Chemistry B Name:
Chapter 13 Notes Per.
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Chapter 13: Gases
Introduction
•
Here we look at
the common gases around us
•
We also get a
better look at models used in science
•
Chapter 13 is a
combo of gases and models
•
Pressure, temp,
volume, amount (moles) are the Big Four here, and there’s a lot of math!!!
•
We’ll see how
they all interrelate…
A little background info on gases:
• Gas molecules
move really fast and are really far apart!
• For example, comparing
volume:
1 mol CO2 solid = 28 cm3
•
•
•
•
•
•
•
What is the volume of same amount of CO2 gas? = 25000 cm3
•
• Most of the volume of gas is
empty space! Extra room can be used to compress the gas (e.g.
tires, scuba tanks)
•
Remember in the early days (CHEM A) we saw a gas
fills its container, can be compressed, and mixes
completely with other gases
•
This chapter we emphasize the property of
pressure
13.1 Pressure
• Force versus
Pressure:
Which would hurt more?
• A bowling ball
lying on your head or a bowling ball on a nail sticking you on your
head?
• Other ex: snowshoes,
high-heeled shoes, and bed of nails!
• Gas pressure is like ping-pong balls
smacking the walls; more smacks, higher pressure; fewer smacks, lower pressure
è
Pressure is force / unit of area,
so last block exerts greatest Pressure
•
The collapsing can experiment is a good intro to
Atmospheric pressure
•
The molecules of air can
pound so hard and so often, the can collapses.
•
This atmospheric pressure can be measured with a
barometer…
•
In 1643 Evangelista Torricelli invented the barometer
•
The Hg didn’t flow out b/c of the…
•
Atmospheric pressure; it presses so hard on the
pool, the Hg was forced up the tube
• On average, @ sea level it rises to 760 mm
•
Why is there atmospheric pressure in the
first place?
•
Gravity! It holds the
ocean of gases down like it holds liquid oceans down
•
The different weather (“hi” pressure, “lo”
pressure) can change the height of the column…
•
Altitude changes it, as well
•
High mountains are at the top of the air ocean =
lower pressure pushing on us
•
Go to Mount Everest
(29,035 ft) and your dead in 10 minutes
Units of Pressure
•
Because barometers have a column of Hg, one unit is mm Hg
•
This pressure
device is called a manometer
• h
is the pressure in actual mm Hg
• mm Hg is also known as torr (after Torricelli)
• Another
unit is the standard atmosphere
•
1.000 atm = 760.0 mm
Hg = 760.0 torr
•
We also use the pascal,
the SI unit:
1.000 atmosphere = 101,325 Pa = 101.325 kPa
• But
wait! there’s more!
•
Used in the US
and by engineers is
Pounds per square inch (psi)
•
1.000 atm = 14.69 psi
•
got all that!
Examples!
The mercury
has risen to a height of 729 mm Hg. What is this pressure in kPa & psi?
Change 745
mm Hg to atm.
0.98
atm
Change 97.5
kPa to atm.
0.962
atm
Examples!
• change 28 psi to:
1.
atm
1.9
atm
2.
torr
1.4
x 103 torr
3.
pascals
1.9
x 105 Pa
Examples!
• The
height of the Hg column in a barometer was 732 mm Hg; what is that in…
1. atm?
0.963 atm
2. torr?
732 torr
3. Pa?
9.76 x 104 Pa
13.2 Pressure & Volume: Boyles
Law
•
Robert Boyle in the 1600’s did an
experiment w/ a tube like this
•
The more Hg he added, the greater the Dh,
the tinier the space
hmmmm….
•
Here’s his data; see a pattern?
•
The Cartesian diver is a way to show this
relationship
•
What we see is that as the pressure is
increased the volume decreases
•
This is an inverse
proportion
•
in math: PV = k
•
This is called Boyle’s Law
•
Temperature and amount of gas still play a role,
but we must assume they are constant!
