Step 1: Understanding the Question:
We need to find the ratio of the speed of sound in oxygen gas (O\(_2\)) to that in hydrogen gas (H\(_2\)) under the same temperature conditions.
Step 2: Key Formula or Approach:
The speed of sound \( v \) in an ideal gas is given by the formula:
\[ v = \sqrt{\frac{\gamma RT}{M}} \]
where \( \gamma \) is the adiabatic index (ratio of specific heats), \( R \) is the universal gas constant, \( T \) is the absolute temperature, and \( M \) is the molar mass of the gas.
Step 3: Detailed Explanation:
We are comparing \( v_{O_2} \) and \( v_{H_2} \) at the same temperature \( T \).
Both Oxygen (O\(_2\)) and Hydrogen (H\(_2\)) are diatomic gases. For diatomic gases, \( \gamma = \frac{7}{5} \). So, \( \gamma \) is the same for both.
The gas constant \( R \) is also the same.
Therefore, the speed of sound is inversely proportional to the square root of the molar mass:
\[ v \propto \frac{1}{\sqrt{M}} \]
We can write the ratio as:
\[ \frac{v_{O_2}}{v_{H_2}} = \sqrt{\frac{M_{H_2}}{M_{O_2}}} \]
The molar masses are:
Molar mass of Hydrogen (\( M_{H_2} \)) \(\approx 2\) g/mol.
Molar mass of Oxygen (\( M_{O_2} \)) \(\approx 32\) g/mol.
Substitute these values into the ratio:
\[ \frac{v_{O_2}}{v_{H_2}} = \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}} = \frac{1}{4} \]
The ratio of the speed of sound in Oxygen to that in Hydrogen is 1:4.
Step 4: Final Answer:
The required ratio is \( 1:4 \).