Answer Key, Problem Set 7 Full

Chemistry 121 Mines, Fall 2019

PS7-1

Answer Key, Problem Set 7—Full

1. NT1; 2. 6.43; 3. 6.56; 4. 6.132; 5. 6.72; 6. 6.19; 7. NT2; 8. 6.20; 9. 6.49; 10. 6.89 & 6.90; 11. NT3; 12. 6.70; 13. 6.91 & 6.92; 14. NT4 -----------------------------------------

Pressure—units and measurement with manometer

[In Mastering only]

“Simple” Gas Laws [Not Dalton’s Law]

1. NT1. A cylinder with a moveable piston contains some argon gas at a certain T, P, and V. If the number of moles of

argon is quadrupled (i.e., made to be four times greater than the original value) and the Kelvin temperature is doubled, how must the volume be changed to make the new pressure end up being three times the original pressure? (Assume ideal gas behavior.)

Answer: V must become 3

8or 62.6 times the original value

Reasoning 1:

1

2

1

2

2

1

22

22

11

11

T

T

n

n

P

PVV

Tn

VP

Tn

VP

nT

PVnRTPV 1 x x x constant) (a R 2

If n is quadrupled, then 1

2

n

n 4; if T is doubled, then

1

2

T

T is 2; and

if P2 is to be three times P1, then 3

1

2

1

P

P Thus,

3

82 x 4x

3

1 x 112 VVV , meaning V must be adjusted to eight-thirds of its original value.

Note: This makes sense in that if the T alone were doubled, P would double. And if n alone were quadrupled, P would quadruple. So those two factors would make P become 8 times what it was. Thus, in order for the final P to end up 3 times (and not 8 times) the original value, the V must changed to make the P become “3/8” (if that were the only change). Since P and V are inversely proportional, V should become 8/3 its value to achieve this desired effect.

Reasoning 2:

P2 3P1

T2 2T1 1x

3

8Rx

3

8

3

R24

R

R V

P

Tn

P

Tn

P

TnV

P

TnVnRTPV

1

11

1

11

2

222

n2 4n1

2. 6.43. An automobile tire has a maximum rating of 38.0 psi (gauge pressure). The tire is inflated (while cold) to a

volume of 11.8 L and a gauge pressure of 36.0 psi at a temperature of 12.0 °C. Driving on a hot day, the tire warms to 65.0 °C and its volume expands to 12.2 L. Does the pressure in the tire exceed its maximum rating? (Note: The gauge pressure is the difference between the total pressure and atmospheric pressure. In this case, assume that atmospheric pressure is 14.7 psi.)

Answer: Yes, it will. Final gauge pressure will become 43.5 psi

Important Comments:

In either approach below, you must recognize that the actual initial pressure in the tire is not 36.0

psi, but rather 36.0 + 14.7 50.7 psi (because of the definition of “gauge pressure”: Pgauge = Pgas

– Pbar Pgas = Pgauge + Pbar; see Note at the end of the problem text above). Likewise, once a final (actual) pressure is obtained (here, 58.2 psi), the gauge pressure is that value minus 14.7 psi

( 43.5 psi).

You must also convert the T’s into Kelvin: T1 12.0 °C + 273.15 285.15 K

T2 65.0 °C + 273.15 338.15 K

Answer Key, Problem Set 7

PS7-2

Approach 1 (as in the prior two problems!):

A “new” P is asked for after a change in T and V, with n constant. Use:

1

2

2

1

2

22

1

11

22

22

11

11

T

T

V

VPP

T

VP

T

VP

Tn

VP

Tn

VP

nT

PVnRTPV 1

n x x constant) (a R 2

constant

Substituting in (see “Important Comments” above):

K 51285.

K 51338. x

L 12.2

L 11.8 x psi 50.72P 58.15.. 58.2 psi (actual pressure)

gauge pressure = 58.2 – 14.7 43.5 psi > 38.0 psi (maximum rating)

NOTE: This approach has advantages over using the ideal gas law to find n first, and then using the ideal gas law a second time (“traditional” approach). 1) You do not need to convert the pressures into atm (and back) or the volumes into L, because in this approach, R cancels out (or rather, it is the ratios of P and V that matter, so their units cancel out.) 2) There ends up being fewer overall calculations, since one main setup is all that is needed (rather than two, not counting the conversions).

Approach 2: Use the ideal gas law to find n, then use it again to find P2. Setups are not shown in key at this time, but n turns out to be 1.739 moles, and P2 ends up being 3.9559 atm (actual pressure). This leads to the same result as above, as it must.]

Density and Molar Mass Problems

3. 6.56. A sample of N2O gas has a density of 2.85 g/L at 298 K. What is the pressure of the gas (in mmHg)? Answer: 1.20 x 103 mmHg

Strategy (Mines):

1) Recognize that since no amount of gas is given, you can pick a convenient “amount”. Since density is given, a convenient amount is just 1 L. Not only does this “give” you a “V”, it also “gives” you an amount of gas: 2.85 g.

