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# 13.2: Kinetic Theory of Temperature

Difficulty Level: At Grade Created by: CK-12

## Objectives

The student will:

• Describe the relationship between temperature and kinetic energy.

## Vocabulary

• internal energy:The total kinetic and potential energy associated with the motions and relative positions of the molecules of an object, excluding the kinetic or potential energy of the object as a whole. An increase in internal energy results in a rise in temperature or a change in phase.
• Kinetic Theory: A theory concerning the thermodynamic behavior of matter, especially the relationships among pressure, volume, and temperature in gases. It is based on the dependence of temperature on the kinetic energy of the rapidly moving particles of a substance. According to the theory, energy and momentum are conserved in all collisions between particles, and the average behavior of the particles can be deduced by statistical analysis.

## Introduction

James Clark Maxwell (1831-1879) determined a very important relationship between the motion of molecules and temperature within a gas in the mid-19th century.

Kinetic Theory relies upon the following assumptions:

1. The number of molecules in a gas is very large and their motion is random.
2. The average distance between any two molecules is much greater than the diameter of the any molecule.
3. The molecules interact in accordance with Newton’s laws. Only the kinetic energy of the molecules is considered, and all the interactions between molecules are elastic.

Under these conditions, a gas is referred to as an ideal gas.

When we ask what the temperature is outside, we’re indirectly asking for a measure of the average kinetic energy of the air molecules. Maxwell was able to show that the relationship between the average kinetic energy of a gas and its temperature is K¯¯¯¯¯E¯¯¯¯=12m(v¯¯¯t)2=32kT\begin{align*}\overline{K} \overline{E} =\frac{1}{2}m(\overline{v}_t)^2=\frac{3}{2}kT\end{align*}, where v¯¯¯t\begin{align*}\overline{v}_t\end{align*} is the mean translational velocity of the molecules, m\begin{align*}m\end{align*} is their mass, k\begin{align*}k\end{align*} the Boltzmann constant, is equal to 1.38×1023JK\begin{align*}1.38 \times 10^{-23} \frac{J}{K}\end{align*}, and the temperature T\begin{align*}T\end{align*} is expressed in Kelvin.

The equation states that the average translational kinetic energy, K¯¯¯¯¯E¯¯¯¯\begin{align*}\overline{K} \overline{E}\end{align*} of the molecules is directly proportional to the temperature T\begin{align*}T\end{align*} of the gas.

#### Illustrative Example 1

Determine the average velocity of random motion of an oxygen molecule at 22.0C\begin{align*}22.0^\circ C\end{align*}. The mass of an oxygen molecule is 5.36×1026kg\begin{align*}5.36 \times 10^{-26}kg\end{align*}, and the constant k=1.38×1023Jkg\begin{align*}k=1.38 \times 10^{-23} \frac{J}{kg}\end{align*}.

Answer: Recall that the temperature must be expressed in degrees Kelvin.

TK=TC+27322+273=295K\begin{align*}T_K=T_C+273 \rightarrow 22+273=295K\end{align*}.

K¯¯¯¯¯E¯¯¯¯=12m(v¯¯¯t)2=32kTm(v¯¯¯t)2=3kT(v¯¯¯t)2=3kTm\begin{align*}\overline{K} \overline{E}=\frac{1}{2}m(\overline{v}_t)^2=\frac{3}{2} kT \rightarrow m(\overline{v}_t)^2=3kT \rightarrow (\overline{v}_t)^2=\frac{3kT}{m} \rightarrow\end{align*}

v¯¯¯=3kTm=3(1.38×1023JK)(295K)(5.36×1026kg)=477.34477ms\begin{align*}\overline{v}=\sqrt{\frac{3kT}{m}}=\sqrt{\frac{3\left(1.38 \times 10^{-23} \frac{J}{K}\right)(295K)}{(5.36 \times 10^{-26}kg)}}=477.34 \rightarrow 477 \frac{m}{s}\end{align*}

### Internal Energy

The temperature of a gas is directly proportional to the average kinetic energy of the particles of the gas. But the total kinetic energy of the molecules of a gas is a measure of the internal energy, or thermal energy of the gas.

The internal energy U\begin{align*}U\end{align*} of a gas is, therefore, the product of the number of molecules N\begin{align*}N\end{align*} of the gas and the average kinetic energy of each molecule:

U=N(12mv¯¯¯2)\begin{align*}U=N \left(\frac{1}{2}m\overline{v}^2\right)\end{align*}

But according to the kinetic theory,

12mv¯¯¯2=32kT\begin{align*}\frac{1}{2} m \overline{v}^2=\frac{3}{2}kT\end{align*}

Therefore, the internal energy can be directly related to the temperature of a gas as

U=N(32kT)U=32NkT\begin{align*}U=N \left(\frac{3}{2}kT\right) \rightarrow U=\frac{3}{2}NkT\end{align*}

The internal energy is, therefore, dependent upon the temperature and the number of particles within a substance. This means that a small quantity of substance at a high temperature may have less internal energy than a large quantity of substance at a low temperature. For example, a boiling pot of water has a higher temperature than a large cool mountain lake. But the internal energy of the lake is considerably larger than the internal energy of the boiling water. Even though the average kinetic energy of the molecules of the lake water is smaller than that of the boiling water, there’s a good deal more molecules in the lake.

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Date Created:
Mar 11, 2013