The equation of a state of a gas is given by `p(V-b)=nRT`. If 1 mole of a gas is isothermally expanded from volume V and 2V, the work done during the process is

Options

A. (a) `RT ln|(2V-b)/(V-b)|`
B. (b) `RT ln|(V-b)/(V)|`
C. (c) `RT In|(V-b)/(2V-b)|`
D. (d) `RT In|(V)/(V-b)|`

Explanation

To solve the problem, we need to calculate the work done during the isothermal expansion of 1 mole of gas from volume \( V \) to \( 2V \) using the given equation of state \( p(V-b) = nRT \).

### Step-by-Step Solution:

1. **Identify the equation of state**:
The equation is given as:
\[ p(V - b) = nRT \]
Rearranging this gives us the pressure \( p \):
\[ p = \frac{nRT}{V - b} \]

2. **Set up the work done formula**:
The work done \( W \) during an isothermal expansion is given by the integral:
\[ W = \int_{V}^{2V} p \, dV \]
Substituting the expression for \( p \):
\[ W = \int_{V}^{2V} \frac{nRT}{V - b} \, dV \]

3. **Take constants out of the integral**:
Since \( nRT \) is constant for this process (1 mole of gas at constant temperature), we can factor it out of the integral:
\[ W = nRT \int_{V}^{2V} \frac{1}{V - b} \, dV \]

4. **Evaluate the integral**:
The integral of \( \frac{1}{V - b} \) is:
\[ \int \frac{1}{V - b} \, dV = \ln |V - b| \]
Therefore, we evaluate the definite integral:
\[ W = nRT \left[ \ln |V - b| \right]_{V}^{2V} \]
This gives:
\[ W = nRT \left( \ln |2V - b| - \ln |V - b| \right) \]

5. **Simplify using properties of logarithms**:
Using the property of logarithms \( \ln a - \ln b = \ln \frac{a}{b} \):
\[ W = nRT \ln \left( \frac{2V - b}{V - b} \right) \]

6. **Substituting the number of moles**:
Since we have 1 mole of gas (\( n = 1 \)):
\[ W = RT \ln \left( \frac{2V - b}{V - b} \right) \]

### Final Answer:
The work done during the isothermal expansion of the gas is:
\[ W = RT \ln \left( \frac{2V - b}{V - b} \right) \]