2) Use grams and molar mass to calculate moles (this gives you “n”)

3) Now, use V, n, and T, to calculate P (using PV nRT)

4) Convert from atm to mmHg by multiplying by 760 mmHg/atm

Execution of Strategy:

MM(N2O) = 2(14.01) + 16.00 = 44.02 g/mol n = g 44.02

ON mo 1 x g 2.85 2 0.06474.. mol N2O

atm 1

mmHg 760 x

L 1

K) (298 0.08206mol)(0.06474..

R Kmol

atmL

V

TnPnRTPV 1203.2..mmHg

1.20 x 103 mmHg

NOTE: You could also use Equation 6.6 in your text to solve this problem, but that just means one more equation to memorize (or learn how to derive), when there is really absolutely no need to do so! Since you will not be allowed an “equation sheet” on the exam, and I will not provide you with any equations on the exam, doesn’t it make sense to learn the most basic equations and then learn how to use them to solve problems (rather than memorizing a bunch of [related] equations so you can try to “plug and chug” everything)?

4. 6.132. A quantity of CO gas occupies a volume of 0.48 L at 1.0 atm and 275 K. The pressure of the gas is lowered

and its temperature is raised until its volume is 1.3 L. Find the density of the CO under the new conditions.

Answer: 0.46 g/L

Strategy (notice the parallel to the prior problem. This is a very similar problem in my opinion):


PS7-3

1) Recognize that to find density, you need two things: “grams” and “Liters” (mass and V)

2) They give you the new volume (1.3 L), so the only thing left to do is find the mass of the gas!

3) The initial conditions give you V, P, and T, so you can easily get n (moles) using the ideal gas equation.

4) Since the gas is “known” (it is CO), you can use its molar mass and moles to calculate grams.

NOTE: If you focus on what you “need”, this problem is actually quite straightforward. The only reason I can think of that this is in the “Challenge Problems” section is because you might think you have to do something with the new pressure and new temperature (sort of like “teasers” in this problem). Also, you have to realize that the mass of a gas does not change with T, V, or P if n remains the same. Perhaps this problem might seem harder because you ”can’t” use Equation 6.6 (since you don’t know the new P?!

Execution:

K) (275 0.08206

L 0.48atm 1

KmolatmLRT

PVnnRTPV 0.0212...mol CO

0.0212..mol CO x (12.01 + 16.00) g/mol 0.5957..g CO

g/L 0.46 8..50.4 L 1.3

0.5957..g

V

md

Stoichiometry

5. 6.72. Consider the chemical reaction (represented by): 2 H2O(l) 2 H2(g) + O2(g) What mass of H2O is required to form 1.4 L of O2 at a temperature of 315 K and a pressure of 0.957 atm?

Answer: 1.9 g H2O needed

Strategy:

1) Recognize that the T, P, and V info provided are for O2, and so 2On can be calculated from the

ideal gas equation.

2) Recognize that the question asks about the mass of H2O (not O2!), so this really is a stoichiometry problem not “just” an ideal gas law problem. Use the balanced equation to get the mole ratio between O2 and H2O, then calculate moles of H2O.

3) Use the molar mass of H2O (2(1.008) + 16.00 = 18.02 g/mol) and moles (#2) to calculate mass.

Execution:

K) (315 0.08206

L 1.4atm 0.957

KmolatmLRT

PVnnRTPV O2

0.05183...mol O2

0.05183...mol O2 x OH mol 1

g 18.02 x

O mol 1

OH mol 2

22

2 1.869..1.9 g H2O needed

Kinetic Molecular Theory (including Speed Distribution Curve Concept)

6. 6.19. What are the basic postulates of kinetic molecular theory (KMT)? How does the concept of pressure follow from

KMT?

NOTE: Your text combines a couple of postulates to end up with only three, but I like to list them as four (even though there is some overlap this way). In any case, you should not have copied, word-for-word the postulates from the text on your paper (unless you have actually memorized the postulates word-for-word). If you don’t make the material “your own”, you will likely not learn much.

Answers:


PS7-4

1) Gas particles are in constant motion, moving in straight lines until they collide with each other or a wall of the container. The collisions with the walls are the cause of the pressure exerted by the gas.

2) The volume of a gas particle is negligibly small compared to the size of the container (because the distance between particles is so large). (For an ideal gas, the particles are considered to have no volume at all. Obviously real gas particles have a finite volume.)

3) Collisions are completely elastic, and between collisions, particles exert no force (attractive or repulsive) on each other.

4) The average kinetic energy of a gas particle (in a sample with lots of gas particles) is dependent only on the Kelvin temperature, to which it is directly proportional.

The concept of pressure directly follows from theory because each collision imparts a force on the wall it hits, and so when billions of collisions each second hit a unit area of a wall, the collective “force” per area is just the sum “force” of all the individual collisions. Since P = force / area, the collective force per area associated with the collisions is the pressure of the gas.

7. NT2. A small balloon with some air inside is placed inside of a syringe, and the syringe (which also has air in it) is

placed inside of a large box with the barrel (plunger) of the syringe pulled about halfway out (see sketch to the right). If the air in the box were pumped out, describe what would initially happen to the (i) barrel (Would it move inward, outward, or not move) and (ii) the balloon (Would it expand, contract, or stay the same size?) IF:

(a) The end of the syringe is capped. Give reasoning. (b) The end of the syringe is left open. Give reasoning.

