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

Options

A. 491m/s
B. 941m/s
C. 421m/s
D. 591m/s

Explanation

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

### Step-by-Step Solution:

1. **Convert Temperature to Kelvin:**
The given temperature is 27°C. To convert this to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]
Thus,
\[ T = 27 + 273 = 300 \, K \]

2. **Identify the Molecular Mass of Phosphorus:**
The molecular formula for phosphorus is \( P_4 \) (tetraphosphorus), and the atomic mass of phosphorus (P) is approximately 31 g/mol. Therefore, the molar mass of \( P_4 \) is:
\[ M = 4 \times 31 = 124 \, g/mol \]
We need to convert this to kilograms for our calculations:
\[ M = 124 \, g/mol \times \frac{1 \, kg}{1000 \, g} = 0.124 \, kg/mol \]

3. **Use the RMS Speed Formula:**
The formula for calculating the rms speed (\( v_{rms} \)) of a gas is given by:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
where:
- \( R \) is the universal gas constant, \( R = 8.31 \, J/(mol \cdot K) \)
- \( T \) is the temperature in Kelvin
- \( M \) is the molar mass in kg/mol

4. **Substitute the Values into the Formula:**
Now we can substitute the values into the rms speed formula:
\[ v_{rms} = \sqrt{\frac{3 \times 8.31 \, J/(mol \cdot K) \times 300 \, K}{0.124 \, kg/mol}} \]

5. **Calculate the RMS Speed:**
First, calculate the numerator:
\[ 3 \times 8.31 \times 300 = 7479 \, J/mol \]
Now, divide by the molar mass:
\[ \frac{7479}{0.124} \approx 60323.39 \, m^2/s^2 \]
Finally, take the square root:
\[ v_{rms} = \sqrt{60323.39} \approx 245.52 \, m/s \]

### Final Answer:
The rms speed of phosphorus molecules at 27°C is approximately \( 245.52 \, m/s \).