The specific heat of gas is found to be 0.075 calories at constant volume and its formula weight is 40. The atomicity of the gas would be

Options

A. One
B. Two
C. Three
D. Four

Answer

To solve the problem, we need to determine the atomicity of the gas based on the given specific heat at constant volume (Cv) and the formula weight of the gas. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Specific heat at constant volume (Cv) = 0.075 calories
- Formula weight (molecular weight) = 40

### Step 2: Calculate the total heat (Cv)
The total heat (Cv) can be calculated using the formula:
\[ Cv = \text{Specific Heat} \times \text{Molecular Weight} \]
Substituting the values:
\[ Cv = 0.075 \times 40 = 3 \text{ calories} \]

### Step 3: Use the relation between Cp and Cv
We know the relation between the specific heats at constant pressure (Cp) and constant volume (Cv):
\[ Cp - Cv = R \]
Where R is the gas constant. In calories, R is approximately 2.

### Step 4: Calculate Cp
Now, we can find Cp using the value of Cv and R:
\[ Cp = Cv + R \]
Substituting the values:
\[ Cp = 3 + 2 = 5 \text{ calories} \]

### Step 5: Calculate gamma (γ)
Gamma (γ) is defined as the ratio of Cp to Cv:
\[ \gamma = \frac{Cp}{Cv} \]
Substituting the values we found:
\[ \gamma = \frac{5}{3} \approx 1.67 \]

### Step 6: Determine the atomicity of the gas
The atomicity of the gas can be determined from the value of γ. For a monoatomic gas, γ = 5/3, for a diatomic gas, γ = 7/5, and for a polyatomic gas, γ is generally less than 5/3. Since γ is approximately 1.67, we can conclude that the atomicity of the gas is 1 (indicating it behaves like a monoatomic gas).

### Final Answer
The atomicity of the gas is 1.

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