CHEMISTRY 30S – MODULE 2
GASES AND THE ATMOSPHERE
LESSON 4 – TEMPERATURE and Volume
OUTCOMES: The student will be able to:
 Explain the relationship between the temperature of a gas and it's volume and a
constant pressure.
 To examine the mathematical relationship between temperature and volume of a
gas

Describe the contribution of Jacques Charles and Lord Kelvin to our
understanding of gas behaviour.

Define absolute zero.

Describe the relationship between the temperature of a gas and its volume.

Solve problems involving temperature-volume relationships.

Describe some practical applications of Charles’ Law.
The simulation on the following website measures the volume of a gas as it is heated (or
cooled), while keeping the pressure constant. When the temperature is changed, the
volume will change by pushing out or drawing in the syringe’s plunger.
Go to this site:
http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/gaslaw/c
harles_law.html
Procedure
1. Record the starting temperature and volume in a table.
2. Move the slider to change the temperature to about 0°C. Allow the system to
equilibrate, then record the temperature and volume in a data table.
3. Repeat for 5 more data sets.
Analysis and Questions
1. Draw a graph of volume (mL) vs. temperature (°C). Be sure to include a title,
labelled axes, appropriate scales and units. Based upon the graph and the data is
this a direct or inverse relationship? How do you know?
2. Extrapolate your graph -- Redraw your graph with the x –axis extending to minus
300°C. The line should extend until the theoretical point where the volume of the
gas would be zero, known as “Absolute Zero”. What is the value of absolute zero
according to your data?
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Chem 30S
Method of Evaluation:
This assignment will be worth 10 marks. Evaluation will be based on presentation of
data, interpretation of the data and accuracy of the answers to questions.
Jacques Charles
Since the dawn of civilization, bakers have known about the relationship between
temperature and volume. Heating a gas in a cake or bread makes the cake or bread rise
more. But the precise mathematical relationship was not described until 1699 when it was
discovered by the French physicist Guillaume Amontons (1663-1705). Amontons
developed a thermometer that used the increasing volume of gas with temperature instead
of a liquid. In 1702, he measured temperature in terms of proportional changes in
pressure. Robert Boyle also studied the relationship between temperature and pressure,
but like Amontons, failed to discover what we call Charles’ Law because a temperature
scale did not exist.
In 1714, Daniel Fahrenheit (1686-1737) invented the mercury thermometer and the
temperature scale that is named after him. He proposed 0 to be the coldest temperature in
Western Europe and 100 being the highest temperature. This made the freezing point of
water 32°F and the boiling point of water 212°F.
In 1742, Anders Celsius (1701-1744) invented the centigrade temperature scale that is
often named after him. He used the freezing point of water as 0°C and the boiling point
of water as 100°C.
Ammonton’s experiment was repeated much later by Jacques Charles (1746-1823) in
1787, and much, much later by Joseph Gay-Lussac in 1802. However, Charles did not
publish his findings, but Gay-Lussac did. The relationship is most frequently called
Charles’s Law in the British sphere of influence and Gay-Lussac’s Law in the French
sphere but it was never called Amonton’s Law!
Jacques Charles, a French scientist, only had a basic knowledge in mathematics and
almost no science education. He became interested in non-mathematical, experimental
physics when Benjamin Franklin visited Paris in 1779 as an ambassador for the new
United States. With the popularity of hot-air balloons in his time, investigated the
expansion rates of different gases due to temperature changes. He used an apparatus very
similar to that of Boyle. He took gases trapped in J-tubes and immersed them in water
baths with varying temperatures. Regardless of the gas tested, he found that for every 1
degree Celsius change, the volume changed . When the temperature was increased by
273°C, the volume doubled.
Absolute Zero
If decreasing the temperature causes a decrease in the volume, what would the
temperature need to be in order for the volume of a gas to be zero? In your lab activity,
you extrapolated the line created by the data and should have got an x-intercept between
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Chem 30S
–270°C and –275°C.
In 1848, 61 years after Charles’ discovery, William Thomson (1824-1907) (later known
as Lord Kelvin) recognized the significance of the 273 and created the Kelvin Scale
where –273°C is the lowest temperature possible or Absolute Zero. Based on this, the xintercept on a graph of volume versus temperature for any gas would always be –273°C.
Lord Kelvin further reasoned that at this temperature all molecular motion would cease,
the kinetic energy would be zero, and the volume of a gas, hypothetically, would also be
zero. The actual 0 K is equal to –273.15°C. The advantage of this scale is that there are
no negative numbers.
