A rigid diatomic ideal gas undergoes an adiabatic process at room temperature .The relation between temperature and volume for this process in `TV^x`= constant ,then x is :

Options

A. `2/3`
B. `5/3`
C. `3/5`
D. `2/5`

Explanation

To solve the problem, we need to find the value of \( x \) in the relation \( TV^x = \text{constant} \) for a rigid diatomic ideal gas undergoing an adiabatic process.

### Step-by-Step Solution:

1. **Understand the Adiabatic Process**:
In an adiabatic process, there is no heat exchange with the surroundings. For an ideal gas, the relationship between temperature \( T \) and volume \( V \) can be expressed as:
\[ TV^{\gamma - 1} = \text{constant} \]
where \( \gamma \) (gamma) is the heat capacity ratio \( C_p/C_v \).

2. **Determine the Value of \( \gamma \)**:
For a diatomic ideal gas, the values of specific heat capacities are:
- \( C_v = \frac{5}{2}R \) (for diatomic gases)
- \( C_p = C_v + R = \frac{5}{2}R + R = \frac{7}{2}R \)

Therefore, the ratio \( \gamma \) is given by:
\[ \gamma = \frac{C_p}{C_v} = \frac{\frac{7}{2}R}{\frac{5}{2}R} = \frac{7}{5} \]

3. **Substituting \( \gamma \) into the Adiabatic Equation**:
Now, we substitute \( \gamma \) into the adiabatic relation:
\[ T V^{\gamma - 1} = T V^{\frac{7}{5} - 1} = T V^{\frac{2}{5}} = \text{constant} \]

4. **Identifying the Value of \( x \)**:
From the equation \( TV^{\frac{2}{5}} = \text{constant} \), we can see that \( x = \frac{2}{5} \).

5. **Final Answer**:
Thus, the value of \( x \) is:
\[ x = \frac{2}{5} \]