Find rms speed of oxygen molecules at temperature `27^(@)C`.

Options

A. 483m/s
B. 447m/s
C. 145m/s
D. 254m/s

Explanation

To find the root mean square (rms) speed of oxygen molecules at a temperature of \(27^\circ C\), we can follow these steps:

### Step 1: Convert the temperature from Celsius to Kelvin
The formula to convert Celsius to Kelvin is:
\[ T(K) = T(°C) + 273 \]
For \(27^\circ C\):
\[ T = 27 + 273 = 300 \, K \]

### Step 2: Identify the molar mass of oxygen
The molar mass of oxygen (O₂) is approximately \(32 \, g/mol\). We need to convert this to kilograms per mole for our calculations:
\[ M = 32 \, g/mol \times \frac{1 \, kg}{1000 \, g} = 32 \times 10^{-3} \, kg/mol \]

### Step 3: Use the rms speed formula
The formula for the rms speed (\(v_{rms}\)) of a gas is given by:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
Where:
- \(R\) is the gas constant, \(R = 8.31 \, J/(mol \cdot K)\)
- \(T\) is the temperature in Kelvin (which we found to be \(300 \, K\))
- \(M\) is the molar mass in kg/mol (which we found to be \(32 \times 10^{-3} \, kg/mol\))

### Step 4: Substitute the values into the formula
Now we substitute the values into the rms speed formula:
\[ v_{rms} = \sqrt{\frac{3 \times 8.31 \, J/(mol \cdot K) \times 300 \, K}{32 \times 10^{-3} \, kg/mol}} \]

### Step 5: Calculate the value
Calculating the numerator:
\[ 3 \times 8.31 \times 300 = 7479 \, J/mol \]
Now, calculate the entire expression:
\[ v_{rms} = \sqrt{\frac{7479}{32 \times 10^{-3}}} \]
\[ = \sqrt{\frac{7479}{0.032}} \approx \sqrt{233715.625} \approx 483.3 \, m/s \]

### Final Result
Thus, the rms speed of oxygen molecules at \(27^\circ C\) is approximately:
\[ v_{rms} \approx 483 \, m/s \]