There are two statement of same gas initially under similar initial state. Gases of both the samples are expanded. I s t sample using reversible isothermal process and I I n d sample using reversible adiabatic process till final pressure of both the samples becomes half of initial pressure. Then:

Options

A. work done by gas in `I^(st)` sample `gt` work done by gas in `II^(nd)` gas
B. work done by gas in `II^(nd)` sample `gt` work done by gas in `I^(st)` sample
C. work done by gas in `I^(st)` sample`=` work done by gas in `II^(th)` sample
D. None of these

Answer

To solve the problem, we need to analyze the work done by the gas in both the reversible isothermal and reversible adiabatic processes.

### Step 1: Work Done in Reversible Isothermal Expansion
For the first sample, which undergoes a reversible isothermal expansion, the work done \( W_1 \) can be calculated using the formula:

\[ W_1 = 2.303 nRT \log \left( \frac{P_1}{P_2} \right) \]

Given that the final pressure \( P_2 \) is half of the initial pressure \( P_1 \):

\[ P_2 = \frac{P_1}{2} \]

Substituting this into the equation:

\[ W_1 = 2.303 nRT \log \left( \frac{P_1}{\frac{P_1}{2}} \right) = 2.303 nRT \log (2) \]

### Step 2: Work Done in Reversible Adiabatic Expansion
For the second sample, which undergoes a reversible adiabatic expansion, the work done \( W_2 \) is given by:

\[ W_2 = \frac{P_2 V_2 - P_1 V_1}{\gamma - 1} \]

Since \( P_2 = \frac{P_1}{2} \), we need to find \( V_2 \) using the adiabatic condition \( P_1 V_1^\gamma = P_2 V_2^\gamma \):

\[ P_1 V_1^\gamma = \frac{P_1}{2} V_2^\gamma \implies V_2^\gamma = 2 V_1^\gamma \implies V_2 = V_1 \cdot 2^{\frac{1}{\gamma}} \]

Now substituting \( V_2 \) into the work done equation:

\[ W_2 = \frac{\left(\frac{P_1}{2}\right) \left(V_1 \cdot 2^{\frac{1}{\gamma}}\right) - P_1 V_1}{\gamma - 1} \]

\[ = \frac{\frac{P_1 V_1}{2} \cdot 2^{\frac{1}{\gamma}} - P_1 V_1}{\gamma - 1} \]

Factoring out \( P_1 V_1 \):

\[ W_2 = \frac{P_1 V_1 \left(2^{\frac{1}{\gamma} - 1} - 1\right)}{\gamma - 1} \]

### Step 3: Comparison of Work Done
Now we have expressions for both \( W_1 \) and \( W_2 \):

- \( W_1 = 2.303 nRT \log (2) \)
- \( W_2 = \frac{P_1 V_1 \left(2^{\frac{1}{\gamma} - 1} - 1\right)}{\gamma - 1} \)

To compare \( W_1 \) and \( W_2 \), we note that the value of \( \gamma \) (the heat capacity ratio) affects the work done in the adiabatic process. Since \( \gamma \) is greater than 1 for all real gases, the term \( 2^{\frac{1}{\gamma} - 1} \) will be less than 1, making \( W_2 \) negative.

### Conclusion
Since \( W_1 \) is always positive (as work is done by the gas during expansion) and \( W_2 \) is negative (as work is done on the gas), we conclude that:

- The work done by the gas in the isothermal process \( W_1 \) is greater than the work done in the adiabatic process \( W_2 \).