Source: Physical Chemistry - P.W. Atkins - 4th edition Oxford University Press
Area of a sphere $S=4\pi r^2$ where $r$ is the radius Probabilities - (1) The probability of an event is a number between 0 and 1. - (2) The probability of two independent events is the product of the individual probabilities. - (3) The probability that at least one of two events will happen is the sum of the individual probabilities - (4) The sum of the probabilities of all possible cases of events is equal to unity. Probability density functions - A probability density function $f(x)$ is a function such that $f(x)dx $ is the probability that an event dependent on $x$ lies in the small $dx$ interval of the variable $x$ - The probability density $F(x,y)$ is a function such that $F(x,y)dxdy $ gives the probability that an event dependent on $x$ and $y$ is in the interval $dxdy$ of the variables $x$ and $y$ - If the variables $x$ and $y$ are independent, we have $F(x,y)dxdy = f(x)dx\cdot f(y)dy$ (see (2)) - If all values of $x$ are in an interval $[a, b]$, then for $a\lt c \lt d \lt b $ we have: $ 0\le\, \int_c^d f(x)dx\,\le 1 $ (see (1)) $\int_a^b f(x)dx=1$ (see (4)) Two integrals $\int_{-\infty}^\infty\,e^{-x^2}dx$ $=$ $\sqrt\pi\, (4) $ $\int_{-\infty}^\infty\,x^2e^{-x^2}dx$ $=$ $\frac{\sqrt\pi}{2}\, (5) $ For demonstrations see → here and → here
Speed of a molecule and its components Using Pythagoras theorem like → here we find that the speed $v$ of a single molecule is: $v^2=v_x^2+v_y^2+v_z^2$ where $v_x$, $v_x$ et $v_z$ are its components Mean square speed of the molecules of an ideal gas We saw → here that - the components of the mean square speed $u$ are equal: $u_x=u_y=u_z\, (6)$ and furthermore: $u_x^2+u_y^2+u_z^2=u^2\, (7)$ an still: $u=\sqrt{\frac{3kT}{m}}\, (8)$ where $T$ is the Kelvin temperature $k=1.38\cdot 10^{-23}$ is Boltzmann's constant $m$ is the mass of one molecule
The movement of the molecules of an ideal gas is assumed random, so all the velocity vectors $\vec v$ of given modulus (length) $V$ must have the same probability. As $v^2=v_x^2+v_y^2+v_z^2$ and considering that the velocity components are independent (random movement), the probability density $F$ is solely function of the parameter $v_x^2+v_y^2+v_z^2$ and can be decomposed as the product of probability densities $f$ of three components: $F(v_x^2+v_y^2+v_z^2)dv_xdv_ydv_z = f(v_x^2)dx f(v_y^2)dy f(v_z^2)dz \, (9)$ Only an exponential function can verify this relationship because of his well-known property: $e^{a+b+c}=e^ae^be^c$ So we should have: $f(v_x^2)= Ke^{-\kappa v_x^2}\, (10)$ Let's search for $-\kappa$ and $K$ : 1) Why -$\kappa$? $-\kappa$ is of course negative, because speeds approaching infinity must have probabilities tending to $0$ 2) Elimination of $K$ $f$ is a probability density function, so we