Chemistry 121
Mines, Fall 2016
Answer Key, Problem Set 7—Full
1. MP (5.28); 2. MP (5.30); 3. MP; 4. MP (5.32); 5. MP (5.34); 6. MP (5.36); 7. WO1; 8. WO2; 9. MP (5.56); 10. MP (5.60);
11. MP (5.132); 12. MP; 13. MP (5.72); 14. MP (For Practice 5.12); 15. MP; 16. WO3; 17. WO4; 18. MP; 19. WO5;
20. WO6; 21. WO7; 22. WO8; 23. MP (5.82); 24. MP (5.62); 25. MP (5.70); 26. MP (Conceptual Connection 5.5);
27. WO9; 28. MP; 29. MP (5.64); 30. WO10
-----------------------------------------
Pressure Problems—units and measurement with manometer
1. MP (5.28). 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 5.1 for equivalences
(a) torr
652.5 torr (because 1 torr = 1 mmHg)
(b) atm
652.5 mmHg x
1 atm
 0.85855..  0.8586 atm (recall that 760 here is exact)
760 mmHg
Answer is reasonable. 652 is somewhat less than 760, so P should be somewhat less than 1 atm
(c) in Hg
1in
652.5 mmHg x
1 cm
x
2.54 cm
10 mm
29.92in Hg
 25.687..  25.69 in Hg (or you could multiply by
760 mmHg
)
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
14.7 psi
 12.620..  12.6 psi (your text only gives psi to 3 SF)
760 mmHg
Answer is reasonable. 652 is somewhat less than 760, so P should be somewhat less than 14.7 psi
2. MP (5.30). 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!
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
10 mm
 23 mm  Pgas  751.5 - 23 = 728.5 = 729 mmHg
1 cm
(b) h = 3.3 – (-3.4) cm = 6.7 cmHg x
10 mm
 67 mm  Pgas  751.5 + 67 = 818.5 = 819 mmHg
1 cm
PS7-1
Answer Key, Problem Set 7
“Simple” Gas Laws [i.e., other than Dalton’s Law]
3. MP Solution to this problem not in key at this time.
4. MP (5.32). 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,
ths
Boyle’s Law). So if V becomes (10.3 / 13.9) 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:
PV  nRT 
PV
PV
V
PV
n ,T constant
 R (a constant)  1 1  2 2 
  P1V1  P2V2  P2  P1 x 1
nT
n1T1 n 2T2
V2
Substituting in with values (make sure you carefully assign the values for “State 1” and “State 2”!) yields:
1.22 atm x
13.9 L
 1.646..  1.65 atm
10.3 L
5. MP (5.34). 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:
PV  nRT 
PV
PV
PV
V V
T
n ,P constant
 R (a constant)  1 1  2 2 
  1  2  V2  V1 x 2
nT
n1T1 n2T2
T1 T2
T1
Substituting in with values yields: V2  1.55 mL x (0.0  273.15)K  1.149..  1 . 1 5m L
(95.3  273.15)K
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. MP (5.36). 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
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:
PS7-2
Answer Key, Problem Set 7
PV  nRT 
PV
PV
V
V
n
PV
P ,T constant
 R (a constant)  1 1  2 2 
  1  2  V2  V1 x 2
nT
n1T1 n 2T2
n1 n 2
n1
Substituting in with values yields: V2  253 mL x (0.553  0.365)mol  419.9..  420.mL
0.553 mol
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
7. WO1. 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
8
or 2.66 times the original value
3
Reasoning 1:
PV  nRT 
PV
PV
P n T
PV
 R (a constant)  1 1  2 2  V2  V1 x 1 x 2 x 2
nT
n1T1 n 2T2
P2 n1 T1
If n is quadrupled, then
n2
T
 4; if T is doubled, then 2 is 2; and
n1
T1
if P2 is to be three times P1, then
V2  V1 x
1
8
x 4 x 2  V1
3
3
P1
1

P2 3
Thus,
, 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
n2 4n1
n RT
4n R2T1 8 n1RT1 8
 PV  nRT  V  nRT  V2  2 2  1
 x
 x V1
P
P2
3P1
3
P1
3
8. WO2 5.43. 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
PS7-3
Answer Key, Problem Set 7
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:
PV  nRT 
PV
PV
PV
PV
V T
PV
n constant
 R (a constant)  1 1  2 2 
  1 1  2 2  P2  P1 x 1 x 2
nT
n1T1 n 2T2
T1
T2
V2 T1
Substituting in (see “Important Comments” above):
P2  50.7 psi x
338.15 K
11.8 L
x
 58.15.. 58.2 psi (actual pressure)
12.2L
285.15 K
 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
9. MP (5.56). 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)?
3
Answer: 1.20 x 10 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 = 2.85 g x

1mo N2 O
 0.06474.. mol N2O
44.02 g

Latm
(0.06474..mol)0.08206 mol
(298 K) 760 mmHg
nRT
K
PV  nRT  P 

x
 1203.2..mmHg
V
1L
1 atm
 1.20 x 10 mmHg
3
NOTE: You could also use Equation 5.6 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)?
PS7-4
Answer Key, Problem Set 7
10. MP (5.60). 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:

1 atm 
1L

 753 mmHg x
117 mL x

760 mmHg 
1000 mL 
PV

PV  nRT  n 

 0.003944..mol
Latm
RT
0.08206 mol
(85  273.15K)
K

MM 

0.555 g
 14.7.. 141 g/mol
0.003944..mol
= 358.15 K (uncertainty in the
units place b/c of addition
rule  3 SF)
NOTE: The authors of the Solutions Manual do this problem using Equation 5.6! 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”.
11. MP (5.132). 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):
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 5.6 (since you don’t know
the new P?!
Execution:
PV  nRT  n 
1atm0.48 L  0.0212...mol CO
PV

