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This is a Sequential Activity
Group Activity: Gas Laws
Introduction
Without even knowing what gases are made of (OK,
they’re made of atoms and molecules) we can
understand how they behave on a macroscopic level. Gases
are a form of ordinary matter that is much less dense
than liquids or solids. Because of this they tend to fill
completely any containers they are in and are very
compressible.
Everyone has experience working with gases and taking
advantage of their properties. If you have ever blown up
a balloon, used a straw, inflated a tire, complained
about the weather, or taken a deep breath then you
already know how this topic relates to “real
life”.
There are four mathematical variables used to describe
the behavior of gases. Pressure (P), Volume (V),
Temperature (T), and amount (n). There are some common
units of measurement for each of these variables so your
first task will be to become familiar with them.
One more thing: Using these variables will help you to understand
the ideal gas laws. Why are they called ideal? It’s
not because they are the best possible laws gases could
follow. Nor is it because they are the best scientific laws
anyone ever found. No, it is because the gases we will discuss are
‘idealized’. That is, they are not real gases and no real gases
act the way these equations say they will. But, and this is important, almost
all gases come very, very close to acting exactly according to
the ideal gas laws. So even though they are not perfect, they are very
useful.
The ‘normal’ amount of pressure exerted
by Earth’s atmosphere
Torr
torr
0.00132 atm
0.0193 psi
Atmos. pressure can drop to 740 torr during a storm
lbs. per
square in.
psi
lb/in^{2}
0.0680 atm
51.7 torr
A car tire might be
rated for 35 psi
Temperature
Unit
Symbol
Convert
Example
Fahrenheit
°F
°C · 9/5 + 32
32°F water freezes
212°F water boils
Celsius
°C
(°F -32) · 5/9
0°C water freezes
100°C water boils
Kelvin
K
°C + 273
77 K liquid nitrogen boils
273 K water freezes
373 K water boils
Amount
Unit
Symbol
Convert
Example
Mole
mol
1 mol of C =
6.02 x 10^{23}
atoms of C
1 mol of C atoms weighs
exactly 12 g
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First, a helpful note about using the units given above.
You will need to be able to change a number given in one
unit into another unit. For instance, you might have a
problem to solve in which a volume is given in cubic
meters (m^{3}). But the formula you will end up
using requires that all volumes be given in liters. How
do you go about changing over? Use a technique called
Unit Analysis. Here’s how it works:
1000 L Put the unit you need on top of the unit factor
1.2 m^{3}x -------- = 1,200 L (1.2 x 10^{3} L)
1 m^{3} Put the unit you’re getting rid of on the bottom
OK, what’s a unit factor?
It is just another name for a conversion factor: a
factor you multiply by number in one unit to make it a
number in another unit. The word unit is a bit of a pun in
the phrase ‘unit factor’. First, it means
you’re dealing with units: liters (L), atmospheres
(atm), or whatever. Second, it means that the factor is
equal to the number one. Let me show you what I mean:
1 m
1 m = 100 cm --> divide both sides by 100 cm --> ----- = 1
100 cm
Equivalently:
100 cm
1 m = 100 cm --> divide both sides by 1 m --> ----- = 1
1 m
What happens when you multiply by one? You get the same
number again, right? So using a unit factor does not
change the quantity! Just the units.
The key is picking the right unit factor. The hint you
need is in the first example above. Find out what one
unit (say liters) is equivalent to in another unit (say
milliliters). In this case 1 L = 1,000 mL. That means the
unit factor is either (1 L)/(1,000 mL) or (1,000 mL)/(1
L). If the number you need to convert is in mL and you
want L then use the first factor. If the number you need
to convert is in L and you want mL use the second factor.
Here, try a few problems to see if you get what I’m
talking about:
Unit Analysis
Pressure Units
Give 2.1 atm in torr
Give 778 torr in atm
Give 14.7 psi in atm
Give 35 psi in torr
Volume Units
Give 5.2 L in mL
Give 10 m^{3} in L
Give 1,345 L in
m^{3}
Give 4,321 mL in L
Temperature Units
Give 350 K in °C
Give 35 °F in °C
Give 55 °C in °F
Give 32 °F in K
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Boyle’s Law (1662)
Now that you know a bit about the units you will be using
you are ready to start working with the mathematical laws
that we will use to build up the Ideal Gas Law.
