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(C) The ideal gas equation is given by $PV = nRT$,where $n = \frac{m}{M}$.For oxygen $(O_2)$: $PV = \frac{m}{M_{O_2}} RT_{O_2}$  --- $(i)$For hydrogen

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$16: 1$

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(C) The ideal gas equation is given by $PV = nRT$,where $n = \frac{m}{M}$.For oxygen $(O_2)$: $PV = \frac{m}{M_{O_2}} RT_{O_2}$  --- $(i)$For hydrogen $(H_2)$: $PV = \frac{m}{M_{H_2}} RT_{H_2}$  --- $(ii)$Since $P$,$V$,and $m$ are the same for both gases,we equate the two expressions:$\frac{m}{M_{O_2}} RT_{O_2} = \frac{m}{M_{H_2}} RT_{H_2}$$\frac{T_{O_2}}{M_{O_2}} = \frac{T_{H_2}}{M_{H_2}}$$\frac{T_{O_2}}{T_{H_2}} = \frac{M_{O_2}}{M_{H_2}}$Given $M_{O_2} = 32$ and $M_{H_2} = 2$:$\frac{T_{O_2}}{T_{H_2}} = \frac{32}{2} = \frac{16}{1}$Therefore,the ratio of their absolute temperatures is $16: 1$.

$A$ sample of oxygen gas and a sample of hydrogen gas both have the same mass,same volume,and the same pressure. The ratio of their absolute temperatures is (Molecular weights of $O_2$ and $H_2$ are $32$ and $2$ respectively).

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