Latm
RT
0.08206 mol
(275 K)
K


0.0212..mol CO x (12.01 + 16.00) g/mol  0.5957..g CO
d
m 0.5957..g

 0.458..  0.46 g/L
V
1.3 L
PS7-5
Answer Key, Problem Set 7
Stoichiometry
12. MP Solution to this problem not in key at this time.
13. MP (5.72). 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 nO2 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:
PV  nRT  nO 
2
0.957 atm1.4 L  0.05183...mol O
PV

2
Latm
RT
0.08206 mol
K (315 K)
0.05183...mol O2 x


2 mol H2 O
18.02 g
x
 1.869..1.9 g H2O needed
1mol O 2
1mol H2 O
14. MP (For Practice 5.12) Solution to this problem not in key at this time, but see prior problems in this key, as
well as Example 5.12 in Tro for a similar problem (if you want to see a worked example from the text).
Kinetic Molecular Theory
15. MP Solution to this problem not in key at this time.
16. WO3 5.19. 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:
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
PS7-6
Answer Key, Problem Set 7
“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.
17. WO4. 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.
(a)
(b)
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.
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)).
18. MP Solution to this problem not in key at this time.
19. WO5 5.20. 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.207. I hope you did not just
copy the text answers on your paper. See note in problem 15 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
PS7-7
Answer Key, Problem Set 7
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 #18 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):
P 
force
# collisions
x
collision
sec
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.
PS7-8
Answer Key, Problem Set 7
Thus,
n

P  mv  x  x v 
V

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:

P  mv
2
 x  Vn 


2
2
Since the first expression (mv ) is proportional to average KE (½ mv ), 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 m2v22)
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!
20. WO6 5.49. 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
PV  nRT  P 
nRT
n
n
T constant
 x R x T 

 P 
V
V
V
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.
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.
PS7-9
Answer Key, Problem Set 7
21. WO7 5.89 & 5.90.
5.89. 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).
5.90. T2 is greater. The average speed of the particles is greater at T2 (peak is farther to the right—
2
~1100 m/s vs. ~800 m/s at T1). Since average KE ( ½ mv ) is greater at higher T (KMT), and
m here is fixed (same substance), average v must be higher at the higher T.
22. WO8. 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.
1 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.
st
2 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.
nd
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
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 Pressure, Dalton’s Law
23. MP (5.82). 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?


n
NEITHER. Partial pressure depends only on n of the gas  PA  A x Ptotal  (Okay, so I just
n total


realized that I covered partial pressure last in lecture and put the partial pressure problems later
PS7-10
Answer Key, Problem Set 7
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.
(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”).
24. MP (5.62). 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 PHe 
nHe
x Ptotal to
n total
calculate nHe. That is slightly longer, but equally valid.)
Execution:
PV  nRT  n He

1 atm 
 219 mmHg x
12.0 L 
760 mmHg 
PHeV



 0.1543..mol He
Latm
RT
0.08206 mol
(273 K)
K
0.1543..mol He x x


4.003 g
 0.6178.. 0.618 g He
1mol He
25. MP (5.70). 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
PS7-11
Answer Key, Problem Set 7
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,
n
recall that the partial pressure of a gas A is related to Ptotal by mole fraction: PA  A x Ptotal
n total
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  2.0 g x
MM(He)  4.003 g/mol He  98.0 g x
1 mol
 0.0625 mol O 2
32.00 g
1 mol
 24.48.. mol He
4.003 g
ntotal  0.625 + 24.48..  25.106..mol
PO 
2
nO
2
n total
x Ptotal 
0.0625 mol
x 8.5 atm  0.0216.. 0.022 atm
25.106..mol
26. Conceptual Connection 5.5 (on p. 223) 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
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).
Real vs. Ideal Gas Behavior [Deviations from Ideal Behavior and qualitative explanation of]
27. WO9 5.91 & 5.92. 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 224, 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.
PS7-12
Answer Key, Problem Set 7
That means that in 1 L (1000 mL) of Ar at STP, the sum of the volume of all of the individual
atoms is only 0.1 mL! This is a negligibly small volume (about the volume of 1-2 drops of water)!
However, assuming ideal behavior, if the pressure were made to be 1000 atm, the volume of the
th
gas would become 1/1000 of 1 L, which equals 1 mL. Now the volume of the individual atoms
(still 0.1 mL) would be 10% of the volume of the gas! Clearly this amount of volume is not
negligible. In fact, the volume would not end up being “only” 1 mL—it would be somewhat larger
than that (see Figure 5.24, which is even better than my example since it is at elevated T which
minimizes the effects of intermolecular forces, which I conveniently ignored.)
(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. 5.25). 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.
28. MP Solution to this problem not in key at this time.
Cumulative Problems
29. 5.64. 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):
475 mL Ar
722. torr
stopcock
275 mL He
752 torr
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)
PS7-13
Answer Key, Problem Set 7
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:
P1V1  P2V2  P2  P1 x
V1
V2
 P2  752. torr x
275 mL
 275.7..torr  276 torr
750 mL
P1V1  P2V2  P2  P1 x
V1
V2
 P2  722. torr x
475 mL
 457.2..torr  457 torr
750 mL
For Ar:
Ptotal  275.7..+ 457.2..  732.96.. 733 torr
30. WO10. 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.
(a)
N2
# particles
with a
given
speed
(b)
C2H2
C2H2
(c)
N2
C2H2
N2
speed 
speed 
speed 
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
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).
PS7-14