A sample of gas is compressed from an initial volume of `2v_(0) " to " v_(0)` using three different processes.
First: Using reversible isothermal
Second: Using reversible adiabatic
Third: Using irreversible adiabatic under a constant external pressure then

Options

A. Final temperature of gas will be highest at the end of third process.
B. Final temperature of gas will be highest at the end of second process
C. Enthalpy change of sample will be highest in isothermal process
D. Final pressure of gas will be highest at the end of second process.

Answer

To solve the problem of determining the work done during the compression of a gas from an initial volume of \(2v_0\) to \(v_0\) using three different processes, we will analyze each process step by step.

### Step 1: Understand the Processes

1. **Reversible Isothermal Process**: In this process, the temperature of the gas remains constant while it is compressed. The work done on the gas can be calculated using the formula:
\[ W_{\text{isothermal}} = -nRT \ln\left(\frac{V_f}{V_i}\right) \]
where \(V_f\) is the final volume, \(V_i\) is the initial volume, \(n\) is the number of moles, \(R\) is the universal gas constant, and \(T\) is the absolute temperature.

2. **Reversible Adiabatic Process**: In this process, there is no heat exchange with the surroundings. The work done can be calculated using the formula:
\[ W_{\text{adiabatic}} = \frac{P_i V_i - P_f V_f}{\gamma - 1} \]
where \(\gamma\) is the heat capacity ratio (\(C_p/C_v\)), and \(P_i\) and \(P_f\) are the initial and final pressures.

3. **Irreversible Adiabatic Process**: In this process, the gas is compressed against a constant external pressure. The work done can be calculated as:
\[ W_{\text{irreversible}} = -P_{\text{ext}}(V_f - V_i) \]
where \(P_{\text{ext}}\) is the constant external pressure.

### Step 2: Calculate Work Done in Each Process

1. **For Reversible Isothermal Process**:
\[ W_{\text{isothermal}} = -nRT \ln\left(\frac{v_0}{2v_0}\right) = nRT \ln(2) \]

2. **For Reversible Adiabatic Process**:
The work done in a reversible adiabatic process is generally less than that in an isothermal process for the same change in volume. The exact value depends on the specific heat capacities and the initial and final states of the gas.

3. **For Irreversible Adiabatic Process**:
The work done is given by:
\[ W_{\text{irreversible}} = -P_{\text{ext}}(v_0 - 2v_0) = P_{\text{ext}}(v_0) \]
Since \(P_{\text{ext}}\) is less than the pressure in the reversible processes, the work done here will be the least.

### Step 3: Compare the Work Done

From the analysis:
- \(W_{\text{isothermal}} > W_{\text{adiabatic}} > W_{\text{irreversible}}\)

### Step 4: Determine Final Temperatures

- In the **reversible isothermal process**, the temperature remains constant.
- In the **reversible adiabatic process**, the temperature decreases as the gas is compressed.
- In the **irreversible adiabatic process**, the temperature also decreases but is generally lower than in the reversible adiabatic process due to the nature of irreversible processes.

### Conclusion

The final temperatures after each process will be:
- Highest in the isothermal process.
- Intermediate in the reversible adiabatic process.
- Lowest in the irreversible adiabatic process.

### Final Answer
The correct option is that the final temperature of the gas will be highest at the end of the isothermal process, followed by the reversible adiabatic, and lowest in the irreversible adiabatic process.