One mole of an ideal gas occupies 12 L at  ${\text{297}}^{0}C$. What is the pressure of the gas?

Given, Volume of gas = 12L 
         mole of gas = 1
 $$\begin{gathered} R=0.0821L-atmmo{l}^{-1}{K}^{-1} \\ T={297}^{0}C+273=570K \end{gathered}$$             
From gas equation,   $pv=nRT$
$p=\frac{nRT}{V}=\frac{1\times 0.0821\times 570}{12}$  = 38.99 atm  (1 atm = 101325 Pa)
= 395066.18 Pa = 395 kPa