18.6.1 The Ideal-Gas Law

Section 18.1 introduced the idea of state variables, those parameters that describe the state of a macroscopic system. The state variables for an ideal gas are the volume $V$ of its container, the number of moles $n$ of the gas present in the container, the temperature $T$ of the gas and its container, and the pressure $p$ that the gas exerts on the walls of the container. These four state parameters are not independent of each other. If you change the value of one—by, say, raising the temperature—then one or more of the others will change as well. Each change of the parameters is a change of state of the system.

Experiments during the 17th and 18th centuries found a very specific relationship between the four state variables. Suppose you change the state of a gas, by heating it or compressing it or doing something else to it, and measure $p$ , $V$ , $n$ , and $T$ . Repeat this many times, changing the state of the gas each time, until you have a large table of $p$ , $V$ , $n$ , and $T$ values.

Then make a graph on which you plot $p V$ , the product of the pressure and volume, on the vertical axis and nT, the product of the number of moles and temperature (in kelvins), on the horizontal axis. The very surprising result is that for any gas, whether it is hydrogen or helium or oxygen or methane, you get exactly the same graph, the linear graph shown in FIGURE 18.8. In other words, nothing about the graph indicates what gas was used because all gases give the same result.

The figure illustrates a graph of p V versus n T for an ideal gas.

FIGURE 18.8 Full Alternative Text

NOTE No real gas could extend to $n T = 0$ because it would condense. But an ideal gas never condenses because the only interactions among the molecules are hard-sphere collisions.

As you can see, there is a very clear proportionality between the quantity $p V$ and the quantity nT. If we designate the slope of the line in this graph as $R$ , then we can write the relationship as

p V = R \times (n T)

It is customary to write this relationship in a slightly different form, namely

p V = n R T (ideal - gas law)

(18.13)

Equation 18.13 is the ideal-gas law. The ideal-gas law is a relationship among the four state variables— $p$ , $V$ , $n$ , and $T$ —that characterize a gas in thermal equilibrium.

The constant $R$ , which is determined experimentally as the slope of the graph in Figure 18.8, is called the universal gas constant. Its value, in SI units, is

R = 8.31 J / mol K

The units of $R$ seem puzzling. The denominator $mol K$ is clear because $R$ multiplies $n T$ . But what about the joules? The left side of the ideal-gas law, $p V$ , has units

Pa m^{3} = \frac{N}{m^{2}} m^{3} = N m = joules

The product $p V$ has units of joules, as shown on the vertical axis in Figure 18.8.

NOTE You perhaps learned in chemistry to work gas problems using units of atmospheres and liters. To do so, you had a different numerical value of $R$ expressed in those units. In physics, however, we always work gas problems in SI units. Pressures must be in Pa, volumes in $m^{3}$ , and temperatures in K.

The surprising fact, and one worth commenting upon, is that all gases have the same graph and the same value of $R$ . There is no obvious reason a very simple atomic gas such as helium should have the same slope as a more complex gas such as methane $(CH_{4})$ . Nonetheless, both turn out to have the same value for $R$ . The ideal-gas law, within its limits of validity, describes all gases with a single value of the constant $R$ .

EXAMPLE 18.4 Calculating a gas pressure

100 g of oxygen gas is distilled into an evacuated $600 {cm}^{3}$ container. What is the gas pressure at a temperature of $150 ° C ?$

MODEL The gas can be treated as an ideal gas. Oxygen is a diatomic gas of $O_{2}$ molecules.

SOLVE From the ideal-gas law, the pressure is $p = n R T / V$ . In Example 18.2 we calculated the number of moles in 100 g of $O_{2}$ and found $n = 3.13 mol$ . Gas problems typically involve several conversions to get quantities into the proper units, and this example is no exception. The SI units of $V$ and $T$ are $m^{3}$ and K, respectively, thus

\begin{matrix} V & = & (600 {cm}^{3}) (\frac{1 m}{100 cm})^{3} = 6.00 \times 10^{- 4} m^{3} \\ T & = & (150 + 273) K = 423 K \end{matrix}

With this information, the pressure is

\begin{matrix} p = \frac{n R T}{V} & = & \frac{(3.13 mol) (8.31 J / mol K) (423 K)}{6.00 \times 10^{- 4} m^{3}} \\ = & 1.83 \times 10^{7} Pa = 181 atm \end{matrix}

In this text we will consider only gases in sealed containers. The number of moles (and number of molecules) will not change during a problem. In that case,

\frac{p V}{T} = n R = constant

(18.14)

If the gas is initially in state i, characterized by the state variables $p_{i}$ , $V_{i}$ , and $T_{i}$ , and at some later time in a final state f, the state variables for these two states are related by

\frac{p_{f} V_{f}}{T_{f}} = \frac{p_{i} V_{i}}{T_{i}} (ideal gas in a sealed container)

(18.15)

This before-and-after relationship between the two states, reminiscent of a conservation law, will be valuable for many problems.

