The equation of state of a gas is given as P(V-b)=nRT, where b is constant ,n is the number of moles and R is the universal gas constant .when 2 moles of this gas undergo reversible isothermal expansion from volume V to 2V ,what is work done by the gas ?

Options

A. The gravitational field intensity at the mid point of any side is `(4Gm)/(3a^(2))`
B. The gravitational field intensity at centroid of the triangle is zero
C. The gravitation field intensity at centroid of the triangle is `(sqrt(3)Gm)/(2a^(2))`
D. The potential energy of system is `(-3Gm^(2))/(a)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Write down the equation of state
The equation of state of the gas is given as:
\[ P(V - b) = nRT \]
where \( P \) is the pressure, \( V \) is the volume, \( b \) is a constant, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature.

### Step 2: Identify the given data
From the problem, we have:
- Number of moles, \( n = 2 \) moles
- Initial volume, \( V_i = V \)
- Final volume, \( V_f = 2V \)
- The process is isothermal, meaning the temperature \( T \) is constant.

### Step 3: Express pressure in terms of volume
From the equation of state, we can express pressure \( P \) as:
\[ P = \frac{nRT}{V - b} \]

### Step 4: Set up the work done integral
The work done \( W \) by the gas during the expansion is given by the integral:
\[ W = \int_{V}^{2V} P \, dV \]
Substituting the expression for \( P \):
\[ W = \int_{V}^{2V} \frac{nRT}{V - b} \, dV \]

### Step 5: Integrate the expression
Since \( nRT \) is constant during the isothermal process, we can take it out of the integral:
\[ W = nRT \int_{V}^{2V} \frac{1}{V - b} \, dV \]
Now, we will integrate:
\[ W = nRT \left[ \ln(V - b) \right]_{V}^{2V} \]
This results in:
\[ W = nRT \left( \ln(2V - b) - \ln(V - b) \right) \]

### Step 6: Simplify using logarithmic properties
Using the property of logarithms, \( \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \):
\[ W = nRT \ln\left(\frac{2V - b}{V - b}\right) \]

### Step 7: Substitute the values
Now, substituting \( n = 2 \):
\[ W = 2RT \ln\left(\frac{2V - b}{V - b}\right) \]

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