Honors chemistry
Name
CH. 14 GAS LAW STATIONS
Directions: Six stations have been developed to help you
understand the behavior of gases. After completing these stations
you should have gained a deeper understanding of the relationships
of gas pressure to volume and temperature. Please use your
textbook as a reference when working. You will not have time to
complete all problems in class. What you do not get to will be
homework.
**Entire completedpacket is due
I!
STATION #1
-Go to lab table marked “station 1”
STATION #2
-Can be done at your desk or at home
STATION #3
-Go to lab table marked “station 3”
STATION #4
Can be done at your desk or at home
-
STATION #5
Can be done at your desk or at home
-
STATION #6
Can be done at your desk or at home
-
/55 points
Station #1: Boyle’s Law
Purpose: The purpose is to determine the relationship between pressure & volume.
Hypothesis:
Part 1: Set the device to have a volume of 35 mL. Record the actual volume to 0.1
rnL. Place a book on top of the device and record the new volume. Ignore the last data
point in which the volume was zero. Plot the data.
Data:
Volume (mL)
# Books
0
7
3
4
5
-
0
>
1
0
2
3
4
5
6
7
8
Pressure (related to # books)
Conclusion:
1.
Describe the graph. What kind of curve do you have’?
2.
What happens to the volume as the pressure increases’?
3.
What happens to the volume as the pressure decreases?
4.
Describe the relationship between pressure & volume.
2
Part II: Play with the Cartesian Diver. Squeeze the bottle see what happens to the
floating object. Explain in word using pressure and volume.
—
ft
As you can see from the above data and curves, Pressure and volume are inversely
related. As the pressure increases, the volume decreases and vice versa.
1’
tv
v
p
Let’s solve some problems
Examples:
1. A gas has a volume of 200.0 mL at a pressure of 800.0 mm Hg. What is the
volume of the same gas at a pressure of 765 mm Hg?
1 =200.0mL
V
1 =800.0mm Hg
P
=?
2
V
=765mmHg
2
P
Using Boyle’s Law equation: P
V
1
2
V
=
200.0 mL x 800.0 mm Hg
765 mm Hg
V and solving for V
2
P
2
=
=
P
V
1
2
p
209 mL
Double check.. Did the volume do what you expected?
.
2. A gas has a volume of 200.0 mL at a pressure of 700.0 mm Hg. What is the
volume of the same gas at a pressure of 765 mm Hg?
1 =200.OmL
V
1 =700.0 mm Hg
P
=?
2
V
=765mmHg
2
P
Using Boyle’s Law equation: P
V
1
=
2
V
=
=
200.0 mL x 700.0 mm Hg
765 mm Hg
V and solving for V
2
P
2
=
P
V
1
7
p
183 mL
Double Check...
3
3. A gas has a volume of 200.0 mL at a pressure of 800.0 mm Hg. What is the
pressure of the gas if the volume is increased to 250.0 mL?
V = 200.0 mL
1
=250.OmL
2
V
P = 800.0 mm Hg
1
=?mmHg
2
P
V
1
Using Boyle’s Law equation: P
2
P
=
800.0 mm Hg x 200.OmL
250.0 mL
=
=
2
V and solving for P
2
P
=
P
V
1
7
V
640.0 mm Hg
Your turn...
I.
A sample of oxygen occupies a volume of 250.0 mL when its pressure in
720.0 mm Hg. What is the volume when the pressure is increased to 750.0
mmHg?
2.
A gas collected when the pressure is 8003 mm Hg has a volume of 380.0 mL.
What is the volume at standard pressure?
3.
A gas has a volume of l00,d mL when the pressure is 735 mm Hg. When the
pressure is decreased to 700.0. what is the volume?
I V
(
4.
__J
I
A gas has a volume of 240 mL at a pressure of 700.0 mm Hg. What pressure
is needed to reduce the volume to 60.0 mL?
4
Station 2: Unit Conversions
1 atm
=
14.7 psi
760mm Hg = 101.3 kPa
All values are standard pressure.
Example #1: Convert 775 mm Hg into atm
Using dimensional analysis:
775 mm Hg x
1 atm
760 mm Hg
=
1.02 atm
Example #2: Convert 825 mm Hg into psi
Using dimensional analysis:
825mmHgx 14.7psi
=16.Opsi
760 mm Hg
Your turn...
Convertpitoatm
2.
Conver 105 kPa to atm
MY
,
3
Convert 640 mm Hg rnto Pa
Temperature in Kelvin = 273
Examples: Kelvin = 273
+
+
V
Temperature in °C
25°C
=
298 K
Your turn...
