In a recipient there is an unknown monoatomic gas that occupies a volume of $\pu{230 cm3}$, at $\pu{300 K}$ and $\pu{1 atm}$. It is known that the atoms occupy a volume of $2\times10^{-4}$. Find the radius of the atoms.

I used the ideal gas law to calculate the number of moles, then I calculated the number of atoms, then divided the volume occupied by the atoms by the number of atoms to find the individual volume of each atom, then applied the formula for the volume of a sphere to find the radius, the problem is that I found 92x10^-10 and the correct answer is 1,25x10^-10, can anybody spot the mistake? Thank you.

Here are the maths I did:

enter image description here

  • $\begingroup$ Your first error is for $n$. The value you have for R is wrong. $R= 82.057338(47)\ \mathrm{cm}^3 \mathrm{atm}\ \mathrm{K}^{−1} \mathrm{mol}^{−1}$. $\endgroup$ – MaxW Jan 10 '19 at 22:25
  • $\begingroup$ Your right, but its still not enough to get to the right answer. $\endgroup$ – Leo Noanh Consoli Jan 11 '19 at 0:30
  • $\begingroup$ Leo - Well fix that part of the math and work the correction through the problem... $\endgroup$ – MaxW Jan 11 '19 at 1:23
  • 1
    $\begingroup$ … and always write the values with the correct units and carry the units through the calculations. $\endgroup$ – user7951 Jan 11 '19 at 19:46

I haven't dived into your math, but I think there are 3 things that are confusing:

  1. intermediate calculations, even though you weren't asked to do any;
  2. missing units and inconsistent notations (e.g. base 10 is not always shown);
  3. ambiguous phrase "atoms occupy a volume of $2\times 10^{-4}$."

I always find it cleaner and more productive to derive the formula first, and only then plug in the numbers minding the proper units. As for the volume issue, I tend to believe that $2\times 10^{-4}$ means the fraction of the volume occupied by gas atoms and it is not the absolute value expressed in cubic centimeters. In fact, in this case you get the correct answer and you don't need to use gas volume ($\pu{230 cm3}$) at all.

Check this out. Radius of an atom $r_\mathrm{a}$ is indeed can be found from the volume of an atom $V_\mathrm{a}$

$$r_\mathrm{a} = \sqrt[3]{\frac{3V_\mathrm{a}}{4π}}.$$

At the same time

$$V_\mathrm{a} = \frac{V_\mathrm{atot}}{N}$$

where $V_\mathrm{atot}$ – volume of all gas atoms depicted with spheres; $N$ – number of the atoms that can be found via Avogadro's number $N_\mathrm{A}$ and amount $n$:

$$N = nN_\mathrm{A}.$$

Finally, you are allowed to use ideal gas law to determine the amount $n$:


where $p$ – pressure; $V$ – volume of the gas (macro scale); $R$ – the molar gas constant; $T$ – temperature. Now, lets's put all the things together assuming that $V_\mathrm{atot}/V = 2\cdot 10^{-4}$:

$$ \begin{align} r_\mathrm{a} &= \sqrt[3]{\frac{3V_\mathrm{atot}RT}{4πpVN_\mathrm{A}}}\\ &= \sqrt[3]{\frac{3\cdot2\cdot10^{-4}\cdot\pu{8.314 J mol-1 K-1}\cdot\pu{300 K}}{4\cdot 3.14\cdot\pu{101325 Pa}\cdot\pu{6.022e23 mol-1}}}\\ &= \pu{1.25e-10 m} \end{align} $$

Now that's the value that actually makes sense:)

  • 1
    $\begingroup$ Thank you, my units were really mesy and I got confused if the volume occupied by the atoms was a pure volume or a fraction of the volume, thank you again. $\endgroup$ – Leo Noanh Consoli Jan 11 '19 at 12:37
  • $\begingroup$ No prob, keep it up:) By the way, as you are probably using $\mathrm{\LaTeX}$ judging from the image you posted, I'd suggest to use siunitx package. It helps to format units and numbers according to SI and ISO standards and is very versatile. $\endgroup$ – andselisk Jan 11 '19 at 12:40

Not the answer you're looking for? Browse other questions tagged or ask your own question.