6.2.2: The Molecular Basis for Temperature

Learning Objectives

• Explain how temperature relates to the motion of atoms.
• State and explain the ideal gas law using Boltzmann's constant.
• Use the ideal gas law to calculate changes in relationships between pressure, volume, temperature, or number of molecules in a gas.

Figure \(\PageIndex{1}\): The air inside this hot air balloon flying over Putrajaya, Malaysia, is hotter than the ambient air. As a result, the balloon experiences a buoyant force pushing it upward. (credit: Kevin Poh, Flickr)

All matter is made up of atoms, and those atoms are always in motion. The motion of these atoms can take several forms: translation, rotation, and vibration. The type of motion that a particular atom has available to it within a substance depends on the type of bonding (if any) that is present between the atoms of those substances. For solids and liquids, this bonding is too complicated to address here. But for gases, we can describe this relationship in greater detail and how it relates to temperature. Monatomic gases can only move via translation: back and forth linear motion.

Because gas phase atoms and molecules have large separations, forces between them can be ignored, except when they collide with each other during collisions. The motion of atoms and molecules (at temperatures well above the boiling temperature) is fast, such that the gas occupies all of the accessible volume and the expansion of gases is rapid. In contrast, in liquids and solids, atoms and molecules are closer together and are quite sensitive to the forces between them.

Figure \(\PageIndex{2}\): Atoms and molecules in a gas are typically widely separated, as shown. Because the forces between them are quite weak at these distances, the properties of a gas depend more on the number of atoms per unit volume and on temperature than on the type of atom.

To get some idea of how pressure, temperature, and volume of a gas are related to one another, consider what happens when you pump air into an initially deflated tire. The tire’s volume first increases in direct proportion to the amount of air injected, without much increase in the tire pressure. Once the tire has expanded to nearly its full size, the walls limit volume expansion. If we continue to pump air into it, the pressure increases. The pressure will further increase when the car is driven and the tires move. Most manufacturers specify optimal tire pressure for cold tires. (See Figure \(\PageIndex{3}\).)

Figure \(\PageIndex{3}\): (a) When air is pumped into a deflated tire, its volume first increases without much increase in pressure. (b) When the tire is filled to a certain point, the tire walls resist further expansion and the pressure increases with more air. (c) Once the tire is inflated, its pressure increases with temperature.

At room temperatures, collisions between atoms and molecules can be ignored. In this case, the gas is called an ideal gas, in which case the relationship between the pressure, volume, and temperature is given by the equation of state called the ideal gas law.

IDEAL GAS LAW

The ideal gas law states that

\[P V=N k T, \nonumber\]

where \(P\) is the absolute pressure of a gas, \(V\) is the volume it occupies, \(N\) is the number of atoms and molecules in the gas, and \(T\) is its absolute temperature. The constant \(k\) is called the Boltzmann constant in honor of Austrian physicist Ludwig Boltzmann (1844–1906) and has the value

\[k=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}. \nonumber \]

The ideal gas law is a combination of a few other gas laws which related binary pairs of pressure, temperature, volume, and number of particles to result in direct or inverse relationships. It is one of these relationships (pressure and temperature) which led to the prediction of absolute zero discussed previously.

Figure \(\PageIndex{4}\): If the relationship between Pressure and Temperature is extrapolated to zero pressure for any gas we arrive at the accepted value of absolute zero temperature.

Let us see how the ideal gas law is consistent with the behavior of filling the tire when it is pumped slowly and the temperature is constant. At first, the pressure \(P\) is essentially equal to atmospheric pressure, and the volume \(V\) increases in direct proportion to the number of atoms and molecules \(N\) put into the tire. Once the volume of the tire is constant, the equation \(P V=N k T\) predicts that the pressure should increase in proportion to the number N of atoms and molecules.

