Question

The density of solid argon $({\text{Ar = 40g/mol)}}$ is \[1.65{\text{ g/ml}}\] at −$40{\text{K}}$. If the argon atom is assumed to be a sphere of radius \[1.54 \times {10^{ - 8}}{\text{cm}}\], the percentage of space in solid argon is:
A.$35.64$
B.$64.36$
C.$74$
D.None of these

Answer 1

Hint: To answer this question, you should recall the concept of the kinetic theory of gases. We shall substitute the appropriate values in the formula given to find the volume occupied by the atoms and use that to calculate the free space.

Formula used:
\[{\text{V}} = \dfrac{4}{3}{{\pi }}{{\text{r}}^3}\] where \[V\] is volume and $r$ is the radius.
${\text{No}}{\text{. of moles = }}\dfrac{{{\text{Mass of the Substance in grams}}}}{{{\text{Molar mass of a Substance}}}} = \dfrac{{{\text{Number of Atoms or Molecules}}}}{{6.022 \times {{10}^{23}}}}$

Complete step by step answer:
The kinetic theory of gases describes a gas as a large number of identical microscopic particles which are in constant, rapid, random motion. Their size is assumed to be much smaller than the average distance between the particles. To calculate the space in the given sample you need to subtract the volume occupied from the total volume.
The volume of one atom is given by \[{\text{V}} = \dfrac{4}{3}{{\pi }}{{\text{r}}^3}\].
Substituting the value of r which is the radius of the particle.
 \[\Rightarrow {\text{V}} = \dfrac{4}{3} \times \dfrac{{22}}{7} \times {(1.54 \times {10^{ - 8}})^3}{\text{c}}{{\text{m}}^{\text{3}}}\].
After solving we get:
\[\Rightarrow {\text{V}} = 1.53 \times {10^{ - 23}}{\text{c}}{{\text{m}}^{\text{3}}}\].
Multiplying this value with the total number of particles present in the given moles of argon, will give us all the volume of all the particles.
The volume of all atoms in \[1.65{\text{ g}}\;Ar\]
\[\Rightarrow {\text{V = }}\dfrac{{1.65}}{{40}} \times 6.02 \times {10^{23}} \times 1.53 \times {10^{ - 23}}{\text{c}}{{\text{m}}^{\text{3}}}\].
We will get the value of \[{{\text{V}}_{{\text{total}}}} = 0.380{\text{c}}{{\text{m}}^{\text{3}}}\]
The volume of solids \[{\text{Ar}}\] contained in solid \[ = 1{\text{c}}{{\text{m}}^{\text{3}}}\].
Space \[ = 1 - 0.380 = 0.620\]
\[\therefore \;\% {\text{space }} = {\text{ }}62\% \].

Hence, the correct answer to this question is option D.

Note:
At 'higher temperature' and 'lower pressure', a gas behaves like an ideal gas, as the potential energy due to intermolecular forces becomes less significant compared with the particles' kinetic energy, and the size of the molecules becomes less significant compared to the space between them.