The initial volume and pressure of a gas are V and P respectively. It is expand by two different processes such that in each process the final volume becomes 2V. If the work done in isothermal change is W1 and the amount of work done in adiabatic change is`W_(2)`, then

Options

A. `W_(1) gt W_(2)`
B. `W_(1) lt W_(2)`
C. `W_(1) = W_(2)`
D. nothing can be said

Answer

To solve the problem regarding the work done in isothermal and adiabatic processes, we can follow these steps:

### Step 1: Understand the Initial Conditions
The initial volume (V) and pressure (P) of the gas are given. The gas expands to a final volume of \(2V\) in both processes.

### Step 2: Work Done in Isothermal Process
For an isothermal process, the work done \(W_1\) can be calculated using the formula:
\[ W_1 = P \cdot V \cdot \ln\left(\frac{V_f}{V_i}\right) \]
Here, \(V_f = 2V\) and \(V_i = V\). Thus, we can substitute these values:
\[ W_1 = P \cdot V \cdot \ln\left(\frac{2V}{V}\right) = P \cdot V \cdot \ln(2) \]

### Step 3: Work Done in Adiabatic Process
For an adiabatic process, the work done \(W_2\) can be calculated using the formula:
\[ W_2 = \frac{1}{\gamma - 1} (P_1 V_1 - P_2 V_2) \]
Where \(P_1\) and \(V_1\) are the initial pressure and volume, and \(P_2\) and \(V_2\) are the final pressure and volume.

Using the adiabatic condition \(PV^\gamma = \text{constant}\), we can express \(P_2\) as:
\[ P_2 = P_1 \left(\frac{V_1}{V_2}\right)^\gamma = P \left(\frac{V}{2V}\right)^\gamma = P \left(\frac{1}{2}\right)^\gamma = \frac{P}{2^\gamma} \]

Now substituting \(P_2\) into the work done formula:
\[ W_2 = \frac{1}{\gamma - 1} \left(PV - \frac{P}{2^\gamma} \cdot 2V\right) \]
\[ = \frac{1}{\gamma - 1} \left(PV - \frac{P \cdot 2V}{2^\gamma}\right) \]
\[ = \frac{PV}{\gamma - 1} \left(1 - \frac{2}{2^\gamma}\right) \]

### Step 4: Compare \(W_1\) and \(W_2\)
From the expressions derived, we can see:
\[ W_1 = PV \ln(2) \]
\[ W_2 = \frac{PV}{\gamma - 1} \left(1 - \frac{2}{2^\gamma}\right) \]

To compare \(W_1\) and \(W_2\), we need to analyze the factors involved. Since the logarithmic function grows slower than polynomial functions, and given that \(\gamma > 1\), it can be inferred that:
\[ W_1 > W_2 \]

### Conclusion
Thus, we conclude that the work done in the isothermal process \(W_1\) is greater than the work done in the adiabatic process \(W_2\):
\[ W_1 > W_2 \]