol O2 Finally, we can use the ideal gas law to calculate the original pressure of the oxygen. Before the Cr2+ solution was added, the O2 occupied the entire container (562 mL = 0.562 L).

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P × 0.562 L = 0.0042071 mol × 62.36 L ⋅ atm/mol ⋅K × 294 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 C3H6 expands from 367 mL to 969 mL (367 mL + 602 mL):

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663 torr × 367 mL = P2 × 969 mLP2 = 251.1 torr

The O2 expands from 602 mL to 969 mL:

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2216 torr × 602 mL = P2 × 969 mLP2 = 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. 2 C3H6 + 9 O2 → 6 CO2 + 6 H2O Initial 251.1 torr 1376.7 torr 0 torr 0 torr Change –251.1 torr –1130.0 torr + 753.3 torr + 753.3 torr End 0 torr 246.7 torr 753.3 torr 753.3 torr As in problem 7, I assumed that C3H6 was the limiting reactant and calculated the amount of O2 that would react:

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251.1 torr C3H6 × 9 torr O2

2 torr C3H6

= 1130.0 torr O2 consumed

Since this amount is less than the initial O2 pressure, my assumption was correct. The amounts of CO2 and H2O are calculated similarly. Finally, we calculate the total pressure in the system: 246.7 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 C3H6 expands from 617 mL to 1550 mL:

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317 torr × 617 mL = P2 × 1550 mLP2 = 126.2 torr

The O2 expands from 933 mL to 1550 mL:

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x torr × 933 mL = P2 × 1550 mLP2 = 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 C3H6 and see what we get… 2 C3H6 + 9 O2 → 6 CO2 + 6 H2O Initial 126.2 torr 0.60194x torr 0 torr 0 torr Change – 126.2 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 CO2 and H2O 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 O2 is the limiting reactant. In this case, all of the “Change” entries are going to contain x, because they will be based on the O2 pressure. For instance, the amount of C3H6 we consume will be:

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0.60194x torr O2 × 2 torr C3H6

9 torr O2

= 0.13376x torr C3H6

Here is the ICE table we get when we assume that O2 is the limiting reactant: 2 C3H6 + 9 O2 → 6 CO2 + 6 H2O Initial 126.2 torr 0.60194x torr 0 torr 0 torr Change – 0.13376x torr – 0.60194x torr + 0.40129x torr + 0.40129x torr End (126.2 – 0.13376x)

torr 0 torr 0.40129x torr 0.40129x torr

Once again, the expressions in the “End” row must add up to 1133 torr. (126.2 – 0.13376x) + 0.40129x + 0.40129x = 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 CO2 and H2O will clearly be positive, but what about the pressure of C3H6? 126.2 – 0.13376(1505) = –75.1 torr

The final pressure of C3H6 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 C3H6 than we have: 0.13376(1505) = 201.3 torr used up, which is greater than the 126.2 torr we started with). Therefore, we can conclude that the initial pressure of O2 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 PV=nRT and then converting moles to grams. Start by calculating the number of moles of CO2 in 1 L:

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855 torr × 1 L = n × 62.36 L ⋅ torr/mol ⋅K × 324 Kn = 0.042317 mol

Then convert this number of moles to grams:

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0.042317 mol × 44.01 g1 mol

= 1.8624 g

Since this is the mass of 1 L of CO2, the density is 1.86 g/L. 12) 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 11.2 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 2 mol C converts them to whole numbers: 2 mol C 3 mol H So the empirical formula of the compound is C2H3. 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.

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675 torr × 1 L = n × 62.36 L ⋅ torr/mol ⋅K × 306.35 Kn = 0.035333 mol

So 0.035333 moles of the compound weighs 1.91 grams. The molar mass of the compound is therefore:

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1.91 g0.035333 mol

= 54.1 g/mol

If the compound were C2H3, the molar mass would be about 27 g/mol. The actual molar mass is twice this number, so the molecular formula must be C4H6.

