Solved: The speed of a molecule in a uniform gas at equilibrium is a random variable whose probability density function is given by

ByDr. Raju Chaudhari

Feb 26, 2021

The speed of a molecule in a uniform gas at equilibrium is a random variable whose probability density function is given by

 \begin{aligned} f(x) &= ax^2 e^{-bx^2}, x\geq 0 \end{aligned}

where $b = m/2kT$ and $k$, $T$, and $m$ denote, respectively, Boltzmann's constant, the absolute temperature of the gas, and the mass of the molecule. Evaluate $a$ in terms of $b$.

Solution

As given $f(x)$ is a probability density function, we have

 \begin{aligned} &\int_0^\infty f(x) \; dx = 1\\ &\Rightarrow \int_0^\infty ax^2 e^{-bx^2} \; dx = 1\\ &\Rightarrow a \int_0^\infty x^{2*(3/2) -1} e^{-bx^2} \; dx = 1\\ &\Rightarrow a \frac{\Gamma(3/2)}{2b^{3/2}} = 1\\ & \quad \quad \text{using } \int_0^\infty x^{2n-1}e^{-ax^2}\; dx =\frac{\Gamma(n)}{2a^n}\\ &\Rightarrow a \Gamma(3/2) =2b^{3/2}\\ &\Rightarrow a \frac{1}{2}\Gamma(1/2) =2b^{3/2}\\ &\Rightarrow a \frac{1}{2}\sqrt{\pi} =2b^{3/2}\\ &\Rightarrow a =\frac{4b^{3/2}}{\sqrt{\pi}} \end{aligned}

Using $\Gamma(n)=(n-1)\Gamma(n-1)$ and $\Gamma(1/2) =\sqrt{\pi}$.