In adiabatic process, the work involved during expansion or compression of an ideal gas is given by :

Options

A. `nC_(v)DeltaT`
B. `(nR)/(gamma-1)(T_(2)-T_(1))`
C. `-nRP_("ext")[(T_(2)P_(1)-T_(1)P_(2))/(P_(1)P_(2))]`
D. `-2.303 " RT log "(V_(2))/(V_(1))`

Explanation

To solve the problem regarding the work involved during the expansion or compression of an ideal gas in an adiabatic process, we will analyze both the reversible and irreversible cases.

### Step-by-Step Solution:

1. **Understand the Adiabatic Process**:
In an adiabatic process, there is no heat exchange with the surroundings. Therefore, we have:
\[ q = 0 \]

2. **Apply the First Law of Thermodynamics**:
The first law states:
\[ \Delta U = q + W \]
Since \( q = 0 \), we can simplify this to:
\[ \Delta U = W \]
This means that the work done \( W \) is equal to the change in internal energy \( \Delta U \).

3. **Relate Internal Energy to Temperature**:
For an ideal gas, the change in internal energy can be expressed as:
\[ \Delta U = n C_v \Delta T \]
where \( n \) is the number of moles, \( C_v \) is the molar heat capacity at constant volume, and \( \Delta T \) is the change in temperature.

4. **Work Done in Reversible Adiabatic Process**:
For a reversible adiabatic process, we can express the work done as:
\[ W = n C_v \Delta T \]
This matches with the first option provided in the question.

5. **Work Done in Irreversible Adiabatic Process**:
For an irreversible adiabatic process, the work done can be expressed as:
\[ W = - P_{\text{ext}} \Delta V \]
where \( P_{\text{ext}} \) is the external pressure and \( \Delta V \) is the change in volume.

6. **Relate Volume Change to Temperature**:
Using the ideal gas law \( PV = nRT \), we can express the volumes at two different states:
\[ V_2 = \frac{nRT_2}{P_2} \quad \text{and} \quad V_1 = \frac{nRT_1}{P_1} \]
Thus, the change in volume \( \Delta V \) can be written as:
\[ \Delta V = V_2 - V_1 = \frac{nR}{P_2}T_2 - \frac{nR}{P_1}T_1 \]

7. **Final Expression for Work Done**:
Substituting \( \Delta V \) into the work done expression for irreversible processes gives:
\[ W = -P_{\text{ext}} \left( \frac{nR}{P_2}T_2 - \frac{nR}{P_1}T_1 \right) \]
After simplification, this matches with the third option provided in the question.

### Summary of Results:
- For reversible adiabatic processes:
\[ W = n C_v \Delta T \]
- For irreversible adiabatic processes:
\[ W = -P_{\text{ext}} \Delta V \]