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:

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n = PVRT

= 1.00 atm × 1 LRT

= 1 L ⋅ atmRT

The temperature is unknown, so we cannot insert a value for T here. We can insert a value for R (we would use 0.08206 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 (32.00 g/mol):

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1 L ⋅ atmRT

× 32.00 g1 mol

= 32 g ⋅L ⋅ atm/molRT

This expression tells us that the mass (in grams) of O2 in 1 L of the sample is 32 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 N2 in a 1 L sample is also 32/RT. Let’s convert that to moles, using the molar mass of N2 (28.02 g/mol):

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32 g ⋅L ⋅ atm/molRT

× 1 mol28.02 g

= 1.142 L ⋅ atmRT

This expression tells us that the number of moles of N2 in 1 L of the sample is 1.142 divided by RT. Finally, we can substitute this expression into the ideal gas law to calculate a pressure for the N2:

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P × 1 L = 1.142 L ⋅ atmRT

× R × T

P × 1 L = 1.142 L ⋅ atm

P = 1.142 L ⋅ atm1 L

= 1.14 atm

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 N2 must be higher because N2 is a lighter gas than O2; in order to have equal masses, we need more molecules of N2, which gives us a higher pressure. 14) First, let’s calculate the number of moles of gaseous H2S in the original container, using the ideal gas law:

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169.1 torr × 2.50 L = n × 62.36 L ⋅ torr/mol ⋅K × 286.15 Kn = 0.023691 mol

Next, let’s calculate the number of moles of gaseous H2S in the container after the system has reached equilibrium. The gas volume is now 2.50 L – 0.2000 L = 2.30 L, and the partial pressure of the H2S is 155.9 torr – 11.2 torr = 144.7 torr.

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144.7 torr × 2.30 L = n × 62.36 L ⋅ torr/mol ⋅K × 286.15 Kn = 0.018651 mol

Now we can calculate the number of moles of H2S that dissolved; it is the difference between the initial and final moles of gaseous H2S. 0.023691 mol – 0.018651 mol = 0.005040 mol The molar concentration of H2S in the water is therefore:

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0.005040 mol0.2000 L

= 0.025 M

15) a) For a moving object, KE = 1/2mv2. In the metric system, the unit of energy is the joule, which equals a kg·m2/sec2. 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.

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131 pounds × 454 g1 pound

× 1 kg1000 g

= 59.474 kg

Now we can calculate Janice’s speed. In order to make the units cancel, we must make the substitution J = kg·m2/sec2

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KE = 12 mv2

1.00 J = 12 (59.474 kg)v2

1.00 kg ⋅m2/sec2 = 12 (59.474 kg)v2

0.033628 m2/sec2 = v2

v = 0.033628 m2/sec2 = 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 = 1/2mv2.

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KE = 12 59.474 kg( ) 1.34 m/sec( )

2

= 53.4 kg ⋅m2/sec2

= 53.4 J

16) a) If one mole (6.022 x 1023) atoms has an energy of 6215 joules, then one atom must have 6215 J divided by Avogadro’s number. Writing this as a formal unit conversion looks like this:

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1 atom × 1 mol6.022 × 1023 atoms

× 6215 J1 mol

= 1.032 × 10-20 J

b) Kinetic energy equals 1/2mv2, 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 6215 J, and the mass of 1 mole of neon is 20.18 g = 0.02018 kg.

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6215 J = 12 (0.02018 kg)v2

6215 kg ⋅m2/sec2 = 12 (0.02018 kg)v2

6.160 × 105 m2/sec2 = v2

v = 6.160 × 105 m2/sec2 = 784.8 m/sec

Option 2: do everything based on a single atom of neon. We already know that one neon atom has a kinetic energy of 1.032 x 10-20 J. A single neon atom weighs…

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1 atom × 1 mol6.022 × 1023 atoms

× 20.18 g1 mol

× 1 kg1000 g

= 3.351 × 10-26 kg

Now we can use KE = 1/2mv2.

