A) atm \[{{L}^{2}}\]
B) atm \[{{L}^{2}}mo{{l}^{-1}}\]
C) arm \[{{L}^{2}}mo{{l}^{-2}}\]
D) atm \[{{L}^{-1}}\]
Correct Answer: C
Solution :
\[a=\frac{P{{V}^{2}}}{{{n}^{2}}}\] P = pressure in atmosphere V = volume in litre n = number of moles then, \[a=\frac{Pressure\times {{\left( Volume \right)}^{2}}}{{{\left( Mol \right)}^{2}}}\] \[=\frac{(atm)\,\,{{(L)}^{2}}}{{{(mol)}^{2}}}\] = atm \[{{L}^{2}}mo{{l}^{-2}}\]
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