ANSWERS. (a) the barrel would move outward, and the balloon would expand. (b) the barrel would not initially move, but the balloon would expand.

EXPLANATIONS. (a) When gas is pumped out of the chamber, the pressure on the barrel from the outside decreases,

and thus a net force from the inside (in the outward direction) develops. Thus the barrel moves outward. Once the barrel moves outward, the situation is identical (from the balloon’s perspective) to the situation in my demo (where I pull the barrel out)—the pressure against the outside of the balloon decreases (because air inside syringe expands into a bigger volume, fewer collisions against the outside of the balloon), while the pressure against the inside of the balloon remains the same (temporarily). Thus there is a net force “outward” on the skin of the balloon and the balloon expands.

(b) If the syringe is NOT capped, then the NET FORCE on the barrel does not change, since air

leaves the syringe when air is pumped out of the chamber. In other words, there will be (equally) less pressure on both ends of the barrel because the concentration of gas will decrease not only on the outside, but also on the inside. However, NO GAS CAN ESCAPE FROM THE BALLOON, so it will push out with the same force as always and as the pressure ON IT from the outside decreases, there will be a net force in the outward direction (same as (a)).

Pressure outward moves barrel because there’s gas in syringe; then balloon moves.

Pressure outward only on skin of the balloon because gas in syringe goes out.

(a) (b)


PS7-5

8. 6.20. Explain how Boyle’s Law, Charles’s Law, Avogadro’s Law, and Dalton’s Law all follow from KMT.

Answers: (Note: your authors directly answer these questions on p.240. I hope you did not just copy the text answers on your paper. See note in problem #6 (6.19) above.)

1) Boyles Law: P is inversely proportional to V at constant T and n. Thus, explain why a decrease in V leads to an increase in P (at constant T & n).

Shortest answer: A decrease in V leads to an increased particle concentration and thus an

increase in the “# of collisions per sec” with the walls. More collisions greater pressure (at constant T).

More detailed answer: Since T is fixed, “force per collision” is fixed. Thus, the pressure depends on the “number of collisions per sec” with a unit area of a wall. If V is decreased while n is constant, the concentration of particles goes up, raising the collisional frequency and thus the pressure. Or you could say that the walls are, in effect, being moved closer to the particles, so more of them will hit the walls each second after the V decrease.

2) Charles’s Law: V is proportional to Kelvin T at constant P and n. Thus, explain why an increase in T leads to an increase in V (at constant P & n).

Shortest answer: An increase T leads to an increase in average KE, making speed greater and thus collisional force and frequency greater (initially). The only way to keep P from increasing under such conditions is to increase V, which will decrease collisional frequency despite the increased speed.

More detailed answer: An increase in T leads to an increase in average kinetic energy of particles. Thus, they move faster and hit “harder” (i.e., with greater “force per collision”). Thus, Pgas must initially increase. But in this case, P is constant, which means the vessel can change its volume (if mechanical equilibrium is disturbed). Once Pgas becomes greater than Pexternal, the walls of the container will feel a net force “outward”, so they will move outward increasing the volume. This will decrease Pgas (see Boyles Law explanation above) until it becomes equal to Pexternal. Alternatively, you could say (as the book authors do) that “in order to keep Pgas from increasing when T increases, V must increase.”

3) Avogadro’s Law: V is proportional to n at constant T and P. Thus, explain why an increase in n leads to an increase in V (at constant T & P).

Shortest answer: An increase in n leads to an increased particle concentration (initially) and thus an increase in the “# of collisions per sec” with the walls, and thus pressure. The only way to keep P from increasing under these conditions is to increase the volume proportionately, which will keep the concentration constant.

More detailed answer: An increase in n leads to an increased particle concentration (initially) and thus an increase in the “# of collisions per sec” with the walls, and thus pressure. But if the gas is in a “constant pressure” container like a syringe, the pressure can’t increase! Rather, the increased Pgas will make Pgas temporarily greater than Pexternal, making the walls move outward, increasing the volume. This offsets the initial pressure increase (if V increases proportionally to n, n/V will remain constant, so the collisional frequency will remain constant). Alternatively, you could say that “in order to keep P from increasing under these conditions, V must increase." Since T is constant, “force per collision” remains constant. This one is all about collisional frequency (and concentration).

4) Dalton’s Law: Ptotal is the sum of all partial pressures. In a sense, this is saying that pressure does not depend on the identity of the particles—at the same T and V, P depends only on n, regardless of the type of particle. Thus, explain why no matter what the particle type is, P depends only on n (at constant T & V).

NOTE: I’ll tell you right now that this one is, by far, the most difficult one to explain properly. Even the textbook does not get it quite right. I will answer it correctly here, but I will not hold you to this level of explanation on an exam. What I do want you to know how to explain is why, for a given gas, if you increase n at constant T and V, P increases (see note in #9 (6.49)


PS7-6

below [next problem]). That is more straightforward—increased concentration means greater collisional frequency, and if T remains the same, the “force per collision” remains constant.

“Correct” answer: Pressure in KMT can be thought of as due to two components: “force per collision” and “# collisions per sec” (collisional frequency):

sec

collisions # x

collision

force P

Force per collision is proportional to (average) mv (momentum). The # collisions per sec is proportional to both concentration and velocity.