(This should sound momentous and important.) The first
law we will examine is known as Boyle’s Law and was
first quantified and mathematically modelled by Robert
Boyle is about the year 1662. The law, in words, says the
following:
At a given constant temperature (T) and number of moles of
gas (n),
the pressure and volume
of a gas are inversely proportional.
That is, the higher the pressure, the smaller the volume;
the lower the pressure, the larger the volume. This
relationship can be expressed in the formula:
P · V = constant or more algebraically: P · V = k (only when T and n are constant)
By the way, what is pressure? The key to understanding
pressure is to think about force per unit area. Which
unit expresses that idea most clearly? What do you think
causes the force?
As you work through the following problems try to stop
briefly after finding each solution and imagine what the
answer means physically. Many people say that
chemistry is very abstract, and it is, but with these
problems you can very often think about the answers in a
clear, physical way. If you do this, you will also have a
leg up on making sure your solution is correct. If your
volume is a negative number and you try to imagine a
negative volume, then you will know you did something
wrong!
First, in order for the
formula to be of any use we need to put it into a form
that tells us about changes in pressure and volume. For
example, say you have a balloon filled with air at 1 atm.
Say also that the balloon has a volume of exactly 1 L.
What would the pressure inside the balloon be if you
squished it down to 0.75 L?
First, find the constant k
for the initial P and V.
Now, plug the new volume and
the constant (k) into the expression above (PV = k)
and use those two variables to find the new pressure.
Call the first P and V
values (1 atm and 1 L) P_{1} and
V_{1}. Their product is by definition a
constant value. So the new P you found multiplied by
a V of 0.75 L should equal the same number.
Let’s call these two new values of P and V,
P_{2} and V_{2}. Show arithmetically
that the product P_{1}· V_{1}
equals P_{2}· V_{2}.
What does this bit of
arithmetic allow you to write algebraically? Write
down the new formula which allows you to quickly and
easily find a new V or P given the other three
variables.
Apply this new formula to the
following problem: An inflated balloon has a volume of
0.55 L at sea level (1 atm) and is allowed to rise to a
height of 6.5 km, where the pressure is about 0.40 atm.
Assuming that the temperature remains constant, what is
the final volume of the balloon? For practice with
conversion, convert your answer to cubic meters (1 m
^{3} = 1000 L). (Source:
Chemistry, Raymond Chang)
A sample of chlorine gas
occupies a volume of 946 mL at a pressure of 726 torr.
Calculate the pressure of the gas (in torr) if the volume
is reduced at constant temperature to 154 mL. Convert
your answer to atm (1 atm = 760 torr). (Source:
Chemistry, Raymond
Chang)
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Charles’s Law (1787)
The next important law you will learn is called
Charle’s Law. Mr. Charles was an Englishman and
played around with hot air balloons. He never established a
mathematical law describing the relationship between
temperature and volume but some English historian liked his
name better than the name of the man who did. The man who
actually figured out this law of nature is Mr. Gay-Lussac,
a Frenchman. Whoever figured it out, the law we’re interested in
is:
At a given constant pressure (P) and number of moles of
gas (n),
the volume and temperature
of a gas are directly proportional.
In other words, the more you raise the temperature, the
larger the volume of gas (as long at it remains at the same
pressure and you don’t add or subtract any
molecules). Temperature must be expressed in Kelvins, not
degrees Celsius (x°C = x + 273 K). This relationship
can be expressed in the formula:
V V
— = constant or — = k (only when P and n are constant)
T T
Can you write down the formula
relating T_{1} and V_{1} to T_{2}
and V_{2}? Hint: it is very similar to the
formula you discovered relating pressure and volume. Show
the same sort of steps you worked out in that problem.
