Calculate the rms speed of an ideal diatomic gas having molecular weight 32 gm/mol at `0^@ C`. If the specific heats at constant pressure and volume are respectively 8.3 cal `mol^(-1) K^(-1) and 6.34` cal `mol^(-1) K^(-1)` respectively.

Answer

To calculate the root mean square (RMS) speed of an ideal diatomic gas with a molecular weight of 32 g/mol at 0°C, we will follow these steps:

### Step 1: Convert the given values to appropriate units
- **Molecular weight (M)**: Given as 32 g/mol, we convert it to kg/mol:
\[ M = 32 \, \text{g/mol} = 32 \times 10^{-3} \, \text{kg/mol} \]

- **Temperature (T)**: Given as 0°C, we convert it to Kelvin:
\[ T = 0 + 273 = 273 \, \text{K} \]

### Step 2: Calculate the gas constant (R)
Using Mayer's formula:
\[ C_p - C_v = R \]
Where:
- \(C_p = 8.3 \, \text{cal/mol·K}\)
- \(C_v = 6.34 \, \text{cal/mol·K}\)

Calculating \(R\):
\[ R = C_p - C_v = 8.3 - 6.34 = 1.96 \, \text{cal/mol·K} \]

Now, convert \(R\) from cal to joules:
\[ 1 \, \text{cal} = 4.2 \, \text{J} \]
Thus,
\[ R = 1.96 \, \text{cal/mol·K} \times 4.2 \, \text{J/cal} = 8.232 \, \text{J/mol·K} \]

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

Substituting the values we have:
\[ v_{rms} = \sqrt{\frac{3 \times 8.232 \, \text{J/mol·K} \times 273 \, \text{K}}{32 \times 10^{-3} \, \text{kg/mol}}} \]

Calculating the numerator:
\[ 3 \times 8.232 \times 273 = 6761.556 \, \text{J} \]

Calculating the denominator:
\[ 32 \times 10^{-3} = 0.032 \, \text{kg/mol} \]

Now substituting into the RMS speed formula:
\[ v_{rms} = \sqrt{\frac{6761.556}{0.032}} = \sqrt{211,865.5} \approx 460.7 \, \text{m/s} \]

### Final Answer
The RMS speed of the ideal diatomic gas at 0°C is approximately:
\[ v_{rms} \approx 461.2 \, \text{m/s} \]