Statement-1: For a gaseous system `C_(p)` is always greater than `C_(v)`
Statement-2: Work done by a gas at constant volume is zero.

Options

A. Statement-1 is true, statement-2 is true, statement-2 is a correct explanation for statement-1
B. Statement-1 is true, statement-2 is true, statement-2 is NOT a correct explanation for statement-1
C. Statement-1 is true, statement-2 is false
D. Statement-1 is false, statement-2 is true

Answer

To analyze the statements provided in the question, we will go through them step by step.

### Step 1: Understanding Statement 1
**Statement 1:** For a gaseous system, \( C_p \) is always greater than \( C_v \).

- **Explanation:**
- \( C_p \) is the specific heat capacity at constant pressure, while \( C_v \) is the specific heat capacity at constant volume.
- When heat is added to a gas at constant pressure, the gas does work on its surroundings (expands), which requires additional energy. Therefore, the heat capacity at constant pressure \( C_p \) must be greater than the heat capacity at constant volume \( C_v \).
- Mathematically, this can be expressed as \( C_p = C_v + R \) for an ideal gas, where \( R \) is the gas constant. This shows that \( C_p \) is always greater than \( C_v \).

### Step 2: Understanding Statement 2
**Statement 2:** Work done by a gas at constant volume is zero.

- **Explanation:**
- Work done by a gas is defined as \( W = P \Delta V \), where \( P \) is the pressure and \( \Delta V \) is the change in volume.
- At constant volume, there is no change in volume (\( \Delta V = 0 \)), which means that the work done \( W = P \cdot 0 = 0 \).
- Therefore, the work done by a gas at constant volume is indeed zero.

### Conclusion
Both statements are correct:
- **Statement 1** is correct because \( C_p \) is always greater than \( C_v \).
- **Statement 2** is correct because the work done by a gas at constant volume is zero.

Thus, the correct answer is that both statements are true.

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