Of course this is only theoretical because eventually as a gas is cooled the kinetic energy
will decrease to a point where the forces of attraction between the particles will begin to
take hold and the gas will condense to a liquid, then a solid. An ideal gas will obey
Charles' Law. In the real world, real gases will condense, their particles take up space and
they have forces of attraction.
According to Charles’ Law, there is a direct relationship between the temperature and the
volume of a gas at a constant pressure. That is, if the Kelvin temperature of a gas is
doubled, the volume of the gas would double as well.
The diagram below shows a comparison of the three temperature scales.
To convert from Celsius to Kelvin:
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Chem 30S
Kelvin = Celsius + 273
OR
K = °C + 273
Example 1. Convert –10.0°C to Kelvin.
Solution.
K = °C + 273 = –10.0°C + 273 = 263 K
Note: There is no degree sign for temperatures in Kelvin. We say the temperature is “263
Kelvin”, no degrees.
Example 2. Convert 298 K into degrees Celsius.
Solution.
°C = K – 273 = 298 – 273 = 25°C
Solving Problems Using Charles’ Law
We solve problems using Charles’ Law in much the same manner that we solved the
problems using Boyle’s Law
1. Determine if the change will increase or decrease the temperature or volume.
2. Multiply by the ratio that will produce the desired change.
The temperature must always be in Kelvin.
Example 3. What is the new volume of a gas if 100 mL of the gas at 25°C is cooled
to –25°C?
Solution.
Since the temperature is decreasing from 25°C to –25°C, the volume must also
decrease.
We must FIRST convert all temperatures to Kelvin
25°C + 273 = 298 K
–25°C + 273 = 248 K
Then, we can use this relationship: T1/V1 = T2/V2 to solve for the new volume, V2
298 K = 248 K
100 ml
V2
Cross multiplying to solve for V2,
(298 K)( V2) = (248 K)(100 mL)
we find V2 = 83.2 mL
Another method for doing this is as follows:
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Chem 30S
1. Determine if the change will increase or decrease the temperature or volume.
2. Multiply by the ratio that will produce the desired change.
There are two possible ratios for the changing temperature:
Since the temperature is decreasing, the volume must also decrease, so we multiply by
the ratio that will bring the volume down.
Since the volume must get smaller, we must use the second ratio:
Example 4. If the volume of a gas at –73°C is doubled to 48.0 L, calculate the final
temperature in degrees Celsius.
Solution.
Convert the Celsius temperature to Kelvin.
–73°C + 273 = 200 K
If the volume doubled to 48.0 L, its initial volume must have been half of 48.0 L or 24.0
L. The volume increases, so according to Charles’ Law the temperature must have
increased. There are two possible ratios:
Since the temperature increases, we must use the second ratio.
Or, we can use T1/V1 = T2/V2
200 K = T2
24.0 L 48.0 L
And cross multiply to solve for T2
Convert the Kelvin temperature to Celsius.
K – 273 = 400 – 273 = 127°C
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Chem 30S
Applying Charles’ Law
When you were younger you may have had a helium balloon and then went out into the
cold. What happened to the balloon? To your disappointment, the balloon started to
shrink. The temperature change from room temperature to the cold winter air caused the
volume of the gas in the balloon to decrease.
If you leave a closed can or a bottle of pop in the window of your car on a hot summer
day, there is a good chance that the can or bottle will explode. The gas present at the top
of the can or bottle may expand to such a large extent, that the container may not be able
to hold it.
Exercise
Answer the following questions. Remember that showing your work is good practice.
1. Calculate the change in temperature, in degrees Celsius, when the volume of 1.00
L of a gas, at 25.0°C, is doubled to 2.00 L.
2. Calculate the volume of 100.0 mL gas if its temperature is doubled from 20°C to
40°C.
3. You get a balloon at the circus with a volume of 2.50 L at room temperature
(25.0°C). You then step outside on a cold winter day (–25.0°C). If the pressure
remains constant, what is the balloon’s new volume?
4. 20.0 mL of a gas at 285 K increases in volume to 32.0 mL. What is the new
temperature of the gas?
5. Find the new volume if the following changes occur at a constant pressure:
a. 225 mL of oxygen at 273 K warmed to 300 K.
b. 3.0 L of nitrogen cooled from 90.0°C to –45.0°C.
6. Find the new temperature in degrees Celsius when the following changes occur at
a constant pressure:
a. 5.0 L of air at 45.0°C expands to 22.0 L.
b. 100.0 mL of helium at 300 K compresses to 80.0 mL
7. Would it be possible for a sample of gas to have a volume of zero? Explain.
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Chem 30S
Answer Key
1.
2.
3.
4.
5.
6.
323 C
107 mL
2.08 L
456 K
1.9 L
a) 1126 C
b) – 33 C
7. As the volume of a gas decreases, the space between the gas molecules will decrease.
Eventually the gas will condense to a liquid then compress to become a solid. There is a
point beyond which the sample cannot be compressed because the molecules actually
occupy space.
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