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

Explanation

To find the RMS (Root Mean Square) speed of Argon molecules at a temperature of 27°C, we can follow these steps:

### Step 1: Convert Temperature to Kelvin
The temperature in Celsius needs to be converted to Kelvin using the formula:
\[ T(K) = T(°C) + 273.15 \]

Given:
\[ T(°C) = 27 \]
\[ T(K) = 27 + 273.15 = 300.15 \approx 300 \, K \]

### Step 2: Identify Constants
- The gas constant \( R \) is given as:
\[ R = 8.31 \, \text{J K}^{-1} \text{mol}^{-1} \]
- The molar mass of Argon (Ar) is approximately:
\[ M = 39.9 \, \text{g/mol} \approx 40 \, \text{g/mol} \]
To convert this to kg, we use:
\[ M = 40 \, \text{g/mol} = 40 \times 10^{-3} \, \text{kg/mol} \]

### Step 3: Use the RMS Speed Formula
The formula for RMS speed \( v_{rms} \) is given by:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]

### Step 4: Substitute the Values
Now, substituting the values into the formula:
\[ v_{rms} = \sqrt{\frac{3 \times 8.31 \, \text{J K}^{-1} \text{mol}^{-1} \times 300 \, K}{40 \times 10^{-3} \, \text{kg/mol}}} \]

### Step 5: Calculate the Value
Calculating the numerator:
\[ 3 \times 8.31 \times 300 = 7473 \, \text{J} \]

Now, calculating the denominator:
\[ 40 \times 10^{-3} = 0.04 \, \text{kg/mol} \]

Now, substituting these into the equation:
\[ v_{rms} = \sqrt{\frac{7473}{0.04}} \]

Calculating:
\[ \frac{7473}{0.04} = 186825 \]

Taking the square root:
\[ v_{rms} = \sqrt{186825} \approx 432.1 \, \text{m/s} \]

### Final Answer
The RMS speed of Argon molecules at 27°C is approximately:
\[ v_{rms} \approx 432 \, \text{m/s} \]

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