Find the expression for the work done by a system undergoing isothermal compression (or expansion) from volume `V_(1)` to `V_(2)` at temperature `T_(0)` for a gas which obeys the van der waals equation of state. `(P +an^(2) //V^(2)) (V-bn) =nRT`?

Explanation

To find the expression for the work done by a system undergoing isothermal compression or expansion from volume \( V_1 \) to \( V_2 \) at temperature \( T_0 \) for a gas that obeys the van der Waals equation, we can follow these steps:

### Step 1: Write the van der Waals equation
The van der Waals equation is given by:
\[ \left( P + \frac{a n^2}{V^2} \right)(V - bn) = nRT \]
where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the universal gas constant, \( a \) and \( b \) are van der Waals constants.

### Step 2: Rearrange the equation to solve for pressure \( P \)
We can rearrange the equation to isolate \( P \):
\[ P = \frac{nRT}{V - bn} - \frac{a n^2}{V^2} \]

### Step 3: Write the expression for work done
The work done \( W \) during an isothermal process is given by:
\[ W = \int_{V_1}^{V_2} P \, dV \]

### Step 4: Substitute the expression for \( P \)
Substituting the expression for \( P \) into the work done equation:
\[ W = \int_{V_1}^{V_2} \left( \frac{nRT}{V - bn} - \frac{a n^2}{V^2} \right) dV \]

### Step 5: Separate the integral
We can separate the integral into two parts:
\[ W = \int_{V_1}^{V_2} \frac{nRT}{V - bn} \, dV - \int_{V_1}^{V_2} \frac{a n^2}{V^2} \, dV \]

### Step 6: Solve the first integral
The first integral can be solved using the substitution \( u = V - bn \):
\[ \int \frac{nRT}{V - bn} \, dV = nRT \ln |V - bn| \]
Evaluating from \( V_1 \) to \( V_2 \):
\[ \left[ nRT \ln |V - bn| \right]_{V_1}^{V_2} = nRT \left( \ln |V_2 - bn| - \ln |V_1 - bn| \right) = nRT \ln \left( \frac{V_2 - bn}{V_1 - bn} \right) \]

### Step 7: Solve the second integral
The second integral is:
\[ \int \frac{a n^2}{V^2} \, dV = -\frac{a n^2}{V} \]
Evaluating from \( V_1 \) to \( V_2 \):
\[ \left[-\frac{a n^2}{V}\right]_{V_1}^{V_2} = -\frac{a n^2}{V_2} + \frac{a n^2}{V_1} = a n^2 \left( \frac{1}{V_1} - \frac{1}{V_2} \right) \]

### Step 8: Combine the results
Combining both parts, we have:
\[ W = nRT \ln \left( \frac{V_2 - bn}{V_1 - bn} \right) + a n^2 \left( \frac{1}{V_1} - \frac{1}{V_2} \right) \]

### Final Expression
Thus, the expression for the work done by the system during isothermal compression or expansion is:
\[ W = nRT \ln \left( \frac{V_2 - bn}{V_1 - bn} \right) + a n^2 \left( \frac{1}{V_1} - \frac{1}{V_2} \right) \]