Two identical gases whose adiabatic exponent is `gamma` are filled in two identical containers at equal pressures. In both the containers the volume of gas is doubled. In first container it is done by an isothermal process and in second container it is done by adiabatic process. Find the condition for which the work done by the gas in the two expansion process is same.

Answer

To solve the problem, we need to find the condition under which the work done by the gas in an isothermal expansion is equal to the work done in an adiabatic expansion. Let's break it down step by step.

### Step 1: Define the Initial Conditions
Let the initial pressure and volume of both gases be:
- Initial pressure, \( P_0 \)
- Initial volume, \( V_0 \)

### Step 2: Work Done in Isothermal Expansion
For the isothermal process, the work done \( W_1 \) when the volume is doubled (final volume \( V_f = 2V_0 \)) is given by:
\[ W_1 = nRT \ln\left(\frac{V_f}{V_i}\right) \]
Since \( P_0 V_0 = nRT \), we can substitute \( nRT \) with \( P_0 V_0 \):
\[ W_1 = P_0 V_0 \ln\left(\frac{2V_0}{V_0}\right) = P_0 V_0 \ln(2) \]

### Step 3: Work Done in Adiabatic Expansion
For the adiabatic process, we use the relation \( P V^\gamma = \text{constant} \). The final pressure \( P_f \) can be expressed as:
\[ P_f V_f^\gamma = P_0 V_0^\gamma \]
Substituting \( V_f = 2V_0 \):
\[ P_f (2V_0)^\gamma = P_0 V_0^\gamma \]
This simplifies to:
\[ P_f = P_0 \cdot 2^{-\gamma + 1} \]

The work done \( W_2 \) in the adiabatic process is given by:
\[ W_2 = \frac{P_f V_f - P_0 V_0}{\gamma - 1} \]
Substituting \( P_f \) and \( V_f \):
\[ W_2 = \frac{P_0 \cdot 2^{-\gamma + 1} \cdot 2V_0 - P_0 V_0}{\gamma - 1} \]
This simplifies to:
\[ W_2 = \frac{P_0 V_0 (2 \cdot 2^{-\gamma + 1} - 1)}{\gamma - 1} \]
\[ W_2 = \frac{P_0 V_0 (2^{2 - \gamma} - 1)}{\gamma - 1} \]

### Step 4: Equate the Work Done
To find the condition for which the work done in both processes is the same:
\[ W_1 = W_2 \]
Substituting the expressions we derived:
\[ P_0 V_0 \ln(2) = \frac{P_0 V_0 (2^{2 - \gamma} - 1)}{\gamma - 1} \]
Dividing both sides by \( P_0 V_0 \) (assuming \( P_0 V_0 \neq 0 \)):
\[ \ln(2) = \frac{2^{2 - \gamma} - 1}{\gamma - 1} \]

### Final Condition
Thus, the condition for which the work done by the gas in both expansion processes is the same is:
\[ \ln(2) = \frac{2^{2 - \gamma} - 1}{\gamma - 1} \]