To justify this, note that if concentration is the same, the average distance to a wall is the same, and so an increased velocity will result in a greater number of collisions per sec. And if average velocity were the same, an increase in concentration would mean a smaller average distance to a wall, and thus result in a greater number of collisions per sec.

Thus,

v

V

nmvP x x

Mathematically, this is the simplest way to show you why it is only T that matters. Note that if you “move” the one v from the second expression to the first, you get:

V

nmvP x 2

Since the first expression (mv2) is proportional to average KE (½ mv2), and average KE depends only on T, this shows that even if mparticle changes, P will not change unless n/V changes. And thus at constant T and V, P is proportional only to n. QED (sort of)

Conceptually is where it gets subtle. If mparticle is larger for one gas (let’s say it’s 4x larger, for simplicity) but T is the same, the force / collision actually will become greater for that gas (twice as great in this case!) even though the T is the same.

This is because v will only become half as great (because the average KE’s are the same--½ m1v12 ½ m2v2

2)

But the collisions / sec will become smaller even though the concentration is the same (half as great in this case because the speed is half as great), so the two effects will offset one another, leaving the P the same.

The bottom line is this, for two gases at the same T, the one whose mparticle is larger will have a smaller collisional frequency (because v is smaller), but a proportionately greater force / collision (because m is bigger by more than v decreases). Whew!

9. 6.49. Which gas sample representation has the greatest pressure? Assume that all the samples are at the same

temperature. Explain.

Answer: (b) [the one with 10 gas particles in it]

NOTE: I put this question here in the PS sequence because I wanted you to be able to use “KMT” to explain your answer. However, you can analyze it with either “law” or “theory” as I’ll show:

Reasoning: Gas Laws

V

nPT

V

n

V

nRTPnRTPV T x R x constant

This shows that when T is constant, P is determined only by particle concentration (n/V). Since V is the same in this set of three pictures, the box that has the most particles in it has the greatest concentration and therefore the greatest pressure. Thus, the answer is (b).

Reasoning: Kinetic Molecular Theory

In KMT, pressure is caused by the collisions of the gas particles with the walls of the container. You can imagine that P is proportional to both the “force per collision” and the “# collisions per sec” (see prior problem). For a given gas, if T is constant, the “force per collision” is constant.


PS7-7

But if the concentration of particles is greater, the frequency of collisions with the walls will also be greater, and thus so will be the pressure. So at constant T, P is proportional to concentration, and box (b) has the greatest pressure. Although technically things are a bit more complicated when you compare different gases (see technical note below as well as prior problem), this idea is sufficient for my class (i.e., on an exam).

Technical note: The above argument is not technically correct because, of course, different gases move with different velocity at the same T (this you are responsible for knowing on an exam! See next question, among others!), and so collisional frequencies will not all be identical at the same T. But as noted in the prior problem (subtle part of explanation), the changes in mass, velocity, momentum, and collisional frequency (at constant T) offset themselves such that at constant T, only concentration affects pressure.

10. 6.89 & 6.90.

6.89. Substance A has the greater molar mass. The average speed of A particles is less than that of B (peak for A is at a “further left” velocity—about 500 m/s vs. about 1100 m/s for B). Thus A must have the greater mass per particle (and thus molar mass).

6.90. T2 is greater. The average speed of the particles is greater at T2 (peak is farther to the right—

~1100 m/s vs. ~800 m/s at T1). Since average KE ( ½ mv2) is greater at higher T (KMT), and m here is fixed (same substance), average v must be higher at the higher T.

11. NT3. Consider two different containers, each filled with 2 moles of neon gas. One of the containers is rigid and has

constant volume. The other container is flexible (like a balloon) and is capable of changing its volume to keep the external pressure and internal pressure equal to each other. If you raise the temperature in both containers, what happens to the pressure and density of the gas inside each container? Assume a constant external pressure.

Answers: In the constant-V container, when T is increased, P increases and density stays the same

In the constant P container (Pgas Pexternal), when T is increased, P remains the same and density decreases.

Reasoning, Gas Laws:

In both containers: 1) n is constant since the containers are “sealed” and no chemical reaction is occurring inside. Thus, PV nRT reduces to PV kT in both cases (for now). 2) It also follows that the mass of gas in each container is also constant (2 x 20.18 g/mol = 40.36 g). 3) Lastly, density (d) is defined as mass/V = mgas/Vcontainer.

1st Container: If V (in addition to n) is also held constant, then PV kT reduces to P k’T, and P is directly proportional to T. Thus when T increases P increases. Also, since Vcontainer is constant and mgas is constant, dgas does not change.

2nd Container: If P (in addition to n) is also held constant, then PV kT reduces to V k’T, and V is directly proportional to T. Thus when T increases, V increases and thus dgas decreases. Since, by definition in this container, Pgas always equals Pexternal, Pgas must remain the same.

Reasoning, Kinetic Molecular Theory:

In both containers, when T increases, the average kinetic energy of the gas particles increases, which means that their average velocity increases (since the gas particles’ masses do not change).

In container 1, since the V is kept the same, the result is that there are a greater number of collisions per second with a given area of the container walls, as well as a greater force per collision, resulting in a greater P. Since the same number of gas particles are moving around in the same volume after the T increase, the density does not change (i.e., density is independent of T as long as V is held constant).

