For a gas `(R)/(C_(P)) = 0.4`. For this gas calculate the following (i) Atomicity and degree of freedom (ii) Value of `C_(V)` and `gamma` (iii) Mean gram- molecular kinetic energy at `300K` temperature

Answer

To solve the problem step by step, we will break down each part of the question regarding the gas with the given ratio \( \frac{R}{C_P} = 0.4 \).

### Step 1: Calculate \( C_P \)

Given:
\[ \frac{R}{C_P} = 0.4 \]
We can rearrange this to find \( C_P \):
\[ C_P = \frac{R}{0.4} = \frac{R}{\frac{4}{10}} = \frac{10}{4} R = 2.5 R \]

### Step 2: Calculate \( C_V \)

Using the relation:
\[ C_P = C_V + R \]
We can substitute \( C_P \):
\[ 2.5 R = C_V + R \]
Rearranging gives:
\[ C_V = 2.5 R - R = 1.5 R = \frac{3}{2} R \]

### Step 3: Determine the Degree of Freedom and Atomicity

For a monoatomic gas, the degree of freedom \( f \) is given by:
\[ f = 3 \]
Since the gas is monoatomic, the atomicity is also:
\[ \text{Atomicity} = 1 \]

### Step 4: Calculate \( \gamma \)

The ratio \( \gamma \) (also known as the adiabatic index) is defined as:
\[ \gamma = \frac{C_P}{C_V} \]
Substituting the values we found:
\[ \gamma = \frac{2.5 R}{1.5 R} = \frac{2.5}{1.5} = \frac{5}{3} \]

### Step 5: Calculate Mean Gram-Molecular Kinetic Energy at 300 K

The mean kinetic energy per mole of gas is given by:
\[ \text{Mean Kinetic Energy} = \frac{3}{2} R T \]
Substituting \( T = 300 \, K \):
\[ \text{Mean Kinetic Energy} = \frac{3}{2} R \times 300 = 450 R \]

### Summary of Results
1. **Atomicity**: 1 (monoatomic gas)
2. **Degree of Freedom**: 3
3. **Value of \( C_V \)**: \( \frac{3}{2} R \)
4. **Value of \( \gamma \)**: \( \frac{5}{3} \)
5. **Mean Gram-Molecular Kinetic Energy at 300 K**: \( 450 R \)