POGIL 18 Page 1 of 11
POGIL EXERCISE 18
All You Need to Know About Gas Laws
Each member should assume his or her role at this time. The new manager takes
charge of the POGIL folder and hands out the GRF and RRF to the appropriate
members. The new recorder should record the names of the group members on the
new GRF.
Table 1. Group Member Role Assignments
GROUP TYPE ->
GROUPS OF THREE
MEMBER NO. ->
1
2
3
Manager
+
Reporter
+
Recorder
+
Reflector
+
Technician
Encourager
+
*
SFUC
+
1
GROUPS OF FOUR
2
3
4
+
+
+
+
+
+
*
OBSERVATION IA. As you know, a gas has no shape or volume of its own. The shape of
the gas depends on the container. The condition where the container is open to the
outside environment is known as an open system and the system where the container is
sealed is called a closed system.
The volume of the gas in the container depends on two environmental conditions: the
absolute temperature (K) and the pressure (P). The absolute temperature is an
important variable because the kinetic energy (energy of motion) of the molecules
increases proportionately with the absolute temperature. As the molecules move faster,
they take up more room in the container and thus the volume will tend to increase in an
open system which causes some gas to leave the container. When gas leaves the
container it means that the amount of gas in the container decreases. On the other
hand, in a closed system, the gas cannot escape and so the movement of the molecules
causes the molecules to exert more pressure on the sides of the container.
1. Suppose you have 10 L of a gas in an open system. If the temperature were to
decrease around the container, would the amount of the gas in the container be
higher or lower than the starting conditions? ___________ Explain your answer.
5 MIN
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2. Suppose you have 10 L of a gas in an open system. If the temperature were to
increase around the container, would the volume of the gas in the container be
higher or lower than the starting conditions? ___________ Explain your answer.
Please discuss your answers to Items #1 and #2 with your instructor before continuing.
OBSERVATION II. The overall conclusion to be gathered from Observation I is that the
difference between an open and closed system is that in an open system, the amount
(i.e. number of moles) of gas can vary and, in a closed system, the amount (i.e. number
of moles) of gas is a constant. Thus, there are two basic gas laws to allow quantification
of components of either the open or closed system. The open system can be handled
only by the ideal gas law. Although the closed system can also be handled by the ideal
gas law, it is far more convenient to use the combined gas law to calculate components
in a closed system.
3. Supposed you needed 100 L of hydrogen gas at 25 C and 1 atmosphere (atm) to fill a
small balloon. You know you can generate hydrogen by reacting zinc with an acid.
You need to know how much zinc and acid you need to generate 100 L of hydrogen.
Which of the two basic laws would you use to calculate the amounts of reactants you
need? ____________________________ What is your rational for this answer?
4. Suppose you added the 100 L of hydrogen to the balloon and it rose to an altitude of
20,000 ft. where the pressure was 0.8 atm and the temperature was 5 C and you
wanted to know what the volume of the balloon would be under those conditions.
Which of the two basic laws would you use to calculate the volume?
____________________________ What is your rational for this answer?
Please discuss your answers to Items #6 and #7 with your instructor before continuing.
10 MIN
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OBSERVATION III. Let us now consider the quantitative relationship between pressure
(P), volume (V) and temperature in Kelvin (T) in a closed system. These relationships are
always used when a constant (but not necessarily known) amount of gas at a certain
volume, temperature and pressure is changed to a different volume, temperature and
or pressure. The combined gas law is described by Equation 1 below:
EQ1: VsPs = VfPf
Ts
Tf
OR
TfVsPs = TsVfPf
The subscript, “s,” in EQ1 stands for starting conditions and the subscript, “f,” stands for
the final conditions. Just to emphasize the point: All the parameters can change, but in
a closed system, the amount (i.e. the number of moles) of the gas does not.