In container 2, initially after the T increase, the P temporarily does increase as in container 1. The difference is that in container 2, the walls are not rigid, so that once the Pinternal is greater than Pexternal, there will be a net force “outward” on the walls of the container pushing them outward, increasing the volume. As V increases, the pressure will decrease because the number of collisions per second will


PS7-8

decrease (on average, particles are farther from a wall when V is larger [concentration is smaller]). So eventually, the Pgas will again become equal to Pexternal even though the particles are moving faster. Since the same number of particles are moving around in a greater Vcontainer, dgas decreases.

Gas Mixtures, Partial P, Dalton’s Law

12. 6.70. A heliox deep-sea diving mixture contains 2.0 g of oxygen to every 98.0 g of helium. What is the partial pressure

of oxygen when this mixture is delivered at a total pressure of 8.5 atm?

Answer: 0.022 atm Note: This would result in a dead diver! The problem authors seemingly made a mistake here! I think they thought they were making a mixture that would result in a partial pressure of 0.22 atm (which would be close to “normal”). Oops!!

Strategy:

1) Note that partial pressure can be obtained in multiple ways. In this problem, since Ptotal is given,

recall that the partial pressure of a gas A is related to Ptotal by mole fraction: total

total

AA x P

n

nP

2) Since masses of each gas are given, moles of each can be calculated from molar masses.

3) Sum the moles to get ntotal, and use the equation in (1) to get PO2

Execution:

MM(O2) 2(16.00) 32.00 g/mol O2 2O mol 520.06g 32.00

mol 1 xg 2.0

MM(He) 4.003 g/mol He He mol 8..424.g 4.003

mol 1 xg 98.0

ntotal 0.625 + 24.48.. 25.106..mol

atm 8.5 x 06..mol125.

mol 520.06 x total

total

O

O2

2 P

n

nP 0.0216.. 0.022 atm

Real Gases [Deviations from Ideal Gas Behavior and qualitative explanation of]

13. 6.91 & 6.92. Which postulate of the KMT breaks down under conditions of (a) high pressure?

(b) low temperature? Explain.

(a) At high pressure, the postulate that states that “the volume of a gas particle is negligible with respect to the volume of the container” is no longer a good assumption. Think of “high pressure” as “low volume” here—that is, in very “compressed” states, where concentration (n/V) is very high. When the particles are so close together, their actual volume becomes evident, and the behavior is no longer “ideal”. The fact that the gas particles take up space makes the actual volume of the gas larger than what it would be if it were behaving like an ideal gas.

Think of it this way: On page 238, your textbook authors state that at STP, the sum of the volumes of the individual atoms in a sample of Ar is only about 0.01% of the volume of the gas. That means that for a sample of 1 mol of Ar at STP, whose volume is close to 22.4 L (= 22400 mL), the sum of the volume of all of the individual atoms is only 2.24 mL! This is a negligibly small volume compared to 22400 mL! However, assuming ideal behavior, if the pressure were made to be 1000 atm, the volume of the gas would become 1/1000th of 22.4 L, which equals 22.4 mL. But the volume of the individual atoms would remain 2.24 mL. So now the volume of the Ar atoms themselves would be 10% of the volume of the gas (if it behaved ideally)! Clearly this amount of volume is not negligible, and the behavior is not ideal anymore. As such, the volume of the sample of real (Ar) gas would not end up being “only” 2.24 mL at 1000 atm (the value of an ideal gas)—it would be somewhat larger than that. This can be seen in Figure 6.24, where the


PS7-9

value of the volume of a mole of (real) Ar gas at 1000 atm is about 5.5 mL, while that of a mole of an ideal gas would be about 4.1 mL.

(Note: the reason the volume here for the ideal gas is 4.1 mL rather than 2.24 mL, as in my example above, is because the gas in Fig. 6.24 is at 500 K rather than at 273 K! In other words, all volumes will be 500/273 = 1.83 times larger at 500 K than at 273 K.)

(b) At low temperature, the postulate that states that “there are no forces between the particles between collisions” is no longer a good assumption. The fact of the matter is that there are attractive forces between particles (called “intermolecular forces”).

This should be obvious to everyone when you consider that all gases will turn into liquids or solids if the temperature is made low enough. Why would molecules of liquid water “stick together” in a liquid if there were no forces of attraction between them? They wouldn’t!

But those forces of attraction have decreasing “effect” on the motion of particles when the particles are moving with high kinetic energy (high T). So at very high temperatures for all substances (and lesser temperatures for those substances with the weakest forces, like “typical” gases at room temperature), the forces become negligible relative to the average kinetic energy (which increases with T), and that explains why gases behave ideally at high T. But think about the reverse—as T decreases, the forces (which remain “the same”) become less and less negligible since the average kinetic energy is steadily decreasing. At some point, the forces become “apparent”, and the assumption that they are negligible is not a good one. That’s when gas behavior begins to deviate from ideal behavior.

The fact that gas particles do attract one another makes the pressure with which they strike the walls of their container less than what it would be if the forces were negligible (See Fig. 6.25). In a very real sense, the particles are being attracted “toward the center of the sample” by their mutual forces of attraction, so they don’t hit the walls as hard.

