For a real gas $$\left( {mol.\,mass = 60} \right)$$    if density at critical point is $$0.80\,g/c{m^3}$$   and its $${T_c} = \frac{{4 \times {{10}^5}}}{{821}}K,$$    then van der waals’ constant a $$\left( {{\text{in}}\,atm\,{L^2}\,mo{l^{ - 2}}} \right)$$     is

Solution :
$$\eqalign{ & {V_c} = \frac{{60}}{{0.80}} \cr & = 75c{m^3}\,mo{l^{ - 1}}; \cr & b = \frac{{{V_c}}}{3} = 25c{m^3}\,mo{l^{ - 1}};\, = 0.25\,L\,mo{l^{ - 1}} \cr & \therefore \,\,{T_c} = \frac{{8a}}{{27\,Rb}} \cr & \frac{{4 \times {{10}^5}}}{{821}} \cr & = \frac{{2.45}}{{0.425}} \cr & = 5.76\,g{L^{ - 1}} \cr & \Rightarrow a = 3.375 \cr} $$