Answer:
1 mol gas occupies 22.4 L volume
at S.T.P., i.e., at 273.15 K and 1 atm pressure.
\[PV=nRT\]
\[R=\frac{PV}{nT}=\frac{1\times
22.4}{1\times 273.15}=0.082L\,atm\,{{K}^{-1}}mo{{l}^{-1}}\]
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