• P1V1
= P2V2
•
In the examples that follow you can use P1V1
= P2V2
•
or you can do it the way I’ll show you
•
My way or the equation way, it’s your call
•
I’ll always give you three of these variables;
you can find the fourth…
Example 1
•
A gas sample
occupies 100.0 cm3 when the pressure is 150.0 kPa. When the pressure
is increased to 200.0 kPa, what is the new V?
•
Will answer be
> or < 100.0 cm3?
•
LESS!
Remember: inverse proportion
Example 2
•A gas sample occupies 25 liters when the pressure is 100.0
torr. When the pressure falls to 75.0 torr, what is the new V?
•Will answer be > or < 25 liters?
•GREATER! Remember: inverse
proportion
•
Example 3
•A gas sample occupies 1.00 L when the pressure is 98.0 mm
Hg. What pressure is required to make the sample only 0.250 L?
•Will answer be > or < 98.0 mm Hg?
•GREATER! Remember: V P↓
13.3 Volume & Temperature:
Charles
law
•
After Boyle, work on gases continued
•
Jacques Charles, a famous balloonist,
quantified a gas relationship
•
He found that when one
cools a gas sample it contracts, when heated it expands
•
The strange, unexpected thing is that all gases behaved very similarly…
•
Yes, they all contracted, but their linear relation took them all to
the same zero volume temp!
•
[The solid line =
real data The dashed line =
extrapolated]
•
All end up at
-273oC
•
This temperature at which
a gas volume ® 0 is absolute
zero…
•
This is the Kelvin
scale and we see that as Temperature rises, so does
Volume in a direct proportion
•
e.g. if
Temperature doubles, Volume doubles
•
But Temperature must
be Kelvins!!!
•
All this gives us Charles’s Law
•
V = bT
•
Which can give the more math-y people amongst us
this:
•
V1/T1
= V2/T2
•
Again, I’ll give you three to find 4th
•
Again, I’ll show you another way…
•
[remember: K = oC
+ 273]
Examples
A gas
sample has a volume of 2.25 L at 298 K. What is the new V when heated to
373 K?
•Your predicted answer will be >
2.25 L
A gas occupies 473 mL at 36.0˚C. What will its new
volume be at 94.0˚C?
•CENTIGRADE MAY NOT PLAY!!!
•First change to K!
•36.0 + 273 = 309 K
•94.0 + 273 = 367 K
A gas
occupies 500. cm3 at 27˚C. What will its new volume be at -48˚C?
•CENTIGRADE MAY NOT PLAY!!!
•First change to K!
•27 + 273 = 300 K
•-48 + 273 = 225 K
A gas has a volume of 5.65 L @
27 oC. at what T will its volume be 6.69 L?
82 oC (355 K)
There’s a gas with V of 9.25 L @
47 oC; what’s the T when it is 3.50 L?
-152 oC (121 K)
13.4 Volume & Moles: Avogadro’s
Law
•
This is easiest of all
•
When you blow air into a balloon it gets bigger
(duh!)
•
When the number of moles
(n) increases, so does V
• V and n are directly proportional
•
V = an
•
called Avogadro’s
Law
•
V1/n1 = V2/n2
Example
A gas sample occupying 3.5 L with 2.0 mol of hydrogen has
been pumped up to 7.0 L. How many mol of hydrogen are in the sample now?
Combining the Gas Laws
•
Before we do the Ideal gas law lets talk about
the
Combined Gas Laws
•
aka CharBoyle’s Law
•
What if I change both
the T and the P? what will happen to V now?
An oxygen sample has a
volume of 7.84 cm3, P of 71.8 kPa, and T of 25˚C. What
is the new volume at P = 101 kPa and T = 0˚C?
You can see this is both
a Charles’ Law (V & T) and
a Boyle’s Law (P & V)
Do them both at same
time... Boyles first, then Charles.
5.10 cm3
Example 2
A
1.00 L sample of nitrogen gas has a P of 2.50 atm @ 25˚C.
What will new T be if P is raised to 4.00 atm and V goes to 2.00 L?
