Correct Answer - Option 1 : 1 : 16
Concept:
\({\left( R \right)_{any\;gas}} = \frac{{\bar R}}{{{{\left( {Molecular\;weight} \right)}_{gas}}}}\)
Calculation:
\(R{o_2} = \frac{{\bar R}}{{{{\left( {Mol.wt} \right)}_O}_2}} = \frac{{\bar R}}{{32}}\)
\({R_{{H_2}}} = \frac{{\bar R}}{{{{\left( {Mol.wt} \right)}_{{H_2}}}}} = \frac{{\bar R}}{2}\)
\({R_{{O_2}}}:{R_{{H_2}}}\)
\(\Rightarrow \frac{{\bar R}}{{32}}:\frac{{\bar R}}{2}\)
⇒ 1 : 16