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1.032 × 10-20 J = 12 (3.361 × 10-26 kg)v2

1.032 × 10-20 kg ⋅m2/sec2 = 12 (3.361 × 10-26 kg)v2

6.160 × 105 m2/sec2 = v2

v = 6.160 × 105 m2/sec2 = 784.8 m/sec

We end up with the same velocity, as we must. c) To get the total kinetic energy, we multiply the average KE (in J/mol) by the number of moles of neon. 1.250 g of Ne equals 0.061943 mol, so we have:

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0.061943 mol × 6215 J1 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 3/2RT, where R = 8.314 J/mol·K:

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6215 J/mol = 32 × 8.314 J/mol ⋅K( ) × T

T = 498.4 K (225.2°C)

e) The most probable KE for a gas equals 1/2RT. Now that we know the temperature, we can calculate KEmp:

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KEmp = 12 × 8.314 J/mol ⋅K( ) × 498.4 K

= 2072 J/mol

f) For the root-mean-square speed, we have:

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vrms = 3RTM

= 3 8.314 J/mol ⋅K( ) 498.4 K( )

0.02018 kg/mol

= 6.160 × 105 J/kg

Before we take the final square root, let’s make sense of the unit. A joule is a kg·m2/sec2, so a J/kg equals a m2/sec2 (because the kg units cancel out).

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vrms = 6.160 × 105 m2/sec2

= 784.9 m/sec

g) For the average speed, we have:

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vave = 8RTπM

= 8 8.314 J/mol ⋅K( ) 498.4 K( )3.14159 0.02018 kg/mol( )

= 5.229 × 105 J/kg

= 5.229 × 105 m2/sec2

= 723.1 m/sec

h) For the most probable speed, we have:

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vmp = 2RTM

= 2 8.314 J/mol ⋅K( ) 498.4 K( )

0.02018 kg/mol

= 4.107 × 105 J/kg

= 4.107 × 105 m2/sec2

= 640.8 m/sec

17) a) First, we need to calculate the mass of a single water molecule. One mole of H2O weighs 18.016 g, so one molecule weighs 18.016 g divided by Avogadro’s number, which equals 2.9917 x 10-23 g = 2.9917 x 10-26 kg. Now we can calculate the kinetic energy of the molecule.

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KE = 12 2.9917 × 10-26 kg( ) 426 m/sec( )

2

= 2.71 × 10-21 kg ⋅m2/sec2

= 2.71 × 10-21 J

b) One mole equals 6.022 x 1023 molecules, so the kinetic energy expressed in J/mol is:

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2.7146 × 10-21 J × 6.022 × 1023 mol-1 = 1.63 × 103 J/mol 18) a) The average kinetic energy of a gas depends only on the temperature, so the krypton would have to be at 25ºC (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 25ºC is:

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vave = 8RTπM

= 8(8.314 J/mol ⋅K)(298 K)3.14159(0.03995 kg/mol)

= 397.4 m/sec

Using this speed, we can calculate the necessary temperature for the krypton:

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397.4 m/sec = 8(8.314 J/mol ⋅K)T3.14159(0.08380 kg/mol)

397.4 m/sec( )2 = 8(8.314 J/mol ⋅K)T

3.14159(0.08380 kg/mol)

T = 397.4 m/sec( )

2(3.14159)(0.08380 kg/mol)

8(8.314 J/mol ⋅K) = 625 K (352ºC)

The quick way to answer this question is to recognize that if the molecular speeds are equal, we can say:

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8RTKr

πMKr

= 8RTAr

πMAr

Since the factor 8R/π is the same for both expressions, this reduces to

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TKr

MKr

= TAr

MAr

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 CO2. 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 CO2 weighs more than a mole of CO, so the CO2 weighs more. 20) a) The O2 has the higher average kinetic energy, because it is at a higher temperature. b) For this question, we need to do a quick calculation of average speed, because the two relevant factors (molar mass and temperature) are opposing each other; the O2 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 O2: 480 m/sec For NO: 459 m/sec So the O2 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 O2 has the higher average speed, so O2 must also have the greater fraction of “fast-moving” molecules. Therefore, NO must have the higher fraction of “slow-moving” molecules. e) We need to use PV=nRT 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 O2:

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n = PVRT

= 1 atm × V R × 348 K

= 0.00287 VR

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mass = 0.00287 VR

× 32.00 g1 mol

= 0.0920 VR

For the NO:

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n = PVRT

= 1 atm × V R × 298 K

= 0.00336 VR

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mass = 0.00336 VR

× 30.01 g1 mol

= 0.101VR

Since 0.101V/R is larger than 0.0920V/R, the NO weighs more.

