A gas is compressed isothermally to half its initial volume. The same gas is compressed separately through an adiabatic process untill its volume is again reduced to half. Then

Options

A. Which of the case (whether compression through isothermal or through isothermal or through adiabatic process) requires more work will depend upon the atomicity of the gas
B. Compressing the gas isothermally will require more work done to be done
C. Compressing the gas through adiabatic process will require more work to be done
D. Compressing the gas isothermally or adiabatically will require the same amount of work

Answer

To solve the problem of comparing the work done during isothermal and adiabatic compression of a gas to half its initial volume, we can follow these steps:

### Step 1: Understand the Processes
- **Isothermal Process**: This is a process that occurs at a constant temperature. For an ideal gas, the work done \( W \) during isothermal compression can be calculated using the formula:
\[ W = -nRT \ln\left(\frac{V_f}{V_i}\right) \]
where \( n \) is the number of moles, \( R \) is the universal gas constant, \( T \) is the absolute temperature, \( V_f \) is the final volume, and \( V_i \) is the initial volume.

- **Adiabatic Process**: This is a process where no heat is exchanged with the surroundings. The work done \( W \) during adiabatic compression can be calculated using the formula:
\[ W = \frac{P_i V_i - P_f V_f}{\gamma - 1} \]
where \( \gamma \) is the heat capacity ratio (Cp/Cv), \( P_i \) and \( P_f \) are the initial and final pressures, respectively.

### Step 2: Calculate Work Done in Isothermal Compression
1. Let the initial volume \( V_i = V_1 \) and the final volume \( V_f = \frac{V_1}{2} \).
2. Substitute into the isothermal work formula:
\[ W_{\text{isothermal}} = -nRT \ln\left(\frac{\frac{V_1}{2}}{V_1}\right) = -nRT \ln\left(\frac{1}{2}\right) = nRT \ln(2) \]

### Step 3: Calculate Work Done in Adiabatic Compression
1. For adiabatic compression, we need to relate the initial and final states using the adiabatic condition:
\[ P_i V_i^\gamma = P_f V_f^\gamma \]
2. Since \( V_f = \frac{V_i}{2} \), we can express \( P_f \) in terms of \( P_i \):
\[ P_f = P_i \left(\frac{V_i}{\frac{V_i}{2}}\right)^\gamma = P_i \cdot 2^\gamma \]
3. Substitute \( P_f \) into the work done formula:
\[ W_{\text{adiabatic}} = \frac{P_i V_i - P_f V_f}{\gamma - 1} = \frac{P_i V_i - P_i \cdot 2^\gamma \cdot \frac{V_i}{2}}{\gamma - 1} \]
Simplifying this gives:
\[ W_{\text{adiabatic}} = \frac{P_i V_i (1 - 2^{\gamma - 1})}{\gamma - 1} \]

### Step 4: Compare Work Done in Both Processes
- From the analysis, we can see that:
- The work done in the adiabatic process is generally greater than the work done in the isothermal process because \( \gamma > 1 \) leads to \( 2^{\gamma - 1} > 1 \), making \( W_{\text{adiabatic}} > W_{\text{isothermal}} \).

### Conclusion
- The work done during adiabatic compression is greater than that during isothermal compression when compressing the gas to half its initial volume.