Question

The critical volume of a gas is $0.072lmo{l^{ - 1}}$ . What will be the radius of molecule in $cm$
(A) ${(\dfrac{3}{{4\pi }} \times {10^{ - 23}})^{\dfrac{1}{3}}}$
(B) ${(\dfrac{{4\pi }}{3} \times {10^{ - 23}})^{\dfrac{1}{3}}}$
(C) ${(\dfrac{{3\pi }}{4} \times {10^{ - 23}})^{\dfrac{1}{3}}}$
(D) ${(\dfrac{4}{{3\pi }} \times {10^{ - 23}})^{\dfrac{1}{3}}}$

Answer 1

Hint : In order to determine the radius of a molecule we have to use the relation that says the critical volume of the gas is the volume of the gas at critical temperature and pressure of gas. It is equal to the three times the Van der Waals constant. . Van der Waals constant is given by the formula written below.
$b = \dfrac{{16}}{3}\pi {r^3}{N_A}$

Complete Step By Step Answer:
Critical condition is the condition above which gases have fluidity like liquids and named as supercritical fluid. The supercritical liquid used in solvent exchange methods like $C{O_2}$ is used in extraction of caffeine.
Assuming that the above gas is following Van der Waals gas equation. Van der Waals gas equation is followed by the real gases, this equation is equivalent to the ideal gas equation. In this equation actual pressure and volume is considered as in ideal gas equation attraction between molecules and volume of molecules is neglected that’s why the pressure of real gas is slightly less than the ideal gas and volume of molecules is considered less. Therefore, critical volume is three times of van der Waals gas constant $b$ which signifies the excluded volume or Co-volume.
$V = 3b$
The magnitude of the constant $b$ is given as-
$b = \dfrac{{16}}{3}\pi {r^3}{N_A}$
Thus critical volume is given as-
$V = 3b \Rightarrow \dfrac{{16}}{3}\pi {r^3}{N_A} \times 3$
Substituting values in above equation, we get volume
$V = 0.072 \times 1000 = \dfrac{{16}}{3} \times 6.023 \times {10^{23}} \times \pi \times {r^3} \times 3$
Thus value of $r$ is
$r =$ ${(\dfrac{3}{{4\pi }} \times {10^{ - 23}})^{\dfrac{1}{3}}}$
Hence option (A) is correct.

Note :
The temperature above which gas cannot be liquefied by increasing pressure or the temperature up to which gas can be liquefied by applying external pressure is critical temperature. And the pressure required to liquefy gas at critical temperature condition is critical pressure. The critical temperature is calculated with the help of Andrew’s experiment. The critical temperature is given as
${T_C} = \dfrac{{8a}}{{27Rb}}$
And critical pressure is ${P_C} = \dfrac{a}{{27{b^2}}}$
And critical volume is given as ${V_C} = 3b$ that is volume at critical temperature and critical pressure.