Example \(\PageIndex{1}\): Calculating Pressure Changes Due to Temperature Changes: Tire Pressure

Suppose your bicycle tire is fully inflated, with an absolute pressure of \(7.00 \times 10^{5} \mathrm{~Pa}\) (a gauge pressure of just under \(90.0 \ \mathrm{lb} / \mathrm{in}^{2}\)) at a temperature of \(18.0^{\circ} \mathrm{C}\). What is the pressure after its temperature has risen to 35.0ºC? Assume that there are no appreciable leaks or changes in volume.

Strategy

The pressure in the tire is changing only because of changes in temperature. First we need to identify what we know and what we want to know, and then identify an equation to solve for the unknown.

We know the initial pressure \(P_{0}=7.00 \times 10^{5} \mathrm{~Pa}\), the initial temperature \(T_{0}=18.0^{\circ} \mathrm{C}\), and the final temperature \(T_{\mathrm{f}}=35.0^{\circ} \mathrm{C}\). We must find the final pressure \(P_{\mathrm{f}} \). How can we use the equation \(P V=N k T\)? At first, it may seem that not enough information is given, because the volume V and number of atoms N are not specified. What we can do is use the equation twice: \(P_{0} V_{0}=N k T_{0}\) and \(P_{\mathrm{f}} V_{\mathrm{f}}=N k T_{\mathrm{f}}\). If we divide \(P_{\mathrm{f}} V_{\mathrm{f}}\) by \(P_{0} V_{0}\) we can come up with an equation that allows us to solve for Pf.

\[\frac{P_{\mathrm{f}} V_{\mathrm{f}}}{P_{0} V_{0}}=\frac{N_{\mathrm{f}} k T_{\mathrm{f}}}{N_{0} k T_{0}} \nonumber\]

Since the volume is constant, \(V_{f}\) and \(V_{0}\) are the same and they cancel out. The same is true for \(N_{\mathrm{f}}\) and \(N_{\mathrm{0}}\), and \(k\), which is a constant. Therefore,

\[\frac{P_{\mathrm{f}}}{P_{0}}=\frac{T_{\mathrm{f}}}{T_{0}}. \nonumber\]

We can then rearrange this to solve for \(P_{f}\):

\[P_{\mathrm{f}}=P_{0} \frac{T_{\mathrm{f}}}{T_{0}}, \nonumber\]

where the temperature must be in units of kelvins, because \(T_{0}\) and \(T_{f}\) are absolute temperatures.

Solution

1. Convert temperatures from Celsius to Kelvin.

\[\begin{aligned}
&T_{0}=(18.0+273) \mathrm{K}=291 \mathrm{~K} \\
&T_{\mathrm{f}}=(35.0+273) \mathrm{K}=308 \mathrm{~K}
\end{aligned} \nonumber\]

2. Substitute the known values into the equation.

\[P_{\mathrm{f}}=P_{0} \frac{T_{\mathrm{f}}}{T_{0}}=7.00 \times 10^{5} \mathrm{~Pa}\left(\frac{308 \mathrm{~K}}{291 \mathrm{~K}}\right)=7.41 \times 10^{5} \mathrm{~Pa} \nonumber\]

Discussion

The final temperature is about 6% greater than the original temperature, so the final pressure is about 6% greater as well. Note that absolute pressure and absolute temperature must be used in the ideal gas law.

MAKING CONNECTIONS: TAKE-HOME EXPERIMENT—REFRIGERATING A BALLOON

Inflate a balloon at room temperature. Leave the inflated balloon in the refrigerator overnight. What happens to the balloon, and why?

Section Summary

• The ideal gas law relates the pressure and volume of a gas to the number of gas molecules and the temperature of the gas.
• The ideal gas law can be written in terms of the number of molecules of gas:

  \[P V=N k T, \nonumber\]

  where \(P\) is pressure, \(V\) is volume, \(T\) is temperature, \(N\) is number of molecules, and \(k\) is the Boltzmann constant

  \[k=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}. \nonumber\]

• The ideal gas law is generally valid at temperatures well above the boiling temperature.

Glossary

ideal gas law

the physical law that relates the pressure and volume of a gas to the number of gas molecules or number of gas particles and the temperature of the gas

Boltzmann constant

k , a physical constant that relates energy to temperature; k=1.38×10^–23 J/K