📂Thermal Physics

Expectation Value of Speed and Velocity of Gas Molecules

Formulas¹

Let’s denote the velocity of gas molecules as $\mathbf{v} = (v_{x}, v_{y}, v_{z})$ and their speed as $v = | \mathbf{v} |$. The expected values of velocity and speed of gas molecules are as follows.

$$\begin{align*} \left\langle v_{x} \right\rangle &= 0 \\ \left\langle |v_{x}| \right\rangle &= \sqrt{\dfrac{2 k_{B} T}{\pi m}} \\ \left\langle v_{x} ^{2} \right\rangle &= \dfrac{k_{B} T}{\pi m} \\ \left\langle v \right\rangle &= \sqrt{\dfrac{8 k_{B} T}{\pi m}} \\ \left\langle v^{2} \right\rangle &= \dfrac{3 k_{B} T}{\pi m} \end{align*}$$

Explanation

The proof uses the following generalized Gaussian integral formula.

$$\begin{equation} \int_{-\infty}^{\infty} xe^{-\alpha x^{2}}dx = 0 \end{equation}$$

$$\begin{equation} \int_{0}^{\infty} x e^{-\alpha x^{2}}dx = \dfrac{1}{2 \alpha} \end{equation}$$

$$\begin{equation} \int_{-\infty}^{\infty} x^{2} e^{-\alpha x^{2}}dx = \dfrac{1}{2}\sqrt{\dfrac{\pi}{\alpha^{3}}} \end{equation}$$

$$\begin{equation} \int_{0}^{\infty} x^{3} e^{-\alpha x^{2}}dx = \dfrac{1}{2 \alpha^{2}} \end{equation}$$

Proof

Expectation

When a variable is $x$, and the probability density function of $x$ is $f(x)$, the expectation of $x$ is as follows.

$$\left\langle x \right\rangle = \int x f(x) dx$$

We calculate directly from the definition of expectation.

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The probability density function of velocity is as follows.

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

From the definition of expectation and the Gaussian integral formula $(1)-(4)$, it is easily calculated.

$$\left\langle v_{x} \right\rangle = \int v_{x} g(v_{x})dy_{x} = \sqrt{ \dfrac{m}{2 \pi k_{B} T}} \int_{-\infty}^{\infty} v_{x} e^{ - m v_{x}^2 / 2 k_{B} T}dv_{x} = 0$$

$$\begin{align*} \left\langle \left| v_{x} \right| \right\rangle =&\ \int_{-\infty}^{\infty} \left| v_{x} \right| g(v_{x})dy_{x} \\ =&\ \sqrt{ \dfrac{m}{2 \pi k_{B} T}} \int_{-\infty}^{\infty} \left| v_{x} \right| e^{ - m v_{x}^2 / 2 k_{B} T}dv_{x} \\ =&\ 2\sqrt{ \dfrac{m}{2 \pi k_{B} T}}\int_{0}^{\infty} v_{x} e^{ - m v_{x}^2 / 2 k_{B} T}dv_{x} \\ =&\ \sqrt{ \dfrac{m}{2 \pi k_{B} T}} \dfrac{2 k_{B} T}{m} \\ =&\ \sqrt{\dfrac{2 k_{B} T}{\pi m}} \end{align*}$$

$$\begin{align*} \left\langle v_{x}^{2} \right\rangle =&\ \int_{-\infty}^{\infty} v_{x}^{2} g(v_{x})dy_{x} \\ =&\ \sqrt{ \dfrac{m}{2 \pi k_{B} T}} \int_{-\infty}^{\infty} v_{x}^{2} e^{ - m v_{x}^2 / 2 k_{B} T}dv_{x} \\ =&\ \sqrt{ \dfrac{m}{2 \pi k_{B} T}} \dfrac{\sqrt{\pi}}{2} \left( \sqrt{\dfrac{2 k_{B} T}{m}} \right)^{3} \\ =&\ \dfrac{k_{B} T}{m} \end{align*}$$

The probability density function of speed follows the Maxwell distribution.

$$f(v) = \dfrac{4}{\sqrt{\pi}} \left( \dfrac{m}{2 k_{B} T} \right)^{3/2} v^{2} e^{-m v^{2} /2 k_{B} T}$$

Therefore, the expectation is calculated as follows.

$$\begin{align*} \left\langle v \right\rangle =&\ \int_{0}^{\infty} v f(v) dv \\ =&\ \int_{0}^{\infty} \dfrac{4}{\sqrt{\pi}} \left( \dfrac{m}{2 k_{B} T} \right)^{3/2} v^{3} e^{-m v^{2} /2 k_{B} T} dv \\ =&\ \dfrac{4}{\sqrt{\pi}} \left( \dfrac{m}{2 k_{B} T} \right)^{3/2} \int_{0}^{\infty} v^{3} e^{-m v^{2} /2 k_{B} T} dv \\ =&\ \dfrac{4}{\sqrt{\pi}} \left( \dfrac{m}{2 k_{B} T} \right)^{3/2} \dfrac{1}{2} \left( \dfrac{2 k_{B} T}{m} \right)^{2} \\ =&\ \dfrac{4}{\sqrt{\pi}} \sqrt{\dfrac{m}{2 k_{B} T} } \left( \dfrac{m}{2 k_{B} T} \right) \dfrac{1}{2} \left( \dfrac{2 k_{B} T}{m} \right)^{2} \\ =&\ \sqrt{\dfrac{1}{\pi} \dfrac{2^{8}}{2^{5}} \dfrac{m^{3}}{m^{4}} \dfrac{k_{B}^{4}}{k_{B}^{3}} \dfrac{T^{4}}{T^{3}} } \\ =&\ \sqrt{\dfrac{2^{3} k_{B} T}{\pi m }} \\ =&\ \sqrt{\dfrac{8 k_{B} T}{\pi m }} \end{align*}$$

Since expectation is linear, the expectation of the square of speed is as follows.

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

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