📂Thermal Physics

Average Kinetic Energy of Gas Molecules

Formulas¹

The average kinetic energy of gas molecules in a system at a temperature of $T$ is as follows.

$$\left\langle E_{K} \right\rangle = {{3} \over {2}} k_{B} T$$

Description

Calculating the kinetic energy of each gas molecule and then averaging is not only inefficient but also practically impossible. However, according to this formula derived statistically, the kinetic energy depends solely on temperature, making it easier to obtain. The reason why the constant appears strangely as $\dfrac{3}{2}$ is because we assume the world we live in is three-dimensional. The method based on this intuitive understanding is Derivation1, and the method that derives the formula in one mathematical step is Derivation2. Either method essentially says the same thing, but if you understand the derivation process of the Maxwell distribution, both will be easy; otherwise, both will be difficult.

Derivation

As mentioned earlier, the essence of the derivation methods is the same; Derivation1 starts from the result in one dimension to derive the result in three dimensions, showing why the denominator is $3$ during the derivation process. Derivation2 calculates it in three dimensions at once.

Derivation1

If the mass of a gas molecule is $m$ and its velocity is $v$, then the kinetic energy is $E_{K} = \frac{1}{2} m v^{2}$, so the average is as follows.

$$\left\langle E_{K} \right\rangle = {{1} \over {2}} m \left\langle v^{2} \right\rangle$$

If $v = (v_{x} , v_{y} , v_{z})$, since expectation is linear, it follows that:

$$\left\langle v^{2} \right\rangle = \left\langle v_{x}^{2} + v_{y}^{2} + v_{z}^{2} \right\rangle = \left\langle v_{x}^{2} \right\rangle + \left\langle v_{y}^{2} \right\rangle + \left\langle v_{z}^{2} \right\rangle$$

The speed of gas molecules follows a distribution with the following probability density function with respect to the $x$ axis.

$$g(v_{x}) = \sqrt{ {m} \over {2 \pi k_{B} T } } e^{ - {{m v_{x}^{2} } \over {2 k_{B} T}} }$$

Therefore, calculating the expectation gives:

$$\left\langle v_{x}^{2} \right\rangle = \int_{-\infty}^{\infty} v_{x}^{2} g(v_{x}) dv_{x} = {{k_{B} T } \over {m}}$$

Thus, calculating for three dimensions gives:

$$\left\langle v^{2} \right\rangle = \left\langle v_{x}^{2} \right\rangle + \left\langle v_{y}^{2} \right\rangle + \left\langle v_{z}^{2} \right\rangle = {{k_{B} T } \over {m}} + {{k_{B} T } \over {m}} + {{k_{B} T } \over {m}} = {{3k_{B} T } \over {m}}$$

Therefore, calculating the expected value of the kinetic energy gives:

$$\left\langle E_{K} \right\rangle = {{1} \over {2}} m \left\langle v^{2} \right\rangle = {{1} \over {2}} m {{3k_{B} T } \over {m}} = {{3} \over {2}} k_{B} T$$

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Derivation2

If the mass of gas molecules is $m$ and its speed is $v$, then the kinetic energy is $E_{K} = \frac{1}{2} m v^{2}$, so averaging yields the following.

$$\left\langle E_{K} \right\rangle = {{1} \over {2}} m \left\langle v^{2} \right\rangle$$

Since the speed of gas molecules follows a Maxwell distribution given by the probability density function $f(v) = \dfrac{4}{\sqrt{ \pi }} \left( \dfrac{m}{2 k_{B} T} \right)^{\frac{3}{2}} v^{2} e^{-\frac{mv^{2}}{2k_{B}} }$, calculating the expectation gives:

$$\left\langle v^{2} \right\rangle = \int_{0}^{\infty} v^{2} f(v) dv = {{3k_{B} T } \over {m}}$$

Therefore, the average kinetic energy is as follows.

$$\left\langle E_{K} \right\rangle = {{1} \over {2}} m \left\langle v^{2} \right\rangle = {{1} \over {2}} m {{3k_{B} T } \over {m}} = {{3} \over {2}} k_{B} T$$

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Especially note that the square root of speed is expressed as $v_{\text{rms}} := \sqrt{ \left\langle v^{2} \right\rangle } = \sqrt{ \dfrac{3k_{B} T}{m} }$.