The rms speed of oxygen molecule in a gas at `27^(@)C` would be given by

Options

A. 483 `ms^(-1)`
B. 966 `ms^(-1)`
C. 4.83 `ms^(-1)`
D. 9.66 `ms^(-1)`

Explanation

To find the root mean square (RMS) speed of an oxygen molecule in a gas at \(27^\circ C\), we can use the formula:

\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]

where:
- \(R\) is the universal gas constant,
- \(T\) is the absolute temperature in Kelvin,
- \(M\) is the molar mass of the gas in kg/mol.

### Step 1: Convert the temperature from Celsius to Kelvin
To convert Celsius to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273 \]

For \(27^\circ C\):

\[ T = 27 + 273 = 300 \, K \]

### Step 2: Determine the molar mass of oxygen
The molecular weight of oxygen (\(O_2\)) is calculated as follows:

\[ M = 2 \times 16 \, g/mol = 32 \, g/mol \]

To convert grams per mole to kilograms per mole, we divide by 1000:

\[ M = 32 \, g/mol = 0.032 \, kg/mol \]

### Step 3: Substitute values into the RMS speed formula
Now we can substitute the values into the RMS speed formula. The universal gas constant \(R\) is:

\[ R = 8.314 \, J/(mol \cdot K) \]

Now substituting \(R\), \(T\), and \(M\) into the formula:

\[ v_{rms} = \sqrt{\frac{3 \times 8.314 \, J/(mol \cdot K) \times 300 \, K}{0.032 \, kg/mol}} \]

### Step 4: Calculate the value inside the square root
Calculating the numerator:

\[ 3 \times 8.314 \times 300 = 7482.6 \, J/mol \]

Now, divide by the molar mass:

\[ \frac{7482.6}{0.032} = 233,828.125 \, m^2/s^2 \]

### Step 5: Take the square root
Now, take the square root to find \(v_{rms}\):

\[ v_{rms} = \sqrt{233,828.125} \approx 483 \, m/s \]

### Final Answer
Thus, the RMS speed of an oxygen molecule in a gas at \(27^\circ C\) is approximately:

\[ \boxed{483 \, m/s} \]