954 K
13.5 The Ideal Gas Law
•
The story so far:
• Boyle’s
Law PV = k
• Charles’s
Law V/T = b
• Avogadro’s
Law V/n = a
•
Combining all the
constants into one, R, and solving for V we get:
V = R(Tn/P)
•
R is called the universal gas constant!
•
For us it has a value of 0.08206 L•atm/mol•K (your good friend!)
•
Now we multiply both sides by P and…
PV = nRT
•
The ideal gas
law, where P is in atm, V in liters, n in
mols, T in Kelvins
•
When we do these problems I will give you 4 of
the values for you to find the 5th
•
[We’ll assume the gas we’re dealing with behaves
ideally (more on that later)]
•
Just remember 2 basic things:
1) Rearrange PV=nRT to solve for what you are
looking for,
2) All variables must be dressed according to
what R says
•
Ready?…
Example 1
What is the pressure of
1.65 g of He gas at 16.0 ˚C with a volume of 3.25 L?
We are solving for P so everything else should be here (V, n, R, T)
But are they properly dressed to go to the Ideal Gas Law? No!
To the dressing room!
Now
in dressing room g à mol, C à K:
1.65 g He à 0.412 mol He
16.0 ˚C à 289 K
3.25 L is OK!
Everything is dressed properly!
NOW,
solve for P = nRT/V
P = (0.412 • 0.08206 • 289) / 3.25
P = 3.01 atm
Example 2
What is the volume in L of
2.5 mol of oxygen gas measured at 25 ˚C and 104.5 kPa?
Are all “dressed” properly? Solve for V
Dressing room…
2.5 mol à OK!
25 ˚C à 298 K
104.5 kPa à 1.031 atm
Solve…
V = nRT/P
V = (2.5 • 0.08206 • 298) / 1.031
V = 59 L
Example 3
At what temperature will 0.0100
mol of Ar gas have a volume of 275
mL at a pressure of 100.0 kPa?
Dressing room...
0.0100
mol à OK!
275 mL à .275
L
100.0 kPa à 0.9869 atm
Solve…
T = PV/nR
T = (0.9869 • 0.275) /
(0.0100 • 0.08206)
T = 331 K
13.6 Dalton’s Law of Partial Pressure
•
There are mixtures of gases all over the place:
atmosphere, scuba gear, lab experiments
•
But, each gas in a mixture acts just as if it
were the only gas
•
One of the first people to work on this was our
old pal, John Dalton
•
He said for a mixture of gases in a
container, the total pressure is equal to the sum of all the partial pressures
• Partial Pressure is the pressure
that a gas would exert if it were all alone
•
called Dalton’s
Law of Partial Pressures
Ptotal = P1 + P2 + P3…
•
Notice that it really doesn’t matter what gas it is, just how many of the little
critters there are (mols)
•
Two important things to notice:
•
The volume of
the individual little atoms or molecules aren’t too important (it doesn’t
matter that Ar is bigger than He)
•
The forces
between the particles aren’t too important neither (at least not now)
•
Which means this: if you know the mol ratios you know the
pressure ratios
•
For this example the
mol fractions are:
N2 = 1.00/1.75 = 0.57
O2 = 0.50/1.75 = 0.29
Ar = 0.25/1.75 = 0.14
•
So each contributes its
share of the total P:
ppN2 = 57% of 8.4 = 4.8 atm
ppO2 = 29% of 8.4 = 2.4 atm
ppAr = 14% of 8.4 = 1.2 atm
•
We use Dalton’s
law when we perform experiments like this
•
We aren’t only collecting the oxygen gas, but
water vapor, as well - we don’t want the water!!!
•
So we have to take the total pressure of the
system and subtract water’s contribution to get the pressure of our gas
•
Thankfully we have a table like this to show us water’s contributions
at different Temperatures
Example
Oxygen
is collected over water @ 22˚C. The total pressure of the system is 754
torr. The water vapor pressure @ 22˚ C is 21 torr. What is the pressure of
the oxygen?
