Question

The critical density of the gas ${\text{C}}{{\text{O}}_{\text{2}}}$ is $0.44{\text{gc}}{{\text{m}}^{{\text{ - 3}}}}$ at a certain temperature. If r is the radius of the molecule, ${{\text{r}}^{\text{3}}}$ in ${\text{c}}{{\text{m}}^{\text{3}}}$ is approximate:
[N is Avogadro number]
A.$\dfrac{{25}}{{{{\pi N}}}}$
B.$\dfrac{{100}}{{{{\pi N}}}}$
C.$\dfrac{6}{{{{\pi N}}}}$
D.$\dfrac{{25}}{{{{4N\pi }}}}$

Answer 1

Hint: The volume occupied by one mole of gas at its critical temperature and critical pressure is known as the critical volume and the corresponding density will be its critical density.
The density of any substance is a measure of its mass per unit volume.

Complete step by step answer:
Given that the critical critical density of the gas ${\text{C}}{{\text{O}}_{\text{2}}}$ is $0.44{\text{gc}}{{\text{m}}^{{\text{ - 3}}}}$ at a certain temperature.
We need to find the value of ${{\text{r}}^{\text{3}}}$ in ${\text{c}}{{\text{m}}^{\text{3}}}$ if r is the radius of the molecule of carbon dioxide.
We have for any substance,
${\text{Density = }}\dfrac{{{\text{Mass}}}}{{{\text{Volume}}}}$
Therefore, the critical density will also be the critical mass divided by the critical volume.
Now, the mass of the carbon dioxide molecule is equal to the sum of the atomic masses of the constituent atoms which are one carbon atom and two oxygen atoms. Thus, mass of carbon dioxide is $ = 12 + 16 \times 2 = 44{\text{g/mol}}$ .
Avogadro’s number is equal to $6.023 \times {10^{23}}$ which is N.
Since gram molecular mass is equal to Avogadro’s number of molecules, thus, mass of N molecules is equal to 44 g of carbon dioxide.
Thus, the mass of one molecule is equal to $\dfrac{{44}}{{\text{N}}}$ .
Here, the volume taken into consideration is critical volume which is equal to 3 times the compressible volume per mole of the gas. Thus, the critical volume is equal to:
$
   = 3 \times \dfrac{4}{3}{{\pi }}{{\text{r}}^{\text{3}}} \\
   = 4{{\pi }}{{\text{r}}^{\text{3}}} \\
 $
Therefore, the critical density is
$
   = \dfrac{{\dfrac{{44}}{{\text{N}}}}}{{{{4\pi }}{{\text{r}}^{\text{3}}}}} \\
   = \dfrac{{11}}{{{{\pi }}{{\text{r}}^{\text{3}}}{\text{N}}}} \\
 $
The critical density is given to be $0.44{\text{gc}}{{\text{m}}^{{\text{ - 3}}}}$ . So, we have:
$
  0.44 = \dfrac{{11}}{{{{\pi }}{{\text{r}}^{\text{3}}}{\text{N}}}} \\
   \Rightarrow {{\text{r}}^{\text{3}}} = \dfrac{1}{{0.04{{\pi N}}}} \\
   \Rightarrow {{\text{r}}^{\text{3}}} = \dfrac{{25}}{{{{\pi N}}}}{\text{c}}{{\text{m}}^3} \\
 $

So, the correct answer is option A.

Note:
The critical temperature of a gas may be defined as that temperature above which the gas cannot be liquefied however high the pressure may be.
When a gas is at its critical temperature and under critical pressure, its physical properties become indistinguishable from those of its liquid state under the same conditions.