# Calculating collision diameter with viscosity, but not having the mass given

I've got a homework question i'm stumped on.

The viscosity of the ethylene gas was measured at $99.8\times10^{-7}\ \mathrm{N\ s\ m^{-2}}$ at $288\ \mathrm K$. Calculate the collision diameter in $\overset{\circ}{\mathrm{A}}$ for this gas molecule.

I started with

$$\eta=\frac{m\overline v}{3\sqrt2\pi d^2}$$

then i said

$$\overline v=\sqrt{\frac{8RT}{\pi M}}$$

Molar mass $M=28.05\ \mathrm{g\ mol^{-1}}$

So my equation looks like

$$\eta=\frac{m}{3\sqrt2\pi d^2}\sqrt{\frac{8RT}{\pi M}}$$

but $m$ is still a variable. What should I do from here? Is there another substitution I can do?

So apparently my confusion about solving this problem, came from a lack of understanding what the $m$ variable was supposed to represent. I had thought that it was the total mass of the system, but it is actually the mass of one molecule.

• I just asked my professor and he speedily said something about taking the molar mass and multiplying by $10^{26} \mathrm{kg}$. – John Snow Sep 16 '14 at 19:23

the value for $(\mathrm{m})$ was supposed to be in $\mathrm{kg}$, and could be found by taking the Molar mass and converting:

$\mathrm{M} \cdot 1.661 x10^{-27}\mathrm{kg\cdot amu^{-1}}$

The solution came out looking like

$\overset{-}{v} = \sqrt{\frac{8x8.314 \mathrm{J \cdot K^{-1} \cdot mol^{-1}} x 288\mathrm{K}}{3.14x28.05x10^{-3}kg \cdot mol^{-1}}}$

$d=\sqrt{\frac{m \overset{-}{v}}{3 \sqrt{2} \pi \eta}}$

$d=\sqrt{\frac{(28.05 \mathrm{amu} \space x \space 1.661x10^{-27} \mathrm{kg} \cdot \mathrm{amu}^{-1}) \space x \space 466.23 \mathrm{m} \cdot \mathrm{s}^{-1}}{3 \sqrt{2} \pi \space x \space 99.8x10^{-7} \mathrm{N} \cdot \mathrm{s} \cdot \mathrm{m}^{-2}}}$

$=4.04x10^{-10} \mathrm{m} = 4.04 \overset {\circ}{A}$

• Your numerical calculation is correct, but your units are wrong. The conversion factor from molar mass $M$ (e.g. in $\mathrm{g/mol}$) to mass $m$ (e.g. in $\mathrm g$) is simply given by the Avogadro constant $N_\mathrm A=6.022\,140\,857(74)\times10^{23}\ \mathrm{mol^{-1}}$. Note that $1/N_\mathrm A=1.661\times10^{-24}\ \mathrm{mol}$, which is equivalent to the numerical value that you have used. You also converted $\mathrm g$ to $\mathrm{kg}$; that’s why your value is $1.661\times10^{-27}\ \mathrm{kg\ mol/g}$. – Loong Nov 7 '16 at 16:34