EXAMPLE 18.5 Calculating a gas temperature

A cylinder of gas is at $0 ° C$ . A piston compresses the gas to half its original volume and three times its original pressure. What is the final gas temperature?

MODEL Treat the gas as an ideal gas in a sealed container.

SOLVE The before-and-after relationship of Equation 18.15 can be written

T_{2} = T_{1} \frac{p_{2}}{p_{1}} \frac{V_{2}}{V_{1}}

In this problem, the compression of the gas results in $V_{2} / V_{1} = \frac{1}{2}$ and $p_{2} / p_{1} = 3$ . The initial temperature is $T_{1} = 0 ° C = 273 K$ . With this information,

T_{2} = 273 K \times 3 \times \frac{1}{2} = 409 K = 136 ° C

REVIEW We did not need to know actual values of the pressure and volume, just the ratios by which they change.

We will often want to refer to the number of molecules $N$ in a gas rather than the number of moles $n$ . This is an easy change to make. Because $n = N / N_{A}$ , the ideal-gas law in terms of $N$ is

p V = n R T = \frac{N}{N_{A}} R T = N \frac{R}{N_{A}} T

(18.16)

$R / N_{A}$ , the ratio of two known constants, is known as Boltzmann’s constant $k_{B} :$

k_{B} = \frac{R}{N_{A}} = 1.38 \times 10^{- 23} J / K

The subscript B distinguishes Boltzmann’s constant from a spring constant or other uses of the symbol $k$ .

Ludwig Boltzmann was an Austrian physicist who did some of the pioneering work in statistical physics during the mid-19th century. Boltzmann’s constant $k_{B}$ can be thought of as the “gas constant per molecule,” whereas $R$ is the “gas constant per mole.” With this definition, the ideal-gas law in terms of $N$ is

p V = N k_{B} T (ideal - gas law)

(18.17)

Equations 18.13 and 18.17 are both the ideal-gas law, just expressed in terms of different state variables.

Recall that the number density (molecules per $m^{3}$ ) was defined as $N / V$ . A re-arrangement of Equation 18.17 gives the number density as

\frac{N}{V} = \frac{p}{k_{B} T}

(18.18)

This is a useful consequence of the ideal-gas law, but keep in mind that the pressure must be in SI units of pascals and the temperature must be in SI units of kelvins.

EXAMPLE 18.6 The distance between molecules

“Standard temperature and pressure,” abbreviated STP, are $T = 0 ° C$ and $p = 1 atm$ . Estimate the average distance between gas molecules at STP.

MODEL Consider the gas to be an ideal gas.

SOLVE Suppose a container of volume $V$ holds $N$ molecules at STP. How do we estimate the distance between them? Imagine placing an imaginary sphere around each molecule, separating it from its neighbors. This divides the total volume $V$ into $N$ little spheres of volume $v_{i}, where i = 1 to N$ . The spheres of two neighboring molecules touch each other, like a crate full of Ping-Pong balls of somewhat different sizes all touching their neighbors, so the distance between two molecules is the sum of the radii of their two spheres. Each of these spheres is somewhat different, but a reasonable estimate of the distance between molecules would be twice the average radius of a sphere.

The average volume of one of these little spheres is

v_{avg} = \frac{V}{N} = \frac{1}{N / V}

That is, the average volume per molecule ( $m^{3}$ per molecule) is the inverse of the number density, the number of molecules per $m^{3}$ . This is not the volume of the molecule itself but the average volume of space that each molecule can claim as its own. We can use Equation 18.18 to calculate the number density:

\frac{N}{V} = \frac{p}{k_{B} T} = \frac{1.01 \times 10^{5} Pa}{(1.38 \times 10^{- 23} J / K) (273 K)} = 2.69 \times 10^{25} m^{- 3}

where we used the definition of STP in SI units. Thus the average volume per molecule is

v_{avg} = \frac{1}{N / V} = 3.72 \times 10^{- 26} m^{3}

The volume of a sphere is $\frac{4}{3} π r^{3}$ , so the average radius of a sphere is

r_{avg} = (\frac{3}{4 π} v_{avg})^{1 / 3} = 2.1 \times 10^{- 9} m = 2.1 nm

The average distance between two molecules, with their spheres touching, is twice $r_{avg}$ . Thus

average distance = 2 r_{avg} \approx 4 nm

This is a simple estimate, so we’ve given the answer with only one significant figure.

REVIEW One of the assumptions of the ideal-gas model is that atoms or molecules are “far apart” in comparison to the sizes of atoms and molecules. Chemistry experiments find that small molecules, such as $N_{2}$ and $O_{2}$ , are roughly 0.3 nm in diameter. For a gas at STP, we see that the average distance between molecules is more than 10 times the size of a molecule. Thus the ideal-gas model works very well for a gas at STP.

NOTE Don’t confuse STP with the standard state of 1 atm, 25°C used in chemistry.