20.4 Temperature

An individual molecule of mass m and velocity v has translational kinetic energy

ϵ = \frac{1}{2} m v^{2}

(20.19)

We’ll use $ϵ$ (lowercase Greek epsilon) to distinguish the energy of a molecule from the system energy E. The average translational kinetic energy of all the molecules in a gas is

\begin{array}{l} ϵ_{avg} & = & average translational kinetic energy of a molecule \\ = & \frac{1}{2} m {(v^{2})}_{avg} = \frac{1}{2} m v_{rms}^{2} \end{array}

(20.20)

We’ve included the word “translational” to distinguish $ϵ$ from rotational kinetic energy, which we will consider later in this chapter. Notice how the rms speed arises naturally in this average.

We can write the gas pressure, Equation 20.18, in terms of the average translational kinetic energy as

p = \frac{2}{3} \frac{N}{V} (\frac{1}{2} m v_{rms}^{2}) = \frac{2}{3} \frac{N}{V} ϵ_{avg}

(20.21)

The pressure is directly proportional to the average molecular translational kinetic energy. This makes sense. More-energetic molecules will hit the walls harder as they bounce and thus exert more force on the walls.

It’s instructive to write Equation 20.21 as

p V = \frac{2}{3} N ϵ_{avg}

(20.22)

We know, from the ideal-gas law, that

p V = N k_{B} T

(20.23)

Comparing these two equations, we reach the significant conclusion that the average translational kinetic energy per molecule is

ϵ_{avg} = \frac{3}{2} k_{B} T (average translational kinetic energy)

(20.24)

where the temperature T is in kelvins. For example, the average translational kinetic energy of a molecule at room temperature $(20 ° C)$ is

ϵ_{avg} = \frac{3}{2} (1.38 \times 10^{- 23} J/K) (293 K) = 6.1 \times 10^{- 21} J

NOTE A molecule’s average translational kinetic energy depends only on the temperature, not on the molecule’s mass. If two gases have the same temperature, their molecules have the same average translational kinetic energy.

Equation 20.24 is especially satisfying because it finally gives real meaning to the concept of temperature. Writing it as

T = \frac{2}{3 k_{B}} ϵ_{a v g}

(20.25)

we can see that, for a gas, this thing we call temperature measures the average translational kinetic energy. A higher temperature corresponds to a larger value of $ϵ_{avg}$ and thus to higher molecular speeds. This concept of temperature also gives meaning to absolute zero as the temperature at which $ϵ_{avg} = 0$ and all molecular motion ceases. (Quantum effects at very low temperatures prevent the motions from actually stopping, but our classical theory predicts that they would.) FIGURE 20.7 summarizes what we’ve learned thus far about the micro/macro connection.

The figure consists of two parts labeled Macro and Micro that illustrate the microscopic and macroscopic connections for pressure and temperature.

FIGURE 20.7 Full Alternative Text

We can now justify our assumption that molecular collisions are perfectly elastic. Suppose they were not. If kinetic energy was lost in collisions, the average translational kinetic energy $ϵ_{avg}$ of the gas would decrease and we would see a steadily decreasing temperature. But that doesn’t happen. The temperature of an isolated system remains constant, indicating that $ϵ_{avg}$ is not changing with time. Consequently, the collisions must be perfectly elastic.

EXAMPLE 20.3 Total microscopic kinetic energy

What is the total translational kinetic energy of the molecules in 1.0 mol of gas at STP?

SOLVE The average translational kinetic energy of each molecule is

\begin{matrix} ϵ_{avg} & = & \frac{3}{2} k_{B} T = \frac{3}{2} (1.38 \times 10^{- 23} J / K) (273 K) \\ = & 5.65 \times 10^{- 21} J \end{matrix}

1.0 mol of gas contains $N_{A}$ molecules; hence the total kinetic energy is

K_{micro} = N_{A} ϵ_{avg} = 3400 J

REVIEW The energy of any one molecule is incredibly small. Nonetheless, a macroscopic system has substantial thermal energy because it consists of an incredibly large number of molecules.

By definition, $ϵ_{avg} = \frac{1}{2} m v_{rms}^{2}$ . Using the ideal-gas law, we found $ϵ_{avg} = \frac{3}{2} k_{B} T$ . By equating these expressions we find that the rms speed of molecules in a gas is

v_{rms} = \sqrt{\frac{3 k_{B} T}{m}}

(20.26)

The rms speed depends on the square root of the temperature and inversely on the square root of the molecular mass. For example, room-temperature nitrogen (molecular mass 28 u) has rms speed

v_{rms} = \sqrt{\frac{3 (1.38 \times 10^{- 23} J/K)(293 K)}{28 (1.66 \times 10^{- 27} kg)}} = 509 m/s

This is in excellent agreement with the experimental results of Figure 20.2.

NOTE Recall that a gas always fills its container. A lower temperature causes the molecules to move more slowly, but they continue to fill the container and to interact only via collisions. Only at the condensation temperature do molecules begin to attract each other and stick together.

EXAMPLE 20.4 Mean time between collisions

Estimate the mean time between collisions for a nitrogen molecule at 1.0 atm pressure and room temperature $(20 ° C)$ .

MODEL Because $v_{rms}$ is essentially the average molecular speed, the mean time between collisions is simply the time needed to travel distance $λ$ , the mean free path, at speed $v_{rms}$ .

SOLVE We found $λ = 2.3 \times 10^{- 7} m$ in Example 20.1 and $v_{rms} = 509 m / s$ above. Thus the mean time between collisions is

{(Δ t)}_{avg} = \frac{λ}{v_{rms}} = \frac{2.3 \times 10^{- 7} m}{509 m/s} = 4.5 \times 10^{- 10} s

REVIEW The air molecules around us move very fast, they collide with their neighbors about two billion times every second, and they manage to move, on average, only about 230 nm between collisions.