Using Vander Waal’s equation calculate the constant ‘a’ in $\mathrm{atm}-{\mathrm{L}}^2/{\mathrm{mol}}^2$ when two moles of a gas is confined in a four litre flask that exerts a pressure of 11.0 atm at a temperature of ${27}^{\circ }\mathrm{C}$ . The value of ‘b’ is 0.05L/mol

Volume = 4.0L Pressure = 11.0 atm
Temperature = 27 + 273 = 300K b =0.05L/ mol 
No of moles $$\begin{gathered} =2.0 \left(\mathrm{P}+\frac{{\mathrm{an}}^2}{{\mathrm{V}}^2}\right)\left(\mathrm{V}-\mathrm{nb}\right)=\mathrm{nRT}\left(\mathrm{P}+\frac{{\mathrm{an}}^2}{{\mathrm{V}}^2}\right)=\frac{\mathrm{nRT}}{\mathrm{V}-\mathrm{nb}} \\ \frac{{\mathrm{an}}^2}{{\mathrm{V}}^2}=\frac{\mathrm{nRT}}{\mathrm{V}-\mathrm{nb}}-\mathrm{p} \mathrm{a}=\frac{{\mathrm{V}}^2}{{\mathrm{n}}^2}\left[\frac{\mathrm{nRT}}{\mathrm{V}-\mathrm{nb}}-\mathrm{p}\right]=\frac{{\left(4\right)}^2}{{\left(2\right)}^2}\left[\frac{2\times 0.0821\times 300}{4-2\times 005}-11\right] \\ 4\left[\frac{2\times 0.0821\times 300}{4-2\times 0.05}\right]-11=4\times 1.63 =6.52 \mathrm{atm}-{\mathrm{L}}^2/{\mathrm{mol}}^2. \end{gathered}$$