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decentfellow
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Homework Statement
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.
Homework Equations
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:-
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:-
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:-
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.