Question

The critical volume of the gas is ${\text{0}}{\text{.072}}\,{\text{lit}}{\text{.mo}}{{\text{l}}^{{\text{ - 1}}}}$. The radius of the molecule will be, in cm:
A.${\left( {\dfrac{3}{{4\pi }} \times {{10}^{ - 23}}} \right)^{\dfrac{1}{3}}}$
B.${\left( {\dfrac{{4\pi }}{3} \times {{10}^{ - 23}}} \right)^{\dfrac{1}{3}}}$
C.${\left( {\dfrac{{3\pi }}{4} \times {{10}^{ - 23}}} \right)^{\dfrac{1}{3}}}$
D.${\left( {\dfrac{4}{{3\pi }} \times {{10}^{ - 23}}} \right)^{\dfrac{1}{3}}}$

Answer 1

The van der Waals gas equation is used for the real gases where two correction terms are used for intermolecular forces and the volume.
In the van der Waals equation the constants $a$ and $b$ are used as a correction term. The constant $a$is used as a correction to the intermolecular forces and the constant $b$ is used for the correction term to the volume.
The critical volume is three times the van der Waals constant$b$.

Formula used: The relation between the critical volume and the van der Waals constant $b$ is given as follows:
${V_c} = 3b$
The van der Waals constant b is given as follows:
$b = \dfrac{{16}}{3}\pi {r^3}{N_A}$

Complete step-by-step answer: The critical constant in terms of the van der Waals constant is as follows:

${V_c} = 3b$……(i)
The value of van der Waals constant is given as follows:
$b = \dfrac{{16}}{3}\pi {r^3}{N_A}$…… (ii)
Here, $b$ is van der Waals constant, $r$ is the radius, and ${N_A}$ is the Avogadro’s constant.
Here, substitutes the value of constant b in equation (i).
${V_c} = 3 \times \dfrac{{16}}{3}\pi {r^3}{N_A}$
Here, the volume is litre per moles converted it into centimetres as follows:

${\text{1}}\,{\text{lit}}{\text{.mo}}{{\text{l}}^{{\text{ - 1}}}}{\text{ = 1000}}\,{\text{cm}}{\text{.mo}}{{\text{l}}^{{\text{ - 1}}}}$
${\text{0}}{\text{.072}}\,{\text{lit}}{\text{.mo}}{{\text{l}}^{{\text{ - 1}}}}{\text{ = }}\dfrac{{{\text{0}}{\text{.072}}\,{\text{lit}}{\text{.mo}}{{\text{l}}^{{\text{ - 1}}}}{{ \times 1000}}\,{\text{cm}}{\text{.mo}}{{\text{l}}^{{\text{ - 1}}}}}}{{{\text{1}}\,{\text{lit}}{\text{.mo}}{{\text{l}}^{{\text{ - 1}}}}}}$
${\text{0}}{\text{.072}}\,{\text{lit}}{\text{.mo}}{{\text{l}}^{{\text{ - 1}}}}{\text{ = 72}}\,\,{\text{cm}}{\text{.mo}}{{\text{l}}^{{\text{ - 1}}}}$
Thus, the volume is ${\text{72}}\,\,{\text{cm}}{\text{.mo}}{{\text{l}}^{{\text{ - 1}}}}$.
Now, substitute ${\text{72}}\,\,{\text{cm}}{\text{.mo}}{{\text{l}}^{{\text{ - 1}}}}$ for ${V_c}$, ${\text{6}}{\text{.023}} \times {\text{1}}{{\text{0}}^{23}}\,{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$ for ${N_A}$, and then rearrange the equation for $r$.
$\Rightarrow 72\,\,{\text{cm}}{\text{.mo}}{{\text{l}}^{{\text{ - 1}}}} = 3 \times \dfrac{{16}}{3}\pi {r^3}\left( {6.023 \times {{10}^{23}}\,{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}} \right)$
$\Rightarrow {\text{72}}\,\,{\text{cm}}{\text{.mo}}{{\text{l}}^{{\text{ - 1}}}}{\text{ = }}16{{\pi }}{{\text{r}}^{\text{3}}}\left( {{\text{6}}{\text{.023}} \times {\text{1}}{{\text{0}}^{23}}\,{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}} \right)$
$\Rightarrow {r^3} = \dfrac{1}{{16\pi }} \times \dfrac{{{\text{72}}\,\,{\text{cm}}{\text{.mo}}{{\text{l}}^{{\text{ - 1}}}}}}{{{\text{6}}{\text{.023}} \times {\text{1}}{{\text{0}}^{23}}\,{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}}}$
Now, take the square root on both sides.
${\text{r}} = {\left( {\dfrac{3}{{{{4\pi }}}} \times {\text{1}}{{\text{0}}^{ - 23}}\,} \right)^{\dfrac{1}{3}}}\,{\text{cm}}$
This is the radius of the molecule.

Therefore, option (A) is the correct answer.

Note: The volume occupied by the substance at the critical point is called critical volume and pressure of the gas at the same point is called critical pressure.
The radius of the molecule can be determined using the values of the critical volume of the gas and the Van der Waals constant $b$.