Let’s take this equation for a ride: A container of nitrogen (or any other gas) with a
volume of 1.58 L at a pressure of 1.21 atm and a temperature of 12.2 C was heated to
32.3 C and the pressure was increased to 1.54 atm. What is the new volume in the
container? _____________________
Looks kind of complicated doesn’t it? OK, let’s take it one step at a time.
a. Because we are changing a gas from one set of conditions to another we know
that we are dealing with a closed system; therefore, we must use the combined
gas law (EQ1) to solve the problem:
VsPs = VfPf
Ts
OR
Tf
TfVsPs = TsVfPf
b. We make out a variable table and fill in the values given in the problem:
VsPs = VfPf
OR
TfVsPs = TsVfPf
Vs =__1.58 L_ Vf = __??____
Ts
Tf
Ps =__1.21 atm___ Pf = _1.54 atm__
Ts =__12.2 C___ Tf = _32.3 C___
c. After filling out the variable table we learn that we are solving Equation 1 for the
volume after the conditions were changed (Vf). So we solve the equation for that
variable before we do anything else.
TfVsPs = TsVfPf; Vf = TfVsPs
TsPf
20 MIN
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d. We notice that the given temperature is in centigrade; however, we know that
the temperature must be in Kelvin work in any gas law. So we convert the
temperatures to their kelvin equivalents using the following equivalence
equation: K = 273 + C.
K = 273 + 12.2 = 285.2 and K = 273 + 32.3 = 305.3
and plug the values into our variable table:
Vf = TfVsPs
TsPf
Vs =__1.58 L_ Vf = __??____
Ps =__1.21 atm___ Pf = _1.54 atm__
Ts =__285.2 K___ Tf = _305.3 K___
e. Using the second form of EQ1, we solve the expression algebraically, then “plug
and chug.”
VsPs = VfPf
Ts
Tf
OR
TfVsPs = TsVfPf
Vs =__1.58 L_ Vf = __??____
Ps =__1.21 atm___ Pf = _1.54 atm__
Vf = TfVsPs = (305.3 K)(1.58 L)(1.21 atm)
TsPf
(285.2 K)(1.54 atm)
Ts =__285.2 K___ Tf = _305.3 K___
= 1.3289 L = 1.33 L
Solve the following problems by using the above example and process
5. Exactly 141 mL of oxygen at 721 torr and 135 K was
subjected to 801 mm Hg pressure with a resulting
volume of 0.152 L. What was the temperature after the
change of conditions? _____________________
a. Fill in all the known values in the spaces provided to
the right.
b. Solve Equation 4 for the unknown parameter, plug in
the values, and solve the problem. Remember to
include units in the calculations.
+35 MIN
Vs =_____ Vf = ______
Ps =_____ Pf = ______
Ts =_____ Tf = ______
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6. Helium, 879 mL, under 5.51 atm and 22.1 C was
increased in volume to 1.05 L and a temperature of
101 F. What was the pressure in the container after the
change? _____________________
a. Fill in all the known values in the spaces provided to
the right.
b. Solve Equation 4 for the unknown parameter, plug in
the values, and solve the problem. Remember to
include units in the calculations.
Vs =_____ Vf = ______
Ps =_____ Pf = ______
Ts =_____ Tf = ______
OBSERVATION IV. The name for EQ1, the combined gas law, is derived from the fact
that not all the relationships incorporated into the equation were known at the same
time. The first special case was described by Boyle and became known as Boyle’s Law.
Through experimentation Boyle determined that the volume of a gas at constant
temperature is inversely proportional to the pressure; i.e., if you increase the pressure
and bleed off the excess kinetic energy to keep the temperature constant, the volume
will decrease proportionally. The mathematical derivation of the special case described
by Boyle’s law is easily accomplished using the combined gas law:
EQ3: VsPs = VfPf
Ts
Tf
Since the temperatures are the same, they cancel out and you have expression for
Boyle’s law (Equation 2).
EQ4: VsPs = VfPf BOYLES LAW.
Note: It is important for the student to know the definition of Boyle’s law and that it
is a special case of a closed system. It is not necessary for the student to remember
the formula since all Boyle’s Law problems can be worked with the combined gas Law.
