Calculate the van der Waals radius for ethane if $b = 0.0493\ \mathrm{dm^3\ mol^{-1}}$.

$$b = 4 \times {4 \over 3}\pi \times r^3 \times N_\mathrm A$$ $$\implies r = \sqrt[3]{3b\over 16\pi N_\mathrm A}= \sqrt[3]{3 0.049 \times 10^{-3}\over 16\pi N_\mathrm A}\ \mathrm m = 1.70 \times 10^{-10}\ \mathrm m$$

The answer given in the book uses $\displaystyle r =\frac12\sqrt[3]{3b\over 4\pi N_\mathrm A} = 1.35 \times 10^{-10}\ \mathrm m$.

Is my solution correct or is the book's solution correct ?

Edit as per martin's suggestion,

Here are the places I read $b = 4 \times V \times N_a$,

1) :- From the book,

enter image description here

2) :- From the Wikipedia,

enter image description here

(The picture is too large, might have to be opened in a separate tab to be readable)


3) :- The solution in the book,

enter image description here

4) :- Wildcard entry

enter image description here


I think I am missing something very crucial but I don't know what it is.

  • $\begingroup$ Could you explain where your formula for $b$ comes from? Did you take is from a book (which?) or derive it yourself (where did you start?)? Adding a bit more context to the question would probably help us better where the underlying problem is. For example, I can only assume that you make the (not unreasonable) assumption that ethane is a sphere and the volume of a sphere is $V=\frac43\pi r^3$. Where does the 4 come from? $\endgroup$ – Martin - マーチン Jan 10 '17 at 4:11
  • $\begingroup$ @Martin-マーチン I got the formula from Peter Atkin's physical chemistry. same can be found on Wikipedia . Is there no agreement on what is the value of $b$ ? At different sources I find different values for $b$, here is another value for $b$ i.e $b = V_{molecule} \times N_a$. hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/waal.html . $\endgroup$ – A---B Jan 10 '17 at 8:53
  • $\begingroup$ Please edit additional information into the post, is the exercise from Atkins book, too? If so, please include a citation. $\endgroup$ – Martin - マーチン Jan 10 '17 at 9:29
  • $\begingroup$ @Martin-マーチン Ok Martin, I will do that after scanning the book. What do you think about my question in previous comment $\endgroup$ – A---B Jan 10 '17 at 9:37

I like yours if the equation for b is correct. Since 16=8*2, you can take the cubed root of 8 out of the radical as 2, as they did. But that leaves 2 under the radical, not 4 as they have in their equation. That means they either made an error or started with an equation of b = 8 * 4/3(pi) * r3 * NA. Also, https://en.wikipedia.org/wiki/Van_der_Waals_constants_(data_page), siting the CRC, gives a b-value of 0.0638 L/mol for ethane, where 1L=1dm3.

| improve this answer | |
  • $\begingroup$ They started out with $$ b = N_a \times {4\over 3}\pi \times (2r)^3$$ $\endgroup$ – A---B Jan 8 '17 at 12:02
  • $\begingroup$ So am I missing something or isn't this the problem? When I use your equation for b I get your answer and when I use your books equation I get the books answer. And the book is using the right equation here. $\endgroup$ – airhuff Jan 8 '17 at 18:35
  • 1
    $\begingroup$ Why ? Isn't $b = \color{red}{4} \times {4 \over 3} \pi r^3 \times N_a$ not $b = \color{red}{8} \times {4 \over 3} \pi r^3 \times N_a$ ???? $\endgroup$ – A---B Jan 9 '17 at 9:21
  • $\begingroup$ It is also written in Wikipedia : en.wikipedia.org/wiki/Van_der_Waals_equation $\endgroup$ – A---B Jan 9 '17 at 9:27
  • $\begingroup$ Sorry to waver on this but I need to retract my sentence "And the book is using the right equation here." The last step of the conventional derivation from the gas law in the link you provided shows where I missed the boat with that statement. Your equation above for b is what you should use. Final answer, I swear ;) Seriously though, sorry for confusing the discussion, and nice job not just assuming the book was right and for defending your answer. $\endgroup$ – airhuff Jan 12 '17 at 6:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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