Sources: Physical Chemistry - P.W. Atkins - 4th edition Oxford University Press (Oxford) Chemical Principles- Steven S. Zumdahl-4th edition- Hougton-Mifflin (London)
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 mean value (expected value) of a random variable $X$ depending on $x$ on the set $I$ of possible values of $x$ is given by: $\lt X \gt=\int_If(x)dx$ Two integrals $\int_{0}^\infty\,x^3e^{-x^2}dx$ $=$ $\frac{1}{2} $ $\int_{0}^\infty\,x^4e^{-x^2}dx$ $=$ $\frac{3}{8}\sqrt \pi $
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 where $T$ is the Kelvin temperature $k=1.38\cdot 10^{-23}$ is the Boltzmann constant $m$ is the mass of one molecule
$\lt v\gt$= $\int_{0}^\infty\,v\mathscr{f}(v)dv=$ $\int_{0}^\infty\,v 4 \pi(\frac{m}{2\pi\cdot k\cdot T})^\frac{3}{2}v^2e^{-\frac{mv^2}{2kT} }dv=$ $4 \pi(\frac{m}{2\pi\cdot k\cdot T})^\frac{3}{2}\int_{0}^\infty\,v^3e^{-\frac{mv^2}{2kT} }dv$ Taking $y^2=\frac{mv^2}{2kT}$ we find: $\lt v\gt$= $4 \pi(\frac{m}{2\pi\cdot k\cdot T})^\frac{3}{2}(\frac{2kT}{m})^2\int_{0}^\infty\,y^3e^{-y^2}dy=$ $4 \pi(\frac{m}{2\pi\cdot k\cdot T})^\frac{3}{2}(\frac{2kT}{m})^2\frac{1}{2}=$ $\sqrt {\frac{8kT}{\pi m}}=$ $\sqrt {\frac{8RT}{\pi M}}$
Average speed of a molecule of an ideal gas: $\lt v\gt =\sqrt {\frac{8RT}{\pi M}}$ where $R = 8.3 \frac{J}{mol\cdot K}$ is the ideal gas constant $M$ is the molar mass of the gas ( $kg$) $T$ is the absolute temperature ( $K$)
Average square velocity: $u=$ $\sqrt{\int_{0}^\infty\,v^2\mathscr{f}(v)dv}= $ $\sqrt{\int_{0}^\infty\,v^2 4 \pi(\frac{m}{2\pi\cdot k\cdot T})^\frac{3}{2}v^2e^{-\frac{mv^2}{2kT} }dv}=$ $\sqrt{\int_{0}^\infty\, 4 \pi(\frac{m}{2\pi\cdot k\cdot T})^\frac{3}{2}v^4e^{-\frac{mv^2}{2kT} }dv}$ Taking $y^2=\frac{mv^2}{2kT}$ we find like above: $u=$ $\sqrt {\frac{3kT}{m}}=$ $\sqrt {\frac{3RT}{M}}$
Average square velocity of a molecule of ideal gas: $u =\sqrt {\frac{3RT}{M}}$ where $R = 8.3 \frac{J}{mol\cdot K}$ is the ideal gas constant $M$ is the molar mass of the gas ( $kg$) $T$ is the absolute temperature ( $K$)
This reproduces the result seen → here
Consider the probability density curve of speed given by the Maxwell-Boltzmann law: Limits: $lim_{v\mapsto \pm\infty}\mathscr{f}(v) = 0$ Maximimum: $\frac{d}{dv}\mathscr{f}(v)=0$ $4 \pi(\frac{m}{2\pi\cdot k\cdot T})^\frac{3}{2}v(2-\frac{m}{kT}v^2)e^{-\frac{mv^2}{2kT}}=0$ $2-\frac{m}{kT}v^2=0$ $v=\sqrt{\frac{2kT}{m}}$ $v=\sqrt{\frac{2RT}{M}}$
Most probable velocity of a molecule of an ideal gas: $v_{max}=\sqrt{\frac{2RT}{M}}$ where $R = 8.3 \frac{J}{mol\cdot K}$ is the ideal gas constant $M$ is the molar mass of the gas ( $kg$) $T$ is the absolute temperature ( $K$)
Let's verify these values: $v_{max}=$
$u=$
$\lt v \gt=$