Cumulative Problems

14. NT4. Which graph below would best represent the distribution of molecular speeds for the gases acetylene (C2H2)

and N2 given the following conditions? Both gases are in the same flask (and thus at the same T) with a total pressure of 750 mm Hg. The partial pressure of N2 is 500 mm Hg. Give your reasoning.

Answer: (a)

Reasoning:

Note: There are several aspects to a (speed) distribution curve: a) the shape, b) the highest point on the curve, whose associated speed is called the “most probable speed”, and c) the overall “height” of the curve (really the area underneath the curve), which is proportional to how many total particles are in the sample (because the y-axis is “number of particles which have a particular speed”). **Tro has chosen to use relative number of particles (rather than actual number) on its y-axes for distribution curves. This differs from the web simulations that I showed you (and what is in the present problem. So be aware of this possible difference. Always check axes!**

SPEED ARGUMENT.

Same T same average kinetic energy, and thus the one whose molecules are less massive will move more quickly. In this case, the molecular mass of C2H2 is 2(12) + 2(1) = 26 amu/molecule and the molecular mass of N2 is 2(14) = 28 amu/molecule. So they are actually VERY CLOSE in mass and should be very close in average speed; but the molecules in the sample of C2H2 should be moving slightly faster (26 amu < 28 amu). That would be indicated on a distribution curve plot by having the peak of the curve for the sample of C2H2 lie just to the right (higher speed) of the peak for

(a) (b) (c) N2

N2

C2H2

C2H2 C2H2

speed speed speed

# particles with a given speed N2


PS7-10

the sample of N2. The only picture that shows this is (a). (b) has the N2 sample having the greater average speed, and (c) indicates that C2H2 is the faster one, but by quite a substantial amount rather than a small amount.

PARTIAL PRESSURE/AMOUNT ARGUMENT.

If you weren’t sure about whether or not it was (a) or (c), you could use the partial pressure information to make it crystal clear. Since the partial pressure of O2 is 500 mm Hg and the total pressure is 750 mm Hg, the partial pressure of C2H2 must be 250 mm Hg. That means that there are TWICE AS MANY O2 MOLECULES in the flask as C2H2. That means that the curve for O2 should have about twice as much area under it (or be about twice as high at every point). The curves in Figure (a) look like this, but not in figure (c).

==================== END OF SET ====================

Problems below are from a prior key. Use these for practice!

6.28. Convert 652.5 mmHg (lowest pressure ever recorded at sea level—inside Typhoon Tip) to:

Answers: (a) 652.5 torr; (b) 0.8586 atm; (c) 25.69 in Hg; (d) 12.6 psi

Note: See Table 6.1 for equivalences

(a) torr 652.5 torr (because 1 torr = 1 mmHg)

(b) atm 652.5 mmHg x mmHg 760

atm 10.85855.. 0.8586 atm (recall that 760 here is exact)

Answer is reasonable. 652 is somewhat less than 760, so P should be somewhat less than 1 atm

(c) in Hg 652.5 mmHg x mmHg 760

Hg in 29.9225.687.. 25.69 in Hg (or you could multiply by

mm 10

cm 1 x

cm 2.54

in 1)

Answer is reasonable. 652 is somewhat less than 760, so P should be somewhat less than 29.92 in Hg

(d) psi 652.5 mmHg x mmHg 760

psi 14.712.620.. 12.6 psi (your text only gives psi to 3 SF)

Answer is reasonable. 652 is somewhat less than 760, so P should be somewhat less than 14.7 psi

6.30. Given a barometric pressure of 751.5 mmHg, calculate the pressure of each gas sample as indicated by the manometer.

Answers: (a) 775 mmHg; (b) 819 mmHg

Strategy:

1) Reason out the proper relationship between Pgas and Pbar. How?

As shown in class, I like to look at it this way: The side that has more Hg in it needs the

“help” (“Pextra”, which equals h, in units of mmHg if the manometer is filled with Hg) to equal the P on the other side.

So in (a), where the height is higher on the “gas” side, Pgas + h Pbar Pgas Pbar h

But in (b), where the height is higher on the “atmosphere” side, Pgas Pbar + h

2) Calculate h ( hhigher – hlower) from the manometer pictures. Note the units!


PS7-11

3) Convert the units (cm) into mm to be consistent with those of Pbar. Then use the equation in (1).

Execution:

(a) h = 1.1 – (-1.2) cm = 2.3 cmHg x cm 1

mm 1023 mm Pgas 751.5 - 23 = 728.5 = 729 mmHg

(b) h = 3.3 – (-3.4) cm = 6.7 cmHg x cm 1

mm 1067 mm Pgas 751.5 + 67 = 818.5 = 819 mmHg

6.32. A sample of gas has an initial volume of 13.9 L at a pressure of 1.22 atm. If the sample is compressed to a volume

of 10.3 L, what will its pressure be? Note: You need to assume constant T and n here, as well as ideal behavior!

Answer: 1.65 atm

Reasoning 1: If n and T are constant, making Vgas smaller makes P bigger (inversely proportional, Boyle’s Law). So if V becomes (10.3 / 13.9)ths of what it was, the P must get (13.9 / 10.3) times what it was.