21) 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 KEmp = 1/2RT, 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 = 2000 J/mol, y ≈ 0.00023. This is the fraction of molecules that have kinetic energies between 1999.5 J/mol and 2000.5 J/mol. (A fraction of 0.00023 equals 0.023%.) d) This value tells us that the fraction of molecules that have kinetic energies between 2000 J/mol and 4000 J/mol is 0.259. Another way to say this is that 25.9% of the molecules have kinetic energies in this range. 22) a) Curve B could represent 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. 23) a) Curve A could represent the speed distribution for H2(g) at –125ºC. As the temperature decreases, the distribution shifts toward lower speeds. b) Curve A could represent the speed distribution for N2(g) at 25ºC. Comparing the distributions for two different gases at the same temperature (H2 and N2 in this case), we expect the heavier gas to have slower speeds. c) Curve B could represent the speed distribution for He(g) at 323ºC. 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: H2 at 25ºC: vave = 1769 m/sec He at 323ºC: vave = 1769 m/sec The two gases have the same average speed! Therefore, they must have exactly the same distribution of speeds. 24) a) The most probable speed is around 240 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.)

b) Since

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vmp = 2RTM

, we can calculate a rough value for the molar mass from the most

probable speed we estimated in part a. Using vmp = 240 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???) c) When x = 200 m/sec, y ≈ 0.0033. This is the fraction of molecules that have speeds between 199.5 J/mol and 200.5 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. 25) We can use Graham’s law of effusion to calculate the molar mass of the gas:

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rate 1rate 2

= M2

M1

Let’s let gas 1 be argon and gas 2 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 M1 = 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 2 = 5.00 mL/370 sec = 0.01351 mL/sec M2 is unknown

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0.01597 mL/sec0.01351 mL/sec

= M2

39.97 g/mol

Squaring both sides gives

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0.01597 mL/sec0.01351 mL/sec

2

= M2

39.97 g/mol

0.01597 mL/sec0.01351 mL/sec

2

× 39.97 g/mol = M2

M2 = 55.9 g/mol

The empirical formula is CH2, which has a molar mass of about 14 g/mol. 55.9/14 ≈ 4 So the molecular formula is C4H8. 26) a) Water has a higher a value because the attraction between H2O molecules is stronger than the attraction between N2 molecules. This agrees with the fact that H2O is a liquid at room temperature while N2 is a gas; the attraction between H2O molecules is strong enough to keep them in contact with one another at room temperature. b) Nitrogen has a higher b value because N2 molecules have a larger volume than H2O 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.)

27) a) Using the ideal gas law gives us:

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150 atm × 253 L = n × 0.08206 L ⋅ atm/mol ⋅K × 523 K

n = 884.26 mol= 884 moles of O2 (to three sig figs)

b) The van der Waals equation for gases is

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P + a nV

2

V - nb( ) = nRT

For oxygen, a = 1.36 atm·L2/mol2 and b = 0.0318 L/mol. We already have n = 884.26 mol, V = 253 L, and T = 523 K. Substituting these values into the equation gives us:

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P + 1.36 atm ⋅L2/mol2 884.26 mol253 L

2

253 L - 884.26 mol × 0.0318 L/mol( ) =

884.26 mol × 0.08206 L ⋅ atm/mol ⋅K × 523 K

Doing all of the arithmetic and canceling units gives us:

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P + 16.6134 atm( ) 224.881 L( ) = 37950.1 L ⋅ atm

P + 16.6134 atm = 37950.1 L ⋅ atm224.881 L

P + 16.6134 atm = 168.756 atm

P = 152.143 atm= 152 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 O2 accordingly. 28) a) The actual pressure is lower than the ideal gas prediction when the volume is greater than about 0.1 L. Over this range, Preal/Pideal is less than 1, which means that Preal is less than Pideal. In this range, the attraction between O2 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 O2 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. 29) 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 = 1/2mv2, 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.