Problem 16
plumleyj 24 Jan 2007 18:43
f(v) is given in problem 15 and takes the form of
(1)\begin{align} f(v)=4\pi v^2 \left(\frac {m} {2\pi kT}\right)^\frac 3 2 e^\frac {-mv^2} {2kT} \end{align}
The most probable value of the molecular speed occurs when the derivative of the function is equal to 0
(2)\begin{align} \begin {flalign*} \frac {\partial f(v)} {\partial v} & = 4\pi \left(\frac {m} {2\pi kT}\right)^\frac 3 2 \left[2v e^\frac {-mv^2} {2kT} + v^2\left(\frac {-mv} {kT} e^\frac {-mv^2} {2kT}\right) \right] \\ & = 4\pi \left(\frac {m} {2\pi kT}\right)^\frac 3 2 e^\frac {-mv^2} {2kT} \left[2v + v^2\left(\frac {-mv} {kT} \right) \right] \\ & = e^\frac {-mv^2} {2kT}\left[2v + v^2\left(\frac {-mv} {kT} \right) \right] \\ & = \left[2+ \left(\frac {-mv^2} {kT} \right) \right] = 0\\ \end {flalign*} \end{align}
Solving for v;
(3)\begin{align} v = \left(\frac {2kT} {m}\right)^{\frac 1 2 } \end{align}
The mean v is determined, as shown below, where $\displaystyle \int_0^\infty x^{2n+1} e^{-ax^2}dx = \frac {n!} {2a^{n+1}}$ is utilized
(4)\begin{align} \begin {flalign*} \langle v \rangle = \displaystyle \int f(v)vdv & = \displaystyle \int_0^\infty v 4\pi v^2 \left(\frac {m} {2\pi kT}\right)^\frac 3 2 e^\frac {-mv^2} {2kT} dv \\ & = 4\pi \left(\frac {m} {2\pi kT}\right)^\frac 3 2 \left[\frac {2k^2T^2} {m^2}\right] \\ & = \left(\frac {8kT} {m \pi} \right)^{\frac 1 2} \end {flalign*} \end{align}
The root-mean-square speed is computed as follows, where $\displaystyle \int_0^\infty x^{2n} e^{-ax^2}dx = \frac{1\cdot 3 \cdot 5 \cdots (2n-1)} {2^{n+1}a^n} \left(\frac \pi a \right) ^\frac 1 2$ is utilized
(5)\begin{align} \begin {flalign*} \langle v^2 \rangle^\frac 1 2 = \left[ \displaystyle \int f(v)v^2dv \right]^\frac 1 2 & = \left[ \displaystyle \int_0^\infty v^2 4\pi v^2 \left(\frac {m} {2\pi kT}\right)^\frac 3 2 e^\frac {-mv^2} {2kT} dv \right]^\frac 1 2 \\ & = \left[4\pi \left(\frac {m} {2\pi kT}\right)^\frac 3 2 \left[\frac 3 2 \left(\frac {2k^5T^5\pi} {m^5}\right)^\frac 1 2 \right] \right]^{\frac 1 2}\\ & =\left(\frac {3kT} {m}\right)^{\frac 1 2} \end {flalign*} \end{align}