# Question e2b05

Oct 5, 2017

The problem is mispresented and unsoluble (impossible) in this form.

#### Explanation:

There is no temperature at which the mode (most probable), the root mean ssquare velocity (${v}_{\text{rms}}$) and the average velocity have the same value in m/s.
In fact the probability distribution of molecular velocities (Maxwell distribution) is asymmetric. Look at this one:

Your teacher should chose one. The easiest one is ${v}_{\text{rms}}$, because, in that case, you could use the simple formula:

${v}_{\text{rms}} = \sqrt{\frac{3 {k}_{B} T}{m}}$

where ${v}_{\text{rms}}$ is the root mean square velocity in $\frac{m}{s}$;
${k}_{B}$ is Botzmann's constant = 1.38064852 × 10^"-23" ""m^2· kg · s^"-2" ·K^"-1"#
$T$ is the unknown temperature in kelvin,
$m$ is the mass of a dioxygen (${O}_{2}$) molecule in kg.

By assuming 1500 m/s is just the root mean square speed, ${v}_{\text{rms}}$, by using molar mass of dioxygen $\left({M}_{{O}_{2}} = 0.031998 \text{kg"/"mol}\right)$, and knowing that:

${k}_{B} = \left(R \text{ perfect gas constant")/(N_A " Avogadro's constant}\right) =$

$= \left(8.31446 \text{ "m^2· kg · s^"-2" ·K^"-1" · mol^"-1")/(6.02214·10^"23"mol^"-1}\right)$,

we can find $T$ with the following calculation:

$T = \left({v}_{\text{rms"^2 · M_(O_2))/(3R)=((1500 m/s)^2 · 0.031998 "kg"/"mol")/(3 · 8.31446" "m^2· kg · s^"-2" · K^"-1" · mol^"-1}}\right) = 2886 K$

In which I have used the fact that a single dioxygen molecule multiplied by the Avogadro's constant yields the molar mass of dioxigen substance, ${M}_{{O}_{2}} = 0.031998 \text{kg"/"mol" or 31.998 g/"mol}$.

The most probable speed ${v}_{\text{mp}}$, average speed, ${v}_{\text{ave}}$ and the rms speed ${v}_{\text{rms}}$ are related in this way:

${v}_{\text{mp"=sqrt(3/2) v_"rms"; " "v_"ave"=2/(sqrt(pi))v_"mp}}$