Reasoning 2: If you do not intuitively see how the inverse proportional relationship between P and V

works, you can either memorize P1V1 P2V2, or derive it from the ideal gas law as shown in class and below:

2

11

T,n

V

VPPVPVP

Tn

VP

Tn

VP

nT

PVnRTPV x constant) (a R 2

constant 2211

22

22

11

11

Substituting in with values (make sure you carefully assign the values for “State 1” and “State 2”!) yields:

atm 1.65 6..41.6L 10.3

L 13.9 x atm 1.22

6.34. A syringe containing 1.55 mL of oxygen gas is cooled from 95.3 °C to 0.0 °C. What is the final volume of oxygen gas? Answer: 1.15 mL

Reasoning 1: If n and P are constant (which must be assumed here!), making Tgas smaller makes V smaller. The relationship is proportional (Charles Law) only if T is in Kelvin). So first convert the T’s to Kelvin by adding 273.15 to each: T1 = 368.45 K; T2 = 273.15 K. So if T becomes

(273.15 K / 368.45 K) ( 0.74134..) times what it was, the V must become 0.74134.. times what it was also (directly proportional). 0.74134.. x 1.55 mL = 1.149... = 1.15 mL

Reasoning 2: If you do not intuitively see how the direct proportional relationship between T and V

works, you can either memorize V1/T1 V2/T2, or derive it from the ideal gas law as shown in class and below:

1

2

2

2

1

1

22

22

11

11

T

TVV

T

V

T

V

Tn

VP

Tn

VP

nT

PVnRTPV 1

P,n x constant) (a R 2

constant

Substituting in with values yields: mL 1.15

9..41.1

K 273.15) 3(95.

K 273.15) 0(0. x mL 51.52V

NOTE: Make sure to always carefully assign values for “State 1” and “State 2”. In particular, here you must realize that 0.365 mol is not n2, because it is the number of moles added. Thus n2 = n1 + 0.365

6.36. A cylinder with a moveable piston contains 0.553 mol of gas and has a volume of 253 mL. What will its volume be

if an additional 0.365 mol of gas is added to the cylinder? (Assume constant temperature and pressure.) Again, you must assume ideal behavior here.

Answer: 420. mL


PS7-12

Reasoning 1: If P and T are constant, making ngas bigger makes V bigger (proportional, Avogadro’s

Law). So if n becomes ((0.553 + 0.365) / 0.553) ( 1.66) times what it was, the V must get 1.66 times what it was also. 1.66 x 253 mL = 420. mL

Reasoning 2: If you do not intuitively see how the direct proportional relationship between n and V

works, you can either memorize V1/n1 V2/n2, or derive it from the ideal gas law as shown in class and below:

1

2

2

2

1

1

22

22

11

11

n

nVV

n

V

n

V

Tn

VP

Tn

VP

nT

PVnRTPV 1

T,P x constant) (a R 2

constant

Substituting in with values yields: mL 420.

.9..941mol 0.553

mol 0.365) (0.553 x mL 2532V

NOTE: Make sure to always carefully assign values for “State 1” and “State 2”. In particular, here you must realize that 0.365 mol is not n2, because it is the number of moles added. Thus n2 = n1 + 0.36

6.60. A sample of gas has a mass of 0.555 g. Its volume is 117 mL at a temperature of 85 °C and a pressure of 753

mmHg. Find the molar mass of the gas.

Answer: 141 g/mol

Strategy (Mines):

1) Recognize that to find molar mass (g/mol), you need two things: “grams” and “moles” (just divide them to get MM)

2) The mass in grams of the sample is already in the problem! So all you need to do is find n, the moles of gas using the ideal gas equation (since V, T, and P are given)! Just remember to get quantities into proper units.

Execution:

K) 273.15 5(8 0.08206

mL 1000

L 1 x mL 117

mmHg 760

atm 1 x mmHg 753

KmolatmLRT

PVnnRTPV 0.003944..mol

MM 4..mol40.0039

g 0.55514.7.. 141 g/mol

NOTE: The authors of the Solutions Manual do this problem using Equation 6.6! In my opinion, that is just plain silly (they actually calculate the density first. Why? Because that formula has density and molar mass it in, so it would seem that you need d to get MM. How “slavish” to formulas this seems! My approach is much more direct. Please try to learn to solve the problem. Don’t try to “find an equation from the book that might work”.

6.82. A flask at room temperature contains exactly equal amounts (in moles) of nitrogen and xenon.

(a) Which of the two gases exerts the greater partial pressure?

NEITHER. Partial pressure depends only on n of the gas

total

total

AA x P

n

nP (Okay, so I just

realized that I covered partial pressure last in lecture and put the partial pressure problems later in the sequence. Sorry this one slipped in.). Since the moles of gases in this problem are “exactly equal”, their partial pressure must also be equal.

(b) The molecules or atoms of which gas have the greater average velocity?

NITROGEN. At the same T, since average kinetic energy is the same, more massive particles move more slowly. Xenon atoms have an average mass of about 131 amu, while nitrogen molecules only have an mass of about 28 amu.

= 358.15 K (uncertainty in the

units place b/c of addition

rule 3 SF)


PS7-13

(c) The molecules or atoms of which gas have the greater average kinetic energy?

