📂Thermal Physics

Derivation of the Ideal Gas Equation through Kinetic Theory of Gases

Definition[^1]

Pressure $p$ of a fluid acting on an area $M$ is defined as the ratio of the force $F$ the fluid exerts perpendicularly on area $M$ to the area $M$ itself.

$$p:=\frac{F}{M} \left[ \mathrm{N/m^{2}} \right]$$

Formula

Let the volume of the gas be $V$, the temperature be $T$, and the number of molecules be $N$. Then, the pressure $p$ of the gas satisfies the following equation.

$$p V = N k_{B}T$$

Here, $k_{B} = 1.3807 \times 10^{-23} J / K$ is the Boltzmann constant.

Explanation

The above formula is referred to as the ideal gas equation. Initially derived from empirical laws, it was later derived mathematically as follows. This resulted from the process of systematically organizing the pressure of a gas.

To understand the derivation process, one must keep in mind that probability and the concept of ratios across a whole are the same. For example, the chance of rolling a sum of 7 with two dice, as can be seen in the following table, is $6$ out of a total of $36$ cases. This ratio directly translates to the probability $\dfrac{7}{36}$ of rolling a sum of $7$ with two dice.

Sum of Dice	2	3	4	5	6	7	8	9	10	11	12	Total
Cases	1	2	3	4	5	6	5	4	3	2	1	36

Derivation

Let’s denote the number of molecules per unit volume as $n = N / V$. The probability density function of the velocity distribution of gas molecules is given as follows.

Maxwell-Boltzmann Distribution

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

$f(v)$ represents the probability of a gas molecule having a speed $v$, meaning the ratio of molecules with speed $v$ out of the total number of molecules. Therefore, the following equation represents the number of molecules with speed $v$ per unit volume.

$$nf (v)$$

Now, let’s fix a specific direction, termed the $z$ axis. Then $\theta$ is equal to variable $(r,\theta, \phi)$ in spherical coordinates.

The ratio of gas molecules within an area with solid angle $\Omega$ is as follows.

$$\dfrac{\Omega}{4\pi} = \dfrac{1}{2}\sin \theta$$

Thus, the following equation represents the number of molecules moving within solid angle $0\sim\Omega$, with speed $v$, per unit volume.

$$\left( n f(v) \right) \left( \dfrac{1}{2} \sin\theta\right)$$

Next, consider a molecule with speed $v$ hitting a wall of area $A$ at angle $\theta$ as shown in the figure below.

Assuming the molecule hits the wall after $t$ seconds. Then, the area swept by the molecules within time $t$ (the shaded area) could be referred to as the number of molecules. Since the area of a parallelogram is base$\times$height, the number of molecules hitting area $A$ within time $t$ is as follows.

$$Avt \cos\theta$$

Therefore, the number of molecules hitting unit area per unit time at angle $\theta$ with speed $v$ is as follows.

$$\left( n f(v) \right) \left( \dfrac{1}{2} \sin\theta\right) \left( v \cos \theta \right)$$

Since force is the rate of change of momentum, let’s calculate the change in momentum of the molecules colliding with the wall. The change in momentum occurs only in the direction perpendicular to the wall. Let’s denote the direction toward the wall as $+$, then it’s as follows.

$$mv\cos\theta - (-mv\cos\theta) = 2mv \cos\theta$$

Thus, the following equation represents the force exerted by molecules moving in direction $\theta$ with speed $v$ on a perpendicular wall per unit time.

$$\left( n f(v) \right) \left( \dfrac{1}{2} \sin\theta\right) \left( v \cos \theta \right) \left( 2mv \cos\theta \right)$$

Hence, integrating over all speeds $v = 0 \sim \infty$ and all angles $\theta = 0 \sim \pi/2$ gives the pressure exerted by the gas on the area.

$$\begin{align*} p &= \int_{v = 0} ^{\infty} \int_{\theta=0}^{\pi/2} \left( n f(v) \right) \left( \dfrac{1}{2} \sin\theta\right) \left( v \cos \theta \right) \left( 2mv \cos\theta \right) dv d\theta \\ &= nm \int_{v = 0} ^{\infty} f(v) v^{2} dv \int_{\theta=0}^{\pi/2} \cos^{2} \theta \sin\theta d\theta \end{align*}$$

The integral over $v$ is the expected value of $v^{2}$. $$\int_{v = 0} ^{\infty} f(v) v^{2} dv = \left\langle v^{2} \right\rangle = \dfrac{3 k_{B} T}{m}$$

The integral over $\theta$, when substituting with $\cos \theta = x$, becomes $-\sin\theta d\theta = dx$, thus it is as follows.

$$\int_{1}^{0} - x^{2} dx = \left. -\dfrac{1}{3}x^{3} \right|_{1}^{0} = \dfrac{1}{3}$$

Therefore, the pressure $p$ is as follows.

$$p = n m \dfrac{3 k_{B} T}{m} \dfrac{1}{3} = n k_{B} T$$

Since $n=N/V$ was the number of molecules per unit volume, the following result is obtained.

$$p=\dfrac{Nk_{B} T}{V} \implies pV = Nk_{B}T$$

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