Step 1: Understanding the Concept:
The Root Mean Square (RMS) velocity is a fundamental statistical measure of the speed of particles in a gas.
In the kinetic theory of gases, we assume that gas molecules are in constant, random motion.
The temperature of a gas is directly proportional to the average kinetic energy of its molecules.
For different gases at different temperatures to have the same RMS velocity, the ratio of their absolute temperature to their molar mass must remain constant.
This stems from the fact that lighter molecules (like Hydrogen) must move faster than heavier molecules (like Oxygen) to maintain the same kinetic energy at a given temperature.
Therefore, if we want the heavy Oxygen molecule to move as fast as the light Hydrogen molecule, the Oxygen must be at a significantly higher temperature.
Step 2: Key Formula or Approach:
The expression for the RMS velocity (\(v_{rms}\)) is derived from the kinetic gas equation:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
Where:
\(R\) is the Universal Gas Constant (\(8.314 \, \text{J/mol}\cdot\text{K}\)).
\(T\) is the absolute temperature measured in Kelvin (\(K\)).
\(M\) is the molar mass of the gas (expressed in \(\text{kg/mol}\) for SI units, though ratios work with \(\text{g/mol}\)).
Step 3: Detailed Explanation:
Let us identify the parameters for both gases provided in the problem.
For Hydrogen (\(H_2\)):
Molar mass \(M_{H_2} = 2 \, \text{g/mol}\).
Given temperature \(t_1 = 27^{\circ}\text{C}\).
First, convert this to the absolute Kelvin scale:
\[ T_1 = 27 + 273.15 \approx 300 \, \text{K} \]
The RMS velocity is given as \(v\). So:
\[ v = \sqrt{\frac{3R \times 300}{2}} \]
For Oxygen (\(O_2\)):
Molar mass \(M_{O_2} = 32 \, \text{g/mol}\).
Let the required temperature be \(T_2\).
The problem states that the RMS velocity for Oxygen must also be \(v\). So:
\[ v = \sqrt{\frac{3R \times T_2}{32}} \]
Equating the two expressions for \(v\):
\[ \sqrt{\frac{3R \times 300}{2}} = \sqrt{\frac{3R \times T_2}{32}} \]
By squaring both sides, we eliminate the square root:
\[ \frac{3R \times 300}{2} = \frac{3R \times T_2}{32} \]
We can cancel the constant \(3R\) from both sides:
\[ \frac{300}{2} = \frac{T_2}{32} \]
Simplifying the left side:
\[ 150 = \frac{T_2}{32} \]
Now, solve for \(T_2\):
\[ T_2 = 150 \times 32 \]
\[ T_2 = 4800 \, \text{K} \]
If we were to convert this temperature to Celsius to check against Option B:
\[ t_2 = 4800 - 273 = 4527^{\circ}\text{C} \]
Since both 4800 K and \(4527^{\circ}\text{C}\) represent the same thermal state, we look at the options. Option (A) is exactly 4800 K, which is a standard way to express gas law solutions. Option (B) is also numerically correct, but standard convention in these exams often prioritizes the Kelvin answer unless specified otherwise.
Step 4: Final Answer:
The required temperature is 4800 K.