Now that you have that formula,
find the new volume in L when a 452 mL sample of fluorine
gas is heated from 22°C to 187°C at constant
pressure. Reminder! The only acceptable temperature unit
is Kelvins!(Source: Chemistry, Raymond Chang)
A sample of carbon monoxide gas
occupies 3.20 L at 125°C. Calculate the temperature
at which the gas will occupy 1.54 L if the pressure
remains constant. Be sure to express your answer in
kelvins. (Source: Chemistry, Raymond Chang)
Under constant pressure
conditions 9.6 L of hydrogen gas initially at 88°C is
cooled to -15°C. What is its volume after the cooling
is complete? (Source: Chemistry, Raymond Chang)
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Avogadro’s Law (1811)
This next law uses the unit Mole (mol) which you may never
have heard of. For now it is enough to know that it refers
to the amount of a substance in terms of the number of
atoms or molecules. We will discuss it in more detail
later. To give you some idea of its application to gases
you should know that 1 mol of any gas (just about) will
have a volume of 22.4 L at 273 K (0 °C) and 1 atm. That
1 mol of gas will contain an incredibly large number of
particles: 6.02 x 10^{23} of them (that 602
followed by 21 zeros). At any rate, the law that Mr.
Avogadro proposed is the following:
At a given constant temperature (T) and constant
pressure (P),
the volume and the number of moles, n, of a
gas are directly proportional.
In other words, the more gas you have, the bigger the
volume (as long as you compare volumes at the same pressure
and temperature). The mathematical expression of this
(obvious) law is:
V V
— = constant or — = k (only when T and P are constant)
n n
Note: you cannot change the number of moles just by
changing the volume! For that you have to add or subtract
gas particles.
This should be very easy by now:
relate V_{1} and n_{1} to V_{2}
and n_{2}. Show your work, please!
One mole of an ideal gas has a
volume of 22.4 L at 1 atm and 273K. (This set of
conditions is known as STP: Standard Temperature and
Pressure). What is the volume of 0.50 mol of gas? Of 0.01
mol of gas?
A balloon containing 2.1 mol of
O_{2} has a volume of 36 L at some temperature
and pressure. You pump out 1.5 mol. What is the new
volume?
As an aside, the word
Mole comes from the German Molekül. Its
abbreviation to ‘mol’ is to save you the
hardship of writing that ‘e’ every time.
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Combined Gas Laws
So what happens if two variables change at once? Say the
temperature and the pressure both change and you need to
find the volume: are you stuck, unable to find an answer?
Fortunately, no. You can use the combined gas laws:
PV = nRT is the combined form of the ideal gas law.
This formula is good for static situations
how about when variables are changing?
This formula combines all the gas laws we have used up to now.
P_{1}V_{1} P_{2}V_{2} P_{1}V_{1} P_{2}V_{2}
R = ------ = ------ or ------ = ------ because moles usually don't change
n_{1}T_{1} n_{2}T_{2} T_{1} T_{2} (ex., n_{1} = n_{2} = 1)
R is called the so-called ‘gas constant’ and
has a value of 0.0821 L·atm/K·mol
Sulfur hexafluoride
(SF_{6}) is a colorless, odorless, very
unreactive gas. It is often used to fill the space
between the panes in double-paned windows. Find the
pressure (in atm) of 1.82 mol of SF_{6} in a
steel vessel of volume 5.43 L at 69.5°C. (Which
expression of the combined gas law is appropriate here?)
What is the volume (in L) of 3.2 mol of N_{2} (nitrogen gas) at 3040 torr and 86°C?
A bubble rising from the bottom of a deep lake starts with a temperature of 8°C and a pressure of 6.4 atm. When it gets to the surface, just before it bursts, the temperature is 25°C and the pressure is 1 atm. If the initial volume was 2.1 mL what is the final volume?
A gas starts with V_{1} = 4,000 mL, P_{1} = 1.2 atm, and T_{1} = 66°C. If no gas is added or removed and the V_{2} = 1.7 L and T_{2} = 42°C, what is P_{2}?
Remember to use only Kelvins in your calculations!