Specific heat of an elementary gas is found to be 0.313 J at constant volume. If the molar mass of the gas is 40 g `mol^(-1)` , what is the atomicity of the gas ?

Explanation

To determine the atomicity of the gas given its specific heat at constant volume (Cv) and molar mass, we can follow these steps:

### Step 1: Identify the given values
- Specific heat at constant volume (Cv) = 0.313 J/g·K
- Molar mass of the gas (M) = 40 g/mol

### Step 2: Calculate Cv in terms of J/mol·K
To convert the specific heat from J/g·K to J/mol·K, we use the formula:
\[ C_v = S_v \times M \]
Where:
- \( S_v \) = specific heat at constant volume (0.313 J/g·K)
- \( M \) = molar mass (40 g/mol)

Calculating:
\[ C_v = 0.313 \, \text{J/g·K} \times 40 \, \text{g/mol} = 12.52 \, \text{J/mol·K} \]

### Step 3: Calculate Cp using the relation Cp = Cv + R
The relation between the specific heats at constant pressure and constant volume is given by:
\[ C_p = C_v + R \]
Where:
- \( R \) = universal gas constant = 8.314 J/mol·K

Calculating:
\[ C_p = 12.52 \, \text{J/mol·K} + 8.314 \, \text{J/mol·K} = 20.834 \, \text{J/mol·K} \]

### Step 4: Calculate the ratio of Cp to Cv (γ)
The ratio of specific heats (γ) is given by:
\[ \gamma = \frac{C_p}{C_v} \]

Calculating:
\[ \gamma = \frac{20.834 \, \text{J/mol·K}}{12.52 \, \text{J/mol·K}} \approx 1.666 \]

### Step 5: Relate γ to atomicity (F)
For an ideal gas, the relationship between γ and the degrees of freedom (F) is given by:
\[ \gamma = 1 + \frac{2}{F} \]

Rearranging this equation to find F:
\[ \gamma - 1 = \frac{2}{F} \]
\[ F = \frac{2}{\gamma - 1} \]

Substituting the value of γ:
\[ F = \frac{2}{1.666 - 1} = \frac{2}{0.666} \approx 3 \]

### Step 6: Determine the atomicity of the gas
The degrees of freedom (F) for a monoatomic gas is 3, which indicates that the gas is monoatomic.

### Final Answer
The atomicity of the gas is **1** (indicating it is a monoatomic gas).

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