The Ideal Gas Law

The Ideal Gas Law

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)

In this section, we continue to explore the thermal behavior of gases. In particular, we examine the characteristics of atoms and molecules that compose gases. (Most gases, for example nitrogen, \({\text{N}}_{2}\), and oxygen, \({\text{O}}_{2}\), are composed of two or more atoms. We will primarily use the term “molecule” in discussing a gas because the term can also be applied to monatomic gases, such as helium.)

Gases are easily compressed. We can see evidence of this in this table, where you will note that gases have the largest coefficients of volume expansion. The large coefficients mean that gases expand and contract very rapidly with temperature changes. In addition, you will note that most gases expand at the same rate, or have the same \(\beta \). This raises the question as to why gases should all act in nearly the same way, when liquids and solids have widely varying expansion rates.

The answer lies in the large separation of atoms and molecules in gases, compared to their sizes, as illustrated in the figure below. Because 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.

Spheres representing atoms and molecules; the spheres are relatively far apart and are distributed randomly.

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 the figure below.)

The figure has three parts, each part showing a pair of tires, and each tire connected to a pressure gauge. Each pair of tires represents the before and after images of a single tire, along with a change in pressure in that tire. In part a, the tire pressure is initially zero. After some air is added, represented by an arrow labeled Add air, the pressure rises to slightly above zero. In part b, the tire pressure is initially at the half-way mark. After some air is added, represented by an arrow labeled Add air, the pressure rises to the three-fourths mark. In part c, the tire pressure is initially at the three-fourths mark. After the temperature is raised, represented by an arrow labeled Increase temperature, the pressure rises to nearly the full mark.

(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

\(\text{PV}=\text{NkT},\)

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

\(k=1\text{.}\text{38}×{\text{10}}^{-\text{23}}\phantom{\rule{0.25em}{0ex}}\text{J}/\text{K}.\)

The ideal gas law can be derived from basic principles, but was originally deduced from experimental measurements of Charles’ law (that volume occupied by a gas is proportional to temperature at a fixed pressure) and from Boyle’s law (that for a fixed temperature, the product \(\text{PV}\) is a constant). In the ideal gas model, the volume occupied by its atoms and molecules is a negligible fraction of \(V\). The ideal gas law describes the behavior of real gases under most conditions. (Note, for example, that \(N\) is the total number of atoms and molecules, independent of the type of gas.)

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 \(\text{PV}=\text{NkT}\) predicts that the pressure should increase in proportion to the number N of atoms and molecules.

Example: Calculating Pressure Changes Due to Temperature Changes: Tire Pressure

Suppose your bicycle tire is fully inflated, with an absolute pressure of \(7\text{.}\text{00}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{Pa}\) (a gauge pressure of just under \(\text{90}\text{.}0\phantom{\rule{0.25em}{0ex}}{\text{lb/in}}^{2}\)) at a temperature of \(\text{18}\text{.}0\text{º}\text{C}\). What is the pressure after its temperature has risen to \(\text{35}\text{.}0\text{º}\text{C}\)? 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 \({P}_{0}=7\text{.00}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{Pa}\), the initial temperature \({T}_{0}=\text{18}\text{.}0ºC\), and the final temperature \({T}_{\text{f}}=35\text{.}0ºC\). We must find the final pressure \({P}_{\text{f}}\). How can we use the equation \(\text{PV}=\text{NkT}\)? 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: \({P}_{0}{V}_{0}={\text{NkT}}_{0}\) and \({P}_{\text{f}}{V}_{\text{f}}={\text{NkT}}_{\text{f}}\). If we divide \({P}_{\text{f}}{V}_{\text{f}}\) by \({P}_{0}{V}_{0}\) we can come up with an equation that allows us to solve for \({P}_{\text{f}}\).

\(\cfrac{{P}_{\text{f}}{V}_{\text{f}}}{{P}_{0}{V}_{0}}=\cfrac{{N}_{\text{f}}{\text{kT}}_{\text{f}}}{{N}_{0}{\text{kT}}_{0}}\)

Since the volume is constant, \({V}_{\text{f}}\) and \({V}_{0}\) are the same and they cancel out. The same is true for \({N}_{\text{f}}\) and \({N}_{0}\), and \(k\), which is a constant. Therefore,

\(\cfrac{{P}_{\text{f}}}{{P}_{0}}=\cfrac{{T}_{\text{f}}}{{T}_{0}}\text{.}\)

We can then rearrange this to solve for \({P}_{\text{f}}\):

\({P}_{\text{f}}={P}_{0}\cfrac{{T}_{\text{f}}}{{T}_{0}},\)

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

Solution

1. Convert temperatures from Celsius to Kelvin.

\(\begin{array}{}{T}_{0}=\left(\text{18}\text{.}0+\text{273}\right)\text{K}=\text{291 K}\\ {T}_{\text{f}}=\left(\text{35}\text{.}0+\text{273}\right)\text{K}=\text{308 K}\end{array}\)

2. Substitute the known values into the equation.

\({P}_{\text{f}}={P}_{0}\cfrac{{T}_{\text{f}}}{{T}_{0}}=7\text{.}\text{00}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{Pa}(\cfrac{\text{308 K}}{\text{291 K}})=7\text{.}\text{41}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{Pa}\)

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?

Example: Calculating the Number of Molecules in a Cubic Meter of Gas

How many molecules are in a typical object, such as gas in a tire or water in a drink? We can use the ideal gas law to give us an idea of how large \(N\) typically is.

Calculate the number of molecules in a cubic meter of gas at standard temperature and pressure (STP), which is defined to be \(0\text{º}\text{C}\) and atmospheric pressure.

Strategy

Because pressure, volume, and temperature are all specified, we can use the ideal gas law \(\text{PV}=\text{NkT}\), to find \(N\).

Solution

1. Identify the knowns.

\(\begin{array}{lll}T& =& 0\text{º}\text{C}=\text{273 K}\\ P& =& 1\text{.}\text{01}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{Pa}\\ V& =& 1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{3}\\ k& =& 1\text{.}\text{38}×{\text{10}}^{-\text{23}}\phantom{\rule{0.25em}{0ex}}\text{J/K}\end{array}\)

2. Identify the unknown: number of molecules, \(N\).

3. Rearrange the ideal gas law to solve for \(N\).

\(\begin{array}{}\text{PV}=\text{NkT}\\ N=\frac{\text{PV}}{\text{kT}}\end{array}\)

4. Substitute the known values into the equation and solve for \(N\).

\(N=\cfrac{\text{PV}}{\text{kT}}=\cfrac{(1\text{.}\text{01}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{Pa})(1\text{.}{\text{00 m}}^{3})}{(1\text{.}\text{38}×{\text{10}}^{-\text{23}}\phantom{\rule{0.25em}{0ex}}\text{J/K})(\text{273 K})}=2\text{.}\text{68}×{\text{10}}^{\text{25}}\phantom{\rule{0.25em}{0ex}}\text{molecules}\)

Discussion

This number is undeniably large, considering that a gas is mostly empty space. \(N\) is huge, even in small volumes. For example, \(1\phantom{\rule{0.25em}{0ex}}{\text{cm}}^{3}\) of a gas at STP has \(2\text{.}\text{68}×{\text{10}}^{\text{19}}\) molecules in it. Once again, note that \(N\) is the same for all types or mixtures of gases.

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