Ptotal
= Poxygen + Pwater
Poxygen
= Ptotal – Pwater
Poxygen = 754 torr - 21 torr
Poxygen = 733 torr
Example
A
balloon contains both helium and nitrogen. The partial pressure of the helium
is 93 torr. The total pressure inside the balloon is 101 torr. What is the pp
of the nitrogen?
Ptotal
= Pnitrogen + Phelium
Pnitrogen
= Ptotal – Phelium
Pnitrogen = 101 torr - 93 torr
Pnitrogen = 8 torr
13.7 Laws and Models:
a review
•
Why is it that gases, when the pressure is low
and the T is high, fit almost perfectly into the ideal gas equation?
but…
•
when the T is low and P is high they deviate,
sometimes a lot!?
•
first a quick look at laws…
•
Remember: laws are like summaries of what
we see, they are not rules that atoms and molecules must obey
•
Laws don’t tell us why these do the things they
do; theories (models) do
•
Remember: looking at microscopic models
can help us see why macroscopic things happen
•
Alright? here’s a theory…
13.8 The Kinetic Molecular Theory of Gases
•
Here is what is called the kinetic molecular theory (it’s all in the
name)
[ aka: KMT ]
•
it’s pretty straightforward, and fairly simple…
•
Remember: kinetic
energy is the Energy of motion, i.e. hot gases are moving faster (#5)
•
No gas fits
these perfectly; but they can
come pretty close
13.9 The Implications of the KMT
•
Here we see how the KMT drives the gas laws and
find out what T is about…
The Meaning of Temperature
•
Temperature, simply put, is just the motion of
the little critters
•
“High” T means they are moving faster on
average, hitting the walls more often
•
“Low” T means slower movement, lower average
speeds, less smacking of the walls
The Relationship Between P and T
•
(Assume a constant V) move the critters faster
and they hit the walls more often = higher P!
•
Therefore higher T means higher P
The relationship between V and T
•
If you want to
make them go faster (T) BUT hit the walls at the same rate (P), you’d
better make the walls farther apart (V)
• \ Charles’ Law!
Examples
•
Using KMT, what will happen to P when the # of
mols goes up? (all else constant)
•
Using KMT, what will happen to V when n is
increased? (all else constant)
•
Using KMT, what will happen to P when V is
decreased? (all else constant)
13.10 Real Gases
•
There is no “ideal” gas, but they can get close
to behaving like one (i.e. zero individual volumes and no attractions)
•
Keep them far apart (low pressure) and moving
fast (high temps) and they can act almost “ideally”
•
This is bringing them closer = bad
•
Now their individual volumes play a role
now they can be attracted to each other
•
Now they risk becoming liquid!
Quick question
• What
if all gases behaved ideally?
•
Bad!
Nothing could condense to a liquid, no water, no rain, no oceans, no bodies, no
life, no good, no duh!
13.11 Gas Stoichiometry
•
Now, because of the cursed n in PV = nRT, we can bring back
stoichiometry for gas rxns!!! yippee!!
•
Remember: stoichiometry is where I give you info
on one substance in a rxn, you give me info on another…
Example
•
Here’s a rxn:
P4(s)
+ 6H2(g) ® 4PH3(g)
•
What volume of hydrogen gas @ 27˙C
and 1.25 atm is required to react with 5.65 g of phosphorus?
•
Remember: they can only talk mol language! you
still must cross mol bridge!
•
You have to change
g P4 ®
mol P4 ®
mol H2 ®
V H2
volume is our destination!
•
mols H2 can be changed to V by
the ideal gas equation!
•
so:
•
Do the old way
until mols
•
Find V from
the ideal gas equation
Example
CaCO3(s)
® CaO(s) + CO2(g)
•
What volume of carbon dioxide gas @ 22¢ªC
and 1.05 atm can be produced from reacting 6.25 g of calcium carbonate?
•
Remember: they can only talk mol language! you
still must cross mol bridge!