1.
What’s the temperature in Kelvin of 100°C?
2
What’s the temperature in °C of —273K
:
—
/
—
)
Station #3: Gay-Lussac’s Law
Purpose: The purpose is to determine the relationship between temperature & pressure
Hypothesis:
Procedure: Plot the data.
Data:
Temperature (K)
273
683
1365
2048
Pressure
0.1 atm
0.25 atm
0.5 atm
0.75 atm
I atm
2730
———
———:z:z:z ZZEZ
,i)
[
4
E
1
1
1
I
—4—
iL_ LL
—4—
0
‘S
—
t)
Pressure (atm)
Conclusion:
Describe the graph
What happens to the temperature as the pressure increases?
What happens to the temperature as the pressure decreases’?
Describe the relationship between pressure & temperature...
5
—
6.
7.
8.
Pump up the pressure in the container. What happens to the temperature? Release the
,4ressure what happens to the temperature now?,,
—
6
As you can see from the above data and lines, Pressure and temperature are directly
related. As the pressure increases, the temperature also increases. Force the particles to
move faster...
P
T
P
t
t
T
Let’s solve some problems...
1. A steel tank contains a gas with a pressure of 12 atm at 27 °C. What is the
pressure of that gas if the temperature is increased to 100 °C?
=27°C
1
T
P=12atm
=lOO°C
2
T
=?
2
P
Using Gay
—
Lussac’s Equation:
1Ej =
or P
2
T
1
1
T
1
T
2
P
7
T
Solving for P
2
T
1
P7=l2atmx (100°C+273)K
‘?7°C±’73 K
=
Remember,
temperature
needs to be in
Kelvin!!!
l0atm
2. A gas has a pressure of 780.0 mm Hg at 20.0 °C. What is the pressure if the
temperature decreases to 0.0 °C?
T = 20.0 °C
1
1 = 780.0 mm Hg
P
2 0.0°C
T
=?mmHg
2
P
Using Gay
Lussac’s Equation: 1
P =
or P
2=P
T
1
1
T
2
1 T
T
2
Solving for P
2 iE.1I2
1
T
2 = 780.0 mm Hg x (0.0 °C + 273) K = 727 mm Hg
P
(20.0 °C + 273) K
—
3. A gas has a pressure of 3.0 atm at 20.0 °C. What is the temperature of the gas if
the pressure is increased to 5.0 atm?
T =20.0°C
1
=?
7
T
Using Gay
Pi=3.Oatm
=5.Oatm
2
P
Lussac’s Equation: 1
P
1
T
Solving for T
2=
1
P
2
T
=
—
(20.0 °C
+
273) x 5.0 atm
3.0 atm
=
or P
2
T
1
=
=
1
T
2
P
2
T
490 K
7
Your turn...
A gas is contained in a 250 mL flask at 17 °C and 780mm Hg is heated to a
temperature of 50 °C. What pressure does the gas exert?
“-
—‘
‘-
I
t
1
-
W
:
3
3
2.
At what temperature will a quantity of gas, which erts
Hg at 25 °C, have a pressure of 700.0 mmHg?
A
=
3.
1
L
—
/
I’
A gas is contained in a 300.0 mL flask at a pressure of 1.00 atm and 30 °C. If
the pressure is doubled, what will the new temperature be’?
=2
—
—-
-
±z—
,
If 4J?
-
=2
4.
A gas has a pressure of 3.5 atm at 37 °C. At what temperature will this gas
exert a pressure of 1.00 atm’?
—
‘1
(
P
/1I
—
3
1
=2
8
___________
_____
Station #4: Graham’s Law of Diffusion
Purpose: The purpose is to determine the relationship between rate of diffusion and
molar mass of the gas.
Hypothesis:
Procedure: Using dropper pipets, students placed about five drops of concentrated HC1
on a cotton swab and five drops of concentrated NH
3 on the other cotton swab. After
several minutes, a white ring will form where the gases meet inside the tube to form one
product, the white compound NH
C1 (ammonium chloride). Using a ruler, accurately
4
measure the distance traveled by each gas from the diagram below measure from the tip
of the arrow to the tip of the arrow. Record your measurements with uncertainty.
—
3
NH
‘---I
2
D
1
D
—
—
.I.__
0,
—
I
—__.z
the distance from HC1 to the product:
HC I
-
f
lO Cfl’L
the distance from NH
3 to the product:__________________
Beginning time was: 9:36 a.m.
Time product formed: 9:53 a.m.
Using the above data:
1.) Calculate the experimental rate of diffusion for each gas.
