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(D) The ideal gas equation is given by $PV = nRT$,where $n = \frac{m}{M}$.For oxygen gas $(O_2)$: $P_1 V_1 = \frac{m_1}{M_1} RT_1$ ... $(i)$For hydrog

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$16: 1$

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(D) The ideal gas equation is given by $PV = nRT$,where $n = \frac{m}{M}$.For oxygen gas $(O_2)$: $P_1 V_1 = \frac{m_1}{M_1} RT_1$ ... $(i)$For hydrogen gas $(H_2)$: $P_2 V_2 = \frac{m_2}{M_2} RT_2$ ... (ii)Given that $P_1 = P_2$,$V_1 = V_2$,and $m_1 = m_2$,dividing equation $(i)$ by (ii) gives:$\frac{P_1 V_1}{P_2 V_2} = \frac{m_1 / M_1}{m_2 / M_2} \cdot \frac{T_1}{T_2}$$1 = \frac{M_2}{M_1} \cdot \frac{T_1}{T_2}$$\frac{T_1}{T_2} = \frac{M_1}{M_2}$Substituting the molar masses $M_1 (O_2) = 32 	ext{ g/mol}$ and $M_2 (H_2) = 2 	ext{ g/mol}$:$\frac{T_1}{T_2} = \frac{32}{2} = \frac{16}{1}$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

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