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Volume occupied a single molecule of water.

  1. Feb 4, 2017 #1
    1. The problem statement, all variables and given/known data
    My problem includes answers from previous problems that are to be used as data in this problem so I will state the previous problems and the answers but not their solutions because I have solved them and they are pretty clear to me. So, I will be only posting the solution to the problem(it is a solved problem) solution of which seems a little doubtful to me.

    Problem 1:- The density of water is ##1000\text{ kg m$^{-3}$}##. The density of water vapor at ##100^{\circ}\text{C}## and ##1\text{ atm}## pressure is ##0.6\text{ kg m$^{-3}$}##. The volume of a molecule multiplied by the total number gives, what is called the molecular volume. Estimate the ratio (or fraction) of the molecular volume to the total volume occupied by the water vapor under the above conditions of temperature and pressure.

    Ans.:- ##\dfrac{\text{Molecular Volume}}{\text{Volume occupied by water vapor}}=6\times 10^{-4}##

    Problem 2:- Estimate the volume of a water molecule using the data in Problem 1.

    Ans.:- Volume of water molecule ##= 3\times 10^{-29} \text{m$^{-3}$}##
    Radius of the water molecule##\approx 2\times 10^{-10} \text{m}= 2 Å##

    Problem 3(The problem which is doubtful to me):-
    What is the average distance between atoms (interatomic distance) in water? Use the data given in Problems 1 & 2.

    2. Relevant equations


    3. The attempt at a solution
    As stated in the starting I will be only posting my working for Problem 3.

    Solution to Problem 3:-

    We assume that the molecules of water are uniformly distributed over the volume occupied by the water vapor and the distance b/w each molecule is same.

    I think this assumption works because even though the interatomic distance is not the same throughout the volume of the water vapor but when the the interatomic distance is averaged over the whole of the volume of water vapor the interatomic distance will be the same as in the model that I proposed for water vapor initially.

    In all the previous problems we were to supposed to consider the density of water as the density of the molecules and that the structure of water was (approximately) same when the molecules would have been arranged touching each other.

    So, the water molecules having radius ##r_{m}## and volume ##V_{rm}## in the liquid state, as assumed, are arranged as shown in the diagram below:-
    geogebra-export.png
    When the waster turns into water vapor then, let's assume that there are virtual water vapor(cocentric with the original ) molecules touching each other, having radius ##r_{vm}## and volume ##V_{vm}##. So, the water vapor molecules would look like this:-

    geogebra-export (1).png

    Since from the answer of Problem 1, we have

    $$V_{vm}=\dfrac{1}{6\times 10^{-4}}V_{m}\implies V_{vm}=\left(1.67\times 10^{3}\right)V_m\\
    \implies V_{vm}=\frac{3\times 10^{-29}}{6\times 10^{-4}}=5\times 10^{-26} \text{m$^{-3}$}$$

    So, we get the radius of the virtual molecule of water as ##r_{vm}=\sqrt[3]{\dfrac{V_{vm}}{\frac{4}{3}\pi}}##
    $$\therefore r_{vm}=2.28\times 10^{-9}\text{ m}\approx 20 Å$$

    As each real molecule is located on the periphery of the circle described by the sphere created by the virtual water vapor molecule, hence the interatomic distance in vapor b/w the water molecules can be approxiamted to ##r_{vm}=20Å##.

    But my answer does not math with that given in the solved problem. The answer of the solved problem is ##40Å##.

    Solution provided by the book:-

    A given mass of the water in vapor state has ##1.67 \times 10^3## times the volume of the same mass of water in liquid state. This is also the increase in the amount of volume of each molecule of water. When volume increases by ##V^{1/3}## or ##10## times, i.e. ##10\times 2Å=20Å##. So the average distance is ##2\times 20Å= 40Å##.

    So what the book assumes the model of the molecules of water in gaseous phase is as follows:-
    geogebra-export (2).png

    So what is wrong in the arrangement of the molecules in the gaseous state that I proposed, and why is the books model correct.

    I have written my thought process that came to my mind as I wrote my solution.

    Thanks for your help.
     
  2. jcsd
  3. Feb 4, 2017 #2
  4. Feb 4, 2017 #3
    Doesn't every real molecule have to be contained within a virtual molecule? And if so, doesn't that make the distance greater than one radius of a virtual molecule?
     
  5. Feb 4, 2017 #4
    That seems to be the case as the book has adopted that pathway to arrive at the solution but it doesn't appear to be the right method to me at the moment. Perhaps you can reason with me as to why is the book's assumption correct.
     
  6. Feb 4, 2017 #5
    Well, the radius was calculated from the volume of a virtual molecule. But in your picture, you have 6 real molecules and only one virtual molecule. So by using 6 real molecules and 1 virtual molecule you have created a different density than what you started with. In other words, by using the molecules as you have shown, more molecules will now fit within a given volume. Another way to word the question: Why is that one molecule allowed more volume than the other 6?
     
  7. Feb 4, 2017 #6

    haruspex

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    I agree with Tom.
    The question itself is a bit misleading though. How is one to define the average distance between the molecules? Taken literally, it would be about equal to the radius of the volume containing the water vapour. It must be more in the sense of nearest neighbours, but in practice they won't be so evenly spaced as in the model. There is a way to work that out, and the answer is rather different.
     
  8. Feb 4, 2017 #7

    haruspex

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    By the way, the question posted seems a bit different. Atoms? Water?
     
  9. Feb 6, 2017 #8
    Actually I am a high school student and the book from which I am reading is the book that is followed mainly in all of India it is printed by one of the govt. organisations NCERT. In the text there always creeps up a error or two here or there in the book but it sure does develop your intuition for the material a high school student is going to encounter. So, I think what the book implies by stating the density of water only and that too when it is the densest at ##4^{\circ}## is that we can approximate the density of the molecules as if they were as close as water. From this what I think it wants to convey is that the first researchers who thought about it must have started to develop their own intuition along these lines.

    I absolutely dont know whether the book wants to teach in that way or not but it sure does make things much more easier for me to understand.
     
  10. Feb 6, 2017 #9
    @TomHart @haruspex I am sorry for the late reply but my net was down for sometime, to be precise since my last visit(it came on and off all the time). I thank you for all the effort that you put into solving my doubt. I also had somehow figured out that what I was doing was reducing the total area of the water vapor by my model.

    Well would you give me a hint as to how that is to be done. If the math goes out of hand for me then I will just be satisfied knowing the theory behind it and would work it out after highschool ends and before the time I have for entering the college that I hopefully get admitted to.
     
  11. Feb 6, 2017 #10

    haruspex

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    Read https://en.m.wikipedia.org/wiki/Mean_inter-particle_distance#Nearest_neighbor_distribution
     
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