Calculate the r.m.s. velocity of argon (atomic mass = 40) at N.T.P.

Explanation

To calculate the root mean square (r.m.s.) velocity of argon gas at normal temperature and pressure (N.T.P.), we can use the formula:

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

Where:
- \( V_{\text{rms}} \) = root mean square velocity
- \( R \) = universal gas constant = 8.314 J/(mol·K)
- \( T \) = absolute temperature in Kelvin
- \( M \) = molar mass in kg/mol

### Step 1: Identify the values needed for the calculation
- The atomic mass of argon is given as 40 g/mol. To convert this to kg/mol:
\[ M = 40 \, \text{g/mol} = 40 \times 10^{-3} \, \text{kg/mol} = 0.040 \, \text{kg/mol} \]
- The temperature at N.T.P. is given as 293 K.

### Step 2: Substitute the values into the formula
Now we can substitute the values into the r.m.s. velocity formula:
\[ V_{\text{rms}} = \sqrt{\frac{3 \times R \times T}{M}} = \sqrt{\frac{3 \times 8.314 \, \text{J/(mol·K)} \times 293 \, \text{K}}{0.040 \, \text{kg/mol}}} \]

### Step 3: Calculate the numerator
Calculating the numerator:
\[ 3 \times 8.314 \times 293 = 7300.662 \, \text{J/mol} \]

### Step 4: Divide by the molar mass
Now divide by the molar mass:
\[ \frac{7300.662}{0.040} = 182516.55 \, \text{m}^2/\text{s}^2 \]

### Step 5: Take the square root
Finally, take the square root to find \( V_{\text{rms}} \):
\[ V_{\text{rms}} = \sqrt{182516.55} \approx 427.43 \, \text{m/s} \]

### Conclusion
The root mean square velocity of argon at N.T.P. is approximately **427.43 m/s**.

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