Question

At ${{100}^{\circ }}C$ and $1$ $atm$, if the density of liquid water is $1.0gc{{m}^{-3}}$ and that of water vapour is $0.0006gc{{m}^{-3}}$ , then the volume occupied by water molecules in one litre of steam at the same temperature is:
A.$6c{{m}^{3}}$
B.$60c{{m}^{3}}$
C.$0.6c{{m}^{3}}$
D.$0.06c{{m}^{3}}$

Answer 1

Hint:At ${{100}^{\circ }}C$, water changes its state from liquid phase to vapor state while still it is in thermodynamic equilibrium.
-We know that the density of a substance is equal to its mass upon volume. In other words the mass of a substance does not change when it undergoes phase transition, so we could use this relationship between mass volume and density to find out the unknown volume of any given substance whose values of density and volume in the other phase are known.

Complete step by step answer:
 If we consider the question, the temperature denoted by the symbol $T$ and pressure which is denoted by $P$ of the water molecules is given as ${{100}^{\circ }}C$ and $1atm$ respectively.
The density of water is denoted by the symbol ${{D}_{1}}$ which has the value $1.0gc{{m}^{-3}}$ as per the given question.
${{D}_{1}}$=${{D}_{liquid}}$ = $1.0gc{{m}^{-3}}$
Similarly, the density of the water vapour is given as ${{D}_{2}}$, which has the value $0.0006gc{{m}^{-3}}$
${{D}_{2}}$ =${{D}_{vapour}}$ = $0.0006gc{{m}^{-3}}$
Since, we know that volume of a liquid or a substance is inversely proportional to the density of the substance, and this statement can be mathematically represented as,
$V\propto \dfrac{1}{D}$
So, we take the ratio of volumes with the ratio of densities of the liquid and the steam and equate both of them on the basis of their masses, meaning they have the same mass, so the ratio of their volume with density will be equal. This can be represented by the following mathematical equation,
$\therefore \dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{{{D}_{2}}}{{{D}_{1}}}$
Where ${{V}_{1}}$ represents the volume of water molecules ; and ${{V}_{2}}$ represents the volume of steam or vapour.
Now we will substitute all the known values which were provided to us in the question, in order to get the unknown value of the volume.
$\Rightarrow $ $\dfrac{{{V}_{1}}}{1}=\dfrac{0.0006}{1}litre$
So after we substitute all the known values, we need to solve the whole equation in order to get the value of volume of water molecule.
 $\Rightarrow $ ${{V}_{1}}$ =$6\times {{10}^{-4}}\times {{10}^{3}}c{{m}^{^{3}}}$ =$0.6c{{m}^{3}}$
Hence, the correct option is (C).

Note:
-During a phase transition, certain properties of the medium change, often discontinuously, as a result of some external condition, as an example, we can look at the solution above.
-The volume of a substance is inversely proportional to the density of that substance, so the ratio of the volume and density of a substance in two different phases can be equated, as it would have the same mass.