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How does a gas behave when
it is heated or cooled?
Introduction
The atmosphere is a gas and to understand how clouds are formed along with many other
atmospheric processes, we need to understand how gases behave when they are heated or cooler;
how they behave when they encounter higher or lower pressures, etc.
Through lecture you are now familiar with the Ideal Gas Law that relates the pressure (p),
volume (V) and temperature (T) of an ideal gas in one compact relationship or equation. Three
names in particular are associated with gas laws, those being Robert Boyle (1627-1691), Jacques
Charles (1746-1823), and J. L. Gay-Lussac (1778-1850). Boyle showed that for a fixed amount
of gas at constant temperature, the pressure and volume are inversely proportional to one another.
In other words, if the pressure increases, then the volume must decrease. Boyle’s Law stated
symbolically is
p V = constant.
Recall density, , is defined as the mass per unit volume or mass / volume or
=m/V
Algebraically solving for V, we get
V = m / .
Replacing the V in Boyle’s Law above with the relationship for volume, mass and density yields
another form of Boyle’s Law. If the mass of the gas remains unchanged during the experiment,
i.e., is kept constant, then we can combine our original constant and the mass to make a new
constant. Thus, we can write
p m /  = constant
or
p /  = constant / m = Constant.
Note that constant and Constant are not equal, but are still unchanging. This statement of Boyle’s
Law relates pressure and density instead of pressure and volume; they are equivalent and equally
valid statements of Boyle’s Law.
In Charles’ Law, it is the pressure that is kept constant. Under this constraint, the volume is
proportional to the temperature. This can be expressed as Charles’ Law.
V / T = constant.
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Or as above with Boyle’s Law, we can rewrite Charles’ Law to relate density and temperature as
 T = constant.
Recall that the temperature in Charles’ Law, and in Gay-Lussac’s Law, below must be the
temperature expressed in Kelvin or Absolute degrees.
Although not discussed in lecture, if the volume is kept constant, it is the pressure of the gas that
is proportional to temperature, i.e., Gay-Lussac’s Law.
P / T = constant.
Note the constants in these equations are different constants. All of the above laws are combined
in the Ideal Gas Law.
An Ideal Gas
What is an ideal gas? An ideal gas has a number of properties; real gases often exhibit behavior
very close to ideal. The properties of an ideal gas are:
•
•
•
•
An ideal gas consists of a large number of identical molecules.
The volume occupied by the molecules themselves is negligible compared to the
volume occupied by the gas, i.e., the space between molecules is large compared to
the size of the molecules.
The molecules obey Newton’s laws of motion, and they move in random motion.
The molecules experience forces only during collisions; any collisions are completely
elastic, and take a negligible amount of time.
Thus, an ideal gas is an idealized model of real gases; real gases follow ideal gas behavior if their
density is low enough that the gas molecules don’t interact much, and when they do interact their
collisions can be approximated as elastic, i.e., with no loss of kinetic energy.
The behavior of an ideal gas, that is, the relationship of pressure, volume and
temperature, can be summarized by the Ideal Gas Law.
p V / T = constant
or
p /  T = constant
Note the constant depends on the gas. For dry air the constant is 287.06
Joules / kg - K.
Part I: Temperature, Volume and Density Experiment
This is a simple experiment that qualitatively demonstrates the effects of
temperature on volume and density as predicted by the Ideal Gas Law. We
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will use a 3-liter bottle, a balloon, an ice bath and a hot water (heat the water to near boiling
using a tea pot) bath.
First, cool the bottle (without the balloon on top) and the air within the bottle by placing the
bottle in the ice bath for a few minutes. After the elapsed time, place the balloon over the top of
the bottle. This seals the air molecules within the bottle and inhibits the exchange of molecules
between the air molecules within the bottle and the environmental air molecules.
Next place the bottle, with the balloon on top, in the hot water bath. Note what happens to the
balloon after a short period of time. Based on the Ideal Gas Law, explain your observations.
What do you think the pressure is in the bottle; greater than, the same as, or less than the pressure
outside the bottle? Explain why you believe your answer.
What would have happened if we had performed this experiment with a stopper placed in the
bottle instead of covered with the balloon? In this case what do you think the pressure would be
in the bottle; greater than, the same as, or less than the pressure outside the bottle?
Next remove the balloon and let environmental air into the bottle. Let the bottle set in the hot
water bath for a few minutes so the air in the bottle will come to equilibrium with the warm
water. Replace the balloon and again place the bottle in the ice bath. However, before doing this,
explain what you expect to occur.
Now place the bottle in the ice bath and compare your observations with your predictions.
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This experiment demonstrates how volume and therefore, density (since the number of
molecules, i.e., the mass of the gas is not changed except when the balloon is removed) changes
with changes in temperature. Recall the discussions in class regarding how side-by-side hot and
cold air columns can induce horizontal pressure variations as one proceeds upward through the
atmosphere. Recall that these pressure variations resulted from temperature induced density
differences.
Part II: Some Ideal Gas Law Calculations
Recalling that pressure always decreases with height and using the Ideal Gas Law, calculate the
pressure at which a balloon would burst if it can only expand to twice its original sea-level size.
For this calculation, assume that the temperature is constant throughout the depth of the
atmosphere. Please show your work.
p = __________________ mb
Calculate the temperature at which a balloon would burst if it can only expand to 1.2 times its
original size. For this calculation, assume that the original temperature of the air in the balloon is
77°F and that the pressure is constant while you are warming the balloon. Please show your
work.
T = __________________ K
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Calculate the pressure in a bottle if the original pressure was 1000 mb and the absolute or Kelvin
temperature of the air in the bottle were cooled to half its original value. Please show your work.
p = __________________ mb
Calculate the density of air in a balloon if the original density was 1.2 kg per m3 and the
temperature of the air in the balloon were cooled from 37°C to 17°C. Please show your work.
 = __________________ kg/m3