Question

Argon has ${T_C} = - {122^\circ }C$, ${P_c} = 48$ atm. What is the radius of the argon atom?

Answer 1

The real gas equation is given as $\left( {P + \dfrac{a}{{{V^2}}}} \right) - (V - b) = nRT$, where a & b are Van-der Waal constant. “a” signifies the attraction between the molecules and “b” is the volume of a single molecule of the gas.
The critical point is the point on $P - V$ plane isotherm where 1st and 2nd derivatives of P w.r.t V is zero (Critical point is a saddle point on $P - V$ isotherm). After solving the got equations and given real gas equation we get
${V_C} = 3nb$, ${T_C} = \left( {\dfrac{{8a}}{{27Rb}}} \right)$ and ${P_C} = \dfrac{a}{{27{b^2}}}$
Where ${P_C}$, ${V_C}$ and ${T_C}$ are critical pressure, critical volume and critical temperature respectively. Substituting the values of ${P_C}$ and ${T_C}$ given in the question in the equation $\dfrac{{{P_c}{V_c}}}{{R{T_c}}} = \dfrac{{3n}}{8}$

Complete step by step answer:
Given in the question are:
Critical temperature, ${T_C} = - {122^\circ }C$
Critical pressure, ${P_c} = 48$ atm
From the formula
$\dfrac{{{P_c}{V_c}}}{{R{T_c}}} = \dfrac{{3n}}{8}$ ..…..(equation 1)
After substituting the value in the equation (1), we get,
${V_C} = \dfrac{{3R{T_C}}}{{8{P_C}}}$
$V = \dfrac{4}{3}\pi {r^3}$
$V = \dfrac{4}{3} \times \dfrac{{22}}{7} \times {r^3}$
$\Rightarrow \dfrac{4}{3} \times \dfrac{{22}}{7} \times {r^3} = \dfrac{{3 \times 0.0821 \times 122}}{{8 \times 48}}$
$\Rightarrow {r^3} = 0.0498m$
$\Rightarrow r = 0.368m$

Therefore, the radius of the argon atom is 0.368 m

Note: This can be done from the Van der Waals Equation of State for real gases. Consider that equation, for one mole of the gas in question:
$\left( {P + \dfrac{a}{{{V^2}}}} \right) - (V - b) = RT$, from which we get, $P = \dfrac{{Rt}}{{(V - b)}} - \dfrac{a}{{{V^2}}}$. We get ${T_c} = \dfrac{{8a}}{{27Rb}}$, and ${P_C} = \dfrac{8}{{27{b^2}}}$