Rate=Distance
//
Time
/
-
‘
‘M
-
/
‘‘ /
7- / /
2.) Calculate the experimental jo between the rate of diffusion of NH
3 to the rate of
diffusion of HC1 using the rates calculated in question #1.
I
——
I-—
4t’
-
L
cV
/
—‘
— -
1
‘
— — --
9
3 to the rate of
3.) Calculate the theoretical ratio of the rate of diffusion of NH
to look Graham’s Law
need
You
(HINT:
Law
diffusion of HC1 using Graham’s
up in your textbook!)
/1
JLfli
/
j
4.) Calculate the percent error for the experimental data and the theoretical
calculation for the ratio of the rates of diffusion.
i,-
/
-1,
AL
I
/
5.) Write a balanced equation (with phases) for the reaction in this experiment.
I
-
,
I
-4
—
1
—
-
/
)
6.) Classify this reaction as one of the basic types of chemical reactions:
10
Station #5: Avagadro’s Law
Purpose: The purpose is to discover Avogadro’s Law.
Hypothesis:
Procedure: Consider the following situation.
1 Liter of 112
M=O.08930g
1 Liter of N
2
M=l.24992g
1 Liter of 02
M=l.42848g
The balloons above are at the same terjpature and jrere.
Calculate the number of moles of each gas.
1
/
I
—
What did you discover?
How many molecules are in each sample?
1)
‘
Write a statement that describes the volumes, temperatures, pressure and # of molecules
of each gas.
1/
11
In the early 1800’s, Avogadro also discovered the same relationship between volume,
pressure, temperature and the number of molecules in a gas. Your statement should
resemble the statement below called Avogadro’s Law:
of8
ases at tile same temyerat’ure anc(yressure
contain an equal numIer ofyartIcles.
Etjualvolumes
Summary: How many liters are in 1 mole of a gas?
/
r
m
‘7’
/
/
This quantity is valid for any as at 0 °C (273 K) and 1 atm pressure. The
conditions are called standard temperature and pressure (STP).
We’ll do lots with this new conversion factor!!
Questions:
1
I
What is the density of HC1 gas at STP9-”-Z
—-——-—
/
,
I I
2.
-
/
—
The mass of 1.0 L of gas is 2.75 g. What is the molar mass of the gas?
—--
3.
What is the mass of 1.00 L of methane gas?
I
I
‘-I
z
1
/
/
1)
12
Station #6: Combined Gas Law
You already know that pressure and volume are indirectly related; temperature and
pressure are directly related; and temperature and volume are directly related. The
combined law takes these three concepts and makes one big law...
Problem solving: It’s the same as all the others. Look at the factors that are changing.
Decide the impact of this change on the unknown variable and set up the ratios of the
changing variable.
Let’s solve some problems...
1. A 5.00 L balloon contains a gas with a pressure of 12.0 atm at 27.0 °C. What is
the volume of that gas if the temperature is increased to 100 °C and the pressure
increases to 14.0 atm?
=5L
1
V
V =?
2
T
=
1
27°C
T2 = 100°C
=l2atm
1
P
P = 14 atm
2
Using the combined gas law fiV =
or P
2=P
T
V
1
1
T
V
2
1
T
2
T
Solving for
1
T
2
P
V = 5.00 L x (100 °C + 273) K x 12.0 atm = 5 L
(27 °C + 273) K x 14.0 atm
Always use this as your
initial conditions.
Your turn. .You have a special balloon. It is inflated to 4 L in Wilmington, DE where
the pressure is 30.27 inches Hg and the temperature was 45.0 °C.
.
1.
You take this balloon scuba diving and are at a depth of 100.0 ft where the
pressure is 7.00 atm and the temperature is 54.2 °C. What is the volume of
the balloon?
/
2.
Next you take the balloon out to Colorado. In Denver the temperature is 37.5
°C and the pressure is 640.0 mm Hg. What is the size of the balloon?
/
13
During the Colorado trip, we hike to the top of Longs Peak (14, 256 ft) where
the pressure is 470.0 mm Hg and the temperature is -20 °C. What is the
volume of the balloon now?
3.
}
jAK
4.
Finally we take the balloon on the airplane and we let it out the window. The
747 is flying at 40,000 ft where the pressure is 80.0 mm Hg and the
temperature -60 °C. What is the volume of the balloon now?
=
LLr
2
‘K
V
5.
The balloon is at last released and soars into the stratosphere where the
pressure has dropped to 0.800 mm Hg and the temperature 0.0 °C. What is
the volume of the balloon just before it pops?
‘
1’
I
/
I
I
I
14