NEITHER. Same T same avg KE!

Don’t mix up “KE” with “velocity”! The nitrogen molecules move more quickly at the same T because they are less massive, not because they have more kinetic energy. In fact, we use the assumption that average KE’s are the same in order to reason out that they must move more quickly!

(d) If a small hole were opened in the flask, which gas would effuse more quickly?

NITROGEN. Effusion and diffusion rates are both directly related to the speed of the particles. If you are moving more quickly, you will collide more frequently. Thus, in the case of effusion, there will be more “collisions” with the “hole” each second, which means more particles per second will escape (i.e., “effuse”).

6.62. A gas mixture with a total pressure of 745 mmHg contains each of the following gases at the indicated partial

pressures: CO2, 125 mmHg; Ar, 214 mmHg; and O2, 187 mmHg. The mixture also contains helium gas. (a) What is the partial pressure of the helium gas? (b) What mass of helium gas is present in a 12.0-L sample of this mixture at 273 K?

(a) Answer: 219 mmHg

Apply Dalton’s Law: Ptotal sum of all Pgas’s: PHe + 125 + 214 + 187 745

PHe 745 – (125 + 214 + 187) 219 mmHg

(b) Answer: 0.618 g He

Strategy:

1) Recognize that the T and V of the gas mixture are also the T and V for each of its components (e.g., He). Thus, nHe can be calculated from the ideal gas law with T, V, and PHe (after converting mmHg to atm).

2) Use moles and molar mass of He (4.003 g/mol) to calculate mHe.

(Alternatively, you could use Ptotal, V, and T to calculate ntotal. Then use total

total

HeHe x P

n

nP to

calculate nHe. That is slightly longer, but equally valid.)

Execution:

K) (273 0.08206

L 12.0mmHg 760

atm 1 x mmHg 219

KmolatmLRT

VPnnRTPV He

He 0.1543..mol He

0.1543..mol He x He mol 1

g 4.003 x 0.6178.. 0.618 g He

Conceptual Connection 6.6 (on p. 237). Nitrogen and hydrogen react to form ammonia according to the following

equation: N2 + 3 H2 → 2 NH3

Consider the following representations of the initial mixture of reactants and the resulting mixture after the reaction has been allowed to react for some time (pictures not shown):

If the volume is kept constant, and nothing is added to the reaction mixture, what happens to the total pressure during the course of the reaction?

Answer: The pressure decreases

Reasoning: Don’t be fooled here. It may initially sound like everything is constant (i.e., T, V, and n [particularly because they say that “nothing was added”!]), but if you actually look at the pictures and count up the molecules (remember PS1 and PS2? Make sure you know the difference between a molecule and


PS7-14

an atom!), you can see that there are fewer afterward. At the same T and V, P is proportional to n! So pressure decreases here. NOTE #1: There is no law of “conservation of molecules” in chemical reactions—only conservation of atoms! NOTE #2: You can see from the balanced chemical equation that as forward reaction occurs, the number of moles of gas must go down, because for every four molecules that get “used up” (dismantled), only two molecules get made (assembled upon repartnering of the atoms).

6.64. A 275 mL flask contains pure helium at a pressure of 752 torr. A second flask with a volume of 475 mL contains pure

argon at a pressure of 722 torr. If the two flasks are connected through a stopcock and the stopcock is opened, what are

the partial pressures of each gas and the total pressure? NOTE: You must assume that the temperature of the two gases is the same to solve the problem.

I think it is helpful to see a sketch of this scenario (usually “roundbottom” flasks are used for these):

Answers: PHe = 276 torr; PAr = 457 torr; Ptotal = 733 torr

Strategy:

1) The key to this problem (in my opinion) is recognizing that EACH GAS will spread out into the FULL VOLUME (of both flasks) once the stopcock is opened, so that V2 for BOTH gases will equal 475 + 275 = 750 mL after the stopcock is opened.

2) Thus, you can use Boyle’s Law (twice) to calculate PAr and PHe after the stopcock is opened. These are the partial pressures asked for!

3) Sum the partial pressures to obtain Ptotal (Dalton’s Law)

NOTE: You could solve this problem by assuming any temperature you want (just make it the same on both sides), and then solving the ideal gas equation (twice) to get the moles of each gas before you open the stopcock. Then you could use ntotal, Vtotal, and Ttotal to calculate Ptotal using the ideal gas equation (a third time). Then you could get the partial pressures using mole fractions. But that would be a lot more work! There is another “simple” way to do this problem, but it involves deriving a formula that I do not believe is “obvious”, so I won’t provide it here (ask me if you wish).

Execution of Strategy:

For He:

2

11

V

VPPVPVP x 2 2211 P2 752. torr x

mL 750

mL 275 275.7..torr 276 torr

For Ar:

2

11

V

VPPVPVP x 2 2211 P2 722. torr x

mL 750

mL 475 457.2..torr 457 torr

Ptotal 275.7..+ 457.2.. 732.96.. 733 torr

475 mL Ar 722. torr

275 mL He 752 torr

stopcock

Answer Key, Problem Set 7 Full

Documents

Transcript of Answer Key, Problem Set 7 Full