+35 MIN
POGIL 18 Page 6 of 11
Let’s look at a problem that utilizes this concept.
7. What would be the resulting pressure if you have 44.5 L
of a gas at 1.22 atm and you expanded the volume to
178 L? _____________________
a. Fill in the data table at right:
b. Which of the four parameters is missing? ______
c. Solve Equation 1 for this parameter and then
substitute the known values. Remember to include
the units as well as the numerical value in the
calculations.
Variable Table
Vs =_____ Vf = ______
Ps =_____ Pf = ______
Ts =_____ Tf = ______
Present your answers to the instructor for validation before continuing with processing.
OBSERVATION IV: The next special case of the combined gas law describes the
relationship in a closed system with a constant amount of gas between temperature and
volume at CONSTANT PRESSURE.
1. This direct proportional relationship of volume to temperature at constant
pressure is known as Charles’s Law. Charles’s law can also be derived from the
combined gas law:
EQ5: VsPs = VfPf
Ts
Tf
EQ6: VS = VF .
TS
TF
leads to…
OR EQ3: VSTF = VFTS
CHARLES’S LAW
This relationship is true only if T is expressed as the absolute temperature (K). In fact, in
all gas laws where temperature is included as operator, the temperature must be
expressed as the absolute temperature (K).
+40 MIN
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Note: It is important for the student to know the definition of Charles’s law and that it
is a special case of a closed system. It is not necessary for the student to remember
the formula since all Charles’s Law problems can be worked with the combined gas
Law.
Let’s now learn to use this relationship solve by solving a problem:
8. What would be the resulting temperature if you had 6.11 Variable Table
L of a gas at 21.2 C and you expanded the volume to
25.90 L? _____________________
OBSERVATION V. The final special case defines the relationship between temperature
and pressure at CONSTANT VOLUME in a closed container with a constant amount of
gas. This direct proportional relationship of pressure to temperature at CONSTANT
VOLUME is known as Gay-Lussac’s Law. Since the amount of gas is constant, GayLussac’s law allows us to calculate the temperature or pressure at any condition. Thus,
this law too can be derived directly from the combined gas law:
EQ7: VsPs = VfPf
Ts
Tf
EQ8: PS = PF .
TS
TF
45 MIN
leads to…
OR PS*TF = PF*TS GAY-LUSSAC’S LAW
POGIL 18 Page 8 of 11
9. What would be the resulting temperature if you had a
Variable Table
gas at 1.00 atm and 37 C and you increased the pressure
to 10 atm and kept the volume the same?
_____________________
OBSERVATION VI. There are many situations in our environment where there are
mixtures of gases – air for example. Dry air at sea level is composed 78% nitrogen, 21%
oxygen, 0.9% argon and 0.04% carbon dioxide. One might ask the question as to the
pressure exerted by each of these gases. Since temperature and volume of the mixture
are constants, this can also be considered a closed system question.
Dalton studied this situation and in the end proposed his Law of Partial Pressure that
says the total pressure of a gaseous mixture is equal to the sum of the pressure exerted
by each gas in the mixture (EQ10).
EQ9: Ttot = Pa + Pb + Pc ….
So if the atmospheric pressure is equal to one atm, it means that the pressure
component of nitrogen (partial pressure of nitrogen) is equal to 0.78 atm (1 atm x
78/100). Let’s use this new law to solve some problems
10. A gas mixture contains each of the following gases with
their partial pressures in parenthesis: N2 (285 torr), O2
(116 torr), He (267 torr). What is the total pressure of
this mixture? ___________
50 MIN
List of Knowns:
POGIL 18 Page 9 of 11
11. A heliox (a commercial mixture of helium and oxygen)
deep-sea diving mixture delivers a partial pressure of
oxygen of 0.30 atm when the total pressure of the
mixture is 11.0 atm. What is the partial pressure of
helium in the mixture? ___________
List of Knowns:
OBSERVATION VI. Let us now consider the open system problems. As stated in
Observation I, these types of problems can only be solved using the Ideal Gas Law. The
primary variable to concentrate on when dealing with these problems is the number of
moles of the gas (the amount of gas) which must be known or calculated to work the
problem. The Ideal Gas Law defines the relationship between the absolute
temperature (T), the pressure (P) which must be expressed in atmospheres, the volume
(V) always expressed in liters and the number of moles (n) of the gas present. This
relationship is seen in Equation 6.
