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 a temperature of 27°C, we can follow these steps:

### Step 1: Convert the temperature to Kelvin
The temperature in Celsius needs to be converted to Kelvin because the gas constant (R) is in terms of Kelvin. The conversion formula is:
\[ T(K) = T(°C) + 273.15 \]
For 27°C:
\[ T = 27 + 273.15 = 300.15 \, K \]
We can approximate this to 300 K for simplicity.

### Step 2: Identify the values required for the RMS speed formula
The formula for the RMS speed (v_rms) of gas molecules is given by:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
Where:
- \( R \) = universal gas constant = 8.31 J/(mol·K)
- \( T \) = absolute temperature in Kelvin = 300 K
- \( M \) = molar mass of the gas in kg/mol = 0.032 kg/mol for oxygen.

### Step 3: Substitute the values into the RMS speed formula
Now, we substitute the values into the formula:
\[ v_{rms} = \sqrt{\frac{3 \times 8.31 \, \text{J/(mol·K)} \times 300 \, K}{0.032 \, \text{kg/mol}}} \]

### Step 4: Calculate the numerator
First, calculate the numerator:
\[ 3 \times 8.31 \times 300 = 7473 \, \text{J/mol} \]

### Step 5: Divide by the molar mass
Now, divide by the molar mass:
\[ \frac{7473}{0.032} = 233,531.25 \, \text{m}^2/\text{s}^2 \]

### Step 6: Take the square root
Finally, take the square root to find the RMS speed:
\[ v_{rms} = \sqrt{233,531.25} \approx 483 \, \text{m/s} \]

### Final Answer
The RMS speed of an oxygen molecule in a gas at 27°C is approximately **483 m/s**.

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