A certain gas has $a=27.0{\mathrm{kPadm}}^6{\mathrm{mol}}^{-2}$ and $b=0.024{\mathrm{dm}}^3{\mathrm{mol}}^{-1}$ Its critical temperature will be about

Given data $a=27.0 k\mathrm{Pa} d{m}^6 {\mathrm{mol}}^{-2}$ and $b=0.024 {\mathrm{dm}}^{3 }{\mathrm{mol}}^{-1}$ Its critical temperature found as, 

 $$\begin{gathered} T_{\mathrm{c}}=\frac{8a}{27Rb} \\ {T}_{\mathrm{c}}=\frac{8\left(27.0{\mathrm{kPadm}}^6{\mathrm{mol}}^{-2}\right)}{27\left(8.314\mathrm{kPa}{\mathrm{dm}}^3{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}\right)\left(0.024{\mathrm{dm}}^3{\mathrm{mol}}^{-1}\right)} \\ {T}_{\mathrm{c}}=40.1\mathrm{K} \end{gathered}$$

Option A is correct.