must have $\int_{-\infty}^\infty f(v_x)dv_x=1$ $\int_{-\infty}^\infty Ke^{-\kappa v_x^2}dv_x=1$ $K\int_{-\infty}^\infty e^{-\kappa v_x^2}dv_x=1$ Taking $\kappa v_x^2=y^2$ it follows: $2\kappa v_xdv_x=2ydy$ $2\kappa \frac{y}{\sqrt \kappa}dv_x=2ydy$ $dv_x=\frac{1}{\sqrt \kappa}dy$ and so: $\frac{K}{\sqrt \kappa}\int_{-\infty}^\infty e^{-y^2}dy=1$ $\frac{K}{\sqrt \kappa}\sqrt \pi=1$ $K=\sqrt \frac{\kappa}{\pi}\, (11)$ and finally: $f(v_x)= \sqrt \frac{\kappa}{\pi}e^{-\kappa v_x^2}$ Détermination de $\kappa$ Let's calculate the average square speed $u_x^2$ of $v_x^2$: $u_x^2$ $=$ $\int_{-\infty}^\infty v_x^2f(v_x)dv_x= \sqrt \frac{\kappa}{\pi}\int_{-\infty}^\infty v_x^2e^{-\kappa v_x^2}dv_x$ Taking again: $\kappa v_x^2=y^2$ we have: $dv_x=\frac{1}{\sqrt \kappa}dy$ and so: $u_x^2$ = $\sqrt \frac{\kappa}{\pi} \frac{1}{\sqrt\kappa}\int_{-\infty}^\infty \frac{y^2}{\kappa} e^{-y^2}dy =$ $\sqrt \frac{\kappa}{\pi} \frac{1}{\sqrt\kappa}\int_{-\infty}^\infty \frac{y^2}{\kappa} e^{-y^2}dy=$ $ \frac{1}{\sqrt\pi\cdot \kappa } \int_{-\infty}^\infty y^2e^{-y^2}dy=$ $\frac{1}{\sqrt\pi\cdot \kappa }\frac{\sqrt\pi}{2 }= $ $\frac{1}{2 \kappa }$ and then by $(6)$ and $(7)$: $u^2=\frac{3}{2 \kappa }$ As we saw by $(8)$ $u^2 = \frac{3kT}{m}$ we find: $\kappa =\frac{m}{2kT}\, (12)$ Probability density of the velocity component along an axis Combining $(10), (11)$ et $(12)$ for the probability density along the axes: $f(v_x)=$ $Ke^{-\kappa v_x^2}=$ $\sqrt \frac{\kappa}{\pi}e^{-\kappa v_x^2 }= $ $\sqrt \frac{m}{2\pi\cdot k\cdot T}e^{-\frac{mv_x^2}{2kT} } \, (13)$ Let's use $(9)$ : $f(v_x^2)f(v_y^2)f(v_z^2)=$ $\sqrt \frac{m}{2\pi\cdot k\cdot T}e^{-\frac{mv_x^2}{2kT} }\sqrt \frac{m}{2\pi\cdot k\cdot T}e^{-\frac{mv_y^2}{2kT} }\sqrt \frac{m}{2\pi\cdot k\cdot T}e^{-\frac{mv_z^2}{2kT} }=$ $(\frac{m}{2\pi\cdot k\cdot T})^\frac{3}{2}e^{-\frac{mv^2}{2kT} }$ Probability density function of the speed of a molecule So far we have assessed the probability $(\frac{m}{2\pi\cdot k\cdot T})^\frac{3}{2}e^{-\frac{mv^2}{2kT} } dv_xdv_ydv_z$ to find the components of the speed $\vec v $ in volume $dv$ delimited by the interval $[v_x,v_x+dv_x]$ ,$[v_y,v_y+dv_y]$ ,$[v_z,v_z+dv_z]$ It remains to determine the probability $\mathscr{f}(v)dv$ so that the speed has a modulus (length) located between $v$ and $v+dv$ :
This probability is the sum of probabilities (see (3)) on all possible volumes $dv_xdv_ydv_z$ ie the volume $4\pi r^2 dv$ represented here: $\mathscr{f}(v)dv=4 \pi(\frac{m}{2\pi\cdot k\cdot T})^\frac{3}{2}v^2e^{-\frac{mv^2}{2kT} } dv$
Law of Maxwell-Boltzmann: Probability density of the speed of a molecule: $\mathscr{f}(v)=4 \pi(\frac{m}{2\pi\cdot k\cdot T})^\frac{3}{2}v^2e^{-\frac{mv^2}{2kT} }$ where $T$ is the Kelvin temperature $k=1.38\cdot 10^{-23}$ is the Boltzmann constant $m$ is the mass of one molecule