Her er nokre utleiingar ein vil få bruk for dersom ein har tenkt å forstå matematikken bak det heile. To integrasjonsresultat ein får bruk for er
\(\newcommand{\into}[0]{\int_0^\infty} \into xe^{-x} dx = 1 \\ \\ \into x^\frac{3}{2}e^{-x} dx = \sqrt{\pi}\)
Gjennomsnittshastigheit til ein gasspartikkel:
\(\newcommand{\closed}[1]{\left(#1\right)} \begin{align*} \langle v \rangle &= \int_0^\infty v P(v) dv \qquad \qquad \text{der} \quad P(v) = \closed{\frac{m}{2\pi kT}}^\frac{3}{2} e^{-\frac{mv^2}{2kT}} 4\pi v^2 \\ &= \int_0^\infty v \closed{\frac{m}{2\pi kT}}^\frac{3}{2} e^{-\frac{mv^2}{2kT}} 4\pi v^2 dv \\ &= \closed{\frac{m}{2\pi kT}}^\frac{1}{2} \int_0^\infty 4\pi v^3 \frac{m}{2\pi kT} e^{-\frac{mv^2}{2kT}} dv \\ &= \closed{\frac{m}{2\pi kT}}^\frac{1}{2} \int_0^\infty 4v \frac{mv^2}{2kT} e^{-\frac{mv^2}{2kT}} dv \\ &= 4 \closed{\frac{m}{2\pi kT}}^\frac{1}{2} \int_0^\infty vue^{-u}\frac{kT}{mv} du\qquad \qquad u = \frac{mv^2}{2kT} \Rightarrow \frac{du}{dv} = \frac{mv}{kT} \Rightarrow dv = \frac{kT}{mv}du \\ &= \frac{4kT}{m} \closed{\frac{m}{2\pi kT}}^\frac{1}{2}\int_0^\infty ue^{-u} du \\ &= \closed{\frac{16k^2T^2}{m^2}\frac{m}{2\pi kT}}^\frac{1}{2} \\ &= \sqrt{\frac{8kT}{m\pi}} \qquad \qquad \square \end{align*} \)
Ideell gasslov:
\(\newcommand{\closed}[1]{\left(#1\right)} \newcommand{\into}[0]{\int_0^\infty} \begin{align*} P &= \frac{1}{3} \into pvn(p) dp \qquad \qquad \text{der} \quad p = mv \Rightarrow v = \frac{p}{m} \\ &= \frac{1}{3} \into p \frac{p}{m} n(p) dp \\ &= \frac{1}{3m} \into p^2 nP(p) dp \\ &= \frac{n}{3m} \into p^2 \closed{\frac{m}{2\pi kT}}^\frac{3}{2} e^{-\frac{p^2}{2mkT}} 4\pi \frac{p^2}{m^2} \frac{1}{m} dp \\ &= \frac{n}{3m} \frac{4\pi}{\pi^\frac{3}{2}} \into \frac{p^4}{m^3} \closed{\frac{m}{2kT}}^\frac{3}{2} e^{-\frac{p^2}{2mkT}} dp \\ &= \frac{4n}{3\sqrt{\pi}m} \into p \closed{\frac{m}{2kT}\frac{p^2}{m^2}}^\frac{3}{2} e^{-\frac{p^2}{2mkT}} dp \\ &= \frac{4n}{3\sqrt{\pi}m} \into p \closed{\frac{p^2}{2mkT}}^\frac{3}{2} e^{-\frac{p^2}{2mkT}} dp \\ &= \frac{4n}{3\sqrt{\pi}m} \into p u^\frac{3}{2} e^{-u} \frac{mkT}{p} du \qquad \qquad u = \frac{p^2}{2mkT} \Rightarrow \frac{du}{dp} = \frac{p}{mkT} \Rightarrow dp = \frac{mkT}{p}du \\ &= \frac{4nkT}{3\sqrt{\pi}} \into u^\frac{3}{2} e^{-u} du \\ &= \frac{4nkT}{3\sqrt{\pi}} \frac{3\sqrt{\pi}}{4} \\ &= nkT \qquad \qquad \square \end{align*} \)
Gjennomsnittsenergi for eit molekyl i ein ideell gass:
\(\newcommand{\closed}[1]{\left(#1\right)} \newcommand{\into}[0]{\int_0^\infty} \begin{align*} \langle E \rangle &= \into f(v) P(v) dv \qquad \qquad \text{der} \quad E = \frac{1}{2}mv^2 \\ &= \into \frac{1}{2}mv^2 \closed{\frac{m}{2\pi kT}}^\frac{3}{2} e^{-\frac{mv^2}{2kT}} 4\pi v^2 dv \\ &= \frac{2\pi m}{\pi^\frac{3}{2}} \into v^4 \closed{\frac{m}{2kT}}^\frac{3}{2} e^{-\frac{mv^2}{2kT}} dv \\ &= \frac{2m}{\sqrt{\pi}} \into v \closed{\frac{mv^2}{2kT}}^\frac{3}{2} e^{-\frac{mv^2}{2kT}} dv \\ &= \frac{2m}{\sqrt{\pi}} \into v u^\frac{3}{2} e^{-u} \frac{kT}{mv} du \qquad \qquad u = \frac{mv^2}{2kT} \Rightarrow \frac{du}{dv} = \frac{mv}{kT} \Rightarrow dv = \frac{kT}{mv}du \\ &= \frac{2kT}{\sqrt{\pi}} \into u^\frac{3}{2} e^{-u} du \\ &= \frac{2kT}{\sqrt{\pi}} \frac{3\sqrt{\pi}}{4} \\ &= \frac{3}{2}kT \qquad \qquad \square \end{align*} \)