EQ17: PV = nRT
R = 8.21 x 10-2 L atm/nK
IDEAL GAS LAW
R is one of the marvels of the universal constants of nature and is the reason why for
our problems the volume, temperature and pressure must be converted to liters, Kelvin,
and atmospheres, respectively.
The ideal gas law can be rearranged to give yet another definition of a mole:
EQ18: n = PV/RT
The ideal gas law can also be used to calculate volume of one mole of any gas under
standard temperature and pressure (STP). STP is 0 C and 1 atm.
PV = nRT; V = nRT = (1 n)( 8.21 x 10-2 L atm/nK)(2.73 x 102 K) = 22.4 L
P
1 atm
55 MIN
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Supposed you needed 100 L of hydrogen gas at 27 C and 1 atmosphere (atm) to fill a
small balloon. You know you can generate hydrogen by reacting zinc with and
hydrochloric acid. You need to know the minimum amount zinc and acid required to
generate 100 L of hydrogen.
a. First you have to recognize that this is a stoichiometric problem that must be
solved from knowing how many moles of molecular hydrogen are in 100 L of the
gas at 27 C and 1 atm. Since it involves moles, you must use the Ideal Gas Law,
rearranging to solve for n and then plug and chug.
PV = nRT; n =
PV . =
(1 atm)(100 L)
. = 4.06 n H2
-2
2
RT
( 8.21 x 10 L atm/nK)(3.00 x 10 K)
b. Write a balanced molecular equation for the reaction:
Zn + 2 HCl  ZnCl2 + H2
c. We convert n hydrogen to moles Zn using the conversion factor from the
balanced equation: 4.06 n H2 x n Zn = 4.06 n Zn
n H2
d. Finally we calculate the grams of Zn we need:
4.06 n Zn x 65.39 g = 16.484 g
n Zn
e. For HCl we first find n HCl using the conversion factor of the balanced equation
4.06 n H2 x 2n HCl = 8.12 n HCl
n H2
f. Finally, since HCl comes as a 12.0 M solution, we have to calculate the minimum
volume of concentrated HCl required:
n = M x L; L = n/M = 8.12 n HCl = 0.677 L 12.0 M HCl
12.0 n/L
Let’s work a problem using the Ideal Gas Law:
60 MIN
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12. What volume of hydrogen at 25 C and 740 mmHg can
be produced from 100 g of magnesium with excess
sulfuric acid?
List of Knowns:
Present your results to the instructor for validation.
Please complete the matching exercise in Table 4.
Table 4. Matching Conditions with Gas Laws
Condition
_____ 13. Relates volume and temperature when pressure
is constant.
_____ 14. Relates partial pressures of a gaseous mixture to
total pressure.
_____ 15. The product of the volume and pressure is a
constant amount of gas.
_____ 16. Relates volume, temperature and pressure in a
closed system.
_____ 17. Relates pressure and volume at constant
temperature
_____ 18. VsPs = VfPf
a.
b.
c.
d.
e.
Gas Law
Boyle’s Law.
Charles’s Law
Gay-Lussac’s Law
Dalton’s Law of
Partial Pressure
Combined Gas Law
_____ 19. TfVsPs = TsVfPf
_____ 20. T = PV/nR
Present your results to the instructor for validation.
EXERCISE END. Managers should collect the GRF and RRF, staple them together, and
place in the back of the left pocket of the folder. The folder should be closed and left
on the table.
NOTIFY INSTRUCTOR WHEN FINISHED. Wait for further instructions.
+65 MIN