Calculate the rms speed of an ideal monatomic gas having molecular weight 28 gm/mol at `27^@ C`. If the specific heats at constant pressure and volume are respectively 6.3 J `mol^(-1) K^(-1) and 3.14 J mol^(-1) K^(-1)` respectively.

Answer

To calculate the root mean square (RMS) speed of an ideal monatomic gas, we can use the formula:

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

where:
- \( R \) is the universal gas constant,
- \( T \) is the absolute temperature in Kelvin,
- \( M \) is the molecular weight in kg/mol.

### Step 1: Convert the temperature to Kelvin
The given temperature is \( 27^\circ C \). To convert this to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273.15 \]

So,

\[ T = 27 + 273.15 = 300.15 \, K \approx 300 \, K \]

### Step 2: Identify the molecular weight in kg/mol
The molecular weight given is \( 28 \, g/mol \). To convert this to kg/mol, we divide by 1000:

\[ M = \frac{28 \, g/mol}{1000} = 0.028 \, kg/mol \]

### Step 3: Use the value of the gas constant
The value of the universal gas constant \( R \) is:

\[ R = 8.314 \, J/(mol \cdot K) \]

### Step 4: Substitute the values into the RMS speed formula
Now we can substitute the values into the RMS speed formula:

\[ v_{rms} = \sqrt{\frac{3 \times R \times T}{M}} \]

Substituting the known values:

\[ v_{rms} = \sqrt{\frac{3 \times 8.314 \, J/(mol \cdot K) \times 300 \, K}{0.028 \, kg/mol}} \]

### Step 5: Calculate the numerator
Calculating the numerator:

\[ 3 \times 8.314 \times 300 = 7482.6 \, J/mol \]

### Step 6: Divide by the molecular weight
Now divide by the molecular weight:

\[ \frac{7482.6}{0.028} = 267,235.71 \, m^2/s^2 \]

### Step 7: Take the square root
Now take the square root to find \( v_{rms} \):

\[ v_{rms} = \sqrt{267,235.71} \approx 516.44 \, m/s \]

### Final Answer
Thus, the RMS speed of the ideal monatomic gas at \( 27^\circ C \) is approximately:

\[ v_{rms} \approx 516 \, m/s \]

---