8.3

The Ideal Gas Law

Chapter contents: Chapter 8: Thermal Physics

A yellow hot-air balloon decorated with a face and sunglasses floats in the sky.
Figure 8.3.1. 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 that is larger than its weight. (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, N2, and oxygen, O2, are composed of two or more atoms. We will primarily use the term “molecule” in discussing a gas, but note that this discussion also applies to monatomic gases, such as helium.)
Gases are easily compressed. 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 β. 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 Figure 8.3.2. 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.
Figure 8.3.2. 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 8.3.3.)
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.
Figure 8.3.3. (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.
In many common circumstances, including, for example, room temperature air, the gas particles have negligible volume and do not interact with each other, aside from perfectly elastic collisions. In such cases, 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
PV=NkBT,
(8.3.1)
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 kB is called the Boltzmann constant in honor of Austrian physicist Ludwig Boltzmann (1844–1906) and has the value
kB=1.38×10−23 J/K.
(8.3.2)
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 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 PV=NkBT predicts that the pressure should increase in proportion to the number N of atoms and molecules.
Example 8.3.1

Calculating Pressure Changes Due to Temperature Changes: Tire Pressure

Suppose your bicycle tire is fully inflated, with an absolute pressure of 7.00×105 Pa (a gauge pressure of just under 90.0lb/in2) at a temperature of 18.0°C. What is the pressure after its temperature has risen to 35.0°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 P0=7.00×105 Pa, the initial temperature T0=18.0°C, and the final temperature Tf=35.0°C. We must find the final pressure Pf. How can we use the equation PV=NkBT? 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: P0V0=NkBT0 and PfVf=NkBTf. If we divide PfVf by P0V0 we can come up with an equation that allows us to solve for Pf.
PfVfP0V0=NfkBTfN0kBT0
Since the volume is constant, Vf and V0 are the same and they cancel out. The same is true for Nf and N0, and kB, which is a constant. Therefore,
PfP0=TfT0.
We can then rearrange this to solve for Pf:
Pf=P0TfT0,
where the temperature must be in units of kelvins, because T0 and Tf are absolute temperatures.
Solution
1. Convert temperatures from Celsius to Kelvin.
T0=(18.0+273) K=291KTf=(35.0+273) K=308K
2. Substitute the known values into the equation.
Pf=P0TfT0=7.00×105 Pa(308K291K)=7.41×105Pa
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 8.3.2

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°C and atmospheric pressure.
Strategy
Because pressure, volume, and temperature are all specified, we can use the ideal gas law PV=NkBT, to find N.
Solution
1. Identify the knowns.
T=0°C=273KP=1.01×105 PaV=1.00 m3kB=1.38×10−23 J/K
2. Identify the unknown: number of molecules, N.
3. Rearrange the ideal gas law to solve for N.
PV=NkBTN=PVkBT
4. Substitute the known values into the equation and solve for N.
N=PVkBT=(1.01×105 Pa)(1.00m3)(1.38×10−23 J/K)(273K)=2.68×1025molecules
Discussion
The calculated number, 2.68×1025, is certainly very large. You might say that the volume of a cubic meter is also large (1m3=1000L), but even in a small volume of 1cm3, which is about size of a thimble (1cm3=10−6m3), a gas at STP has 2.68×1019 molecules in it (still a very large number). Once again, note that N is the same for all types or mixtures of gases.
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