Question

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

Answer 1

Hint: 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 \[{P_C} = \dfrac{8}{{27{b^2}}}\]