The temperature of n-moles of an ideal gas is increased from `T_0` to `2T_0` through a process `p=alpha/T`. Find work done in this process.

Explanation

To find the work done during the process where the temperature of n moles of an ideal gas is increased from \( T_0 \) to \( 2T_0 \) under the condition \( p = \frac{\alpha}{T} \), we can follow these steps:

### Step-by-Step Solution:

1. **Use the Ideal Gas Law**:
The ideal gas law states that:
\[ PV = nRT \]
Given that pressure \( p \) is defined as \( p = \frac{\alpha}{T} \), we can substitute this into the ideal gas equation.

2. **Express Volume in Terms of Temperature**:
Substitute \( p \) into the ideal gas equation:
\[ \frac{\alpha}{T} \cdot V = nRT \]
Rearranging gives:
\[ V = \frac{nRT^2}{\alpha} \]

3. **Differentiate Volume with Respect to Temperature**:
To find \( dV \), differentiate \( V \) with respect to \( T \):
\[ dV = \frac{d}{dT}\left(\frac{nRT^2}{\alpha}\right) = \frac{2nRT}{\alpha} dT \]

4. **Set Up the Work Done Integral**:
The work done \( W \) during a process is given by:
\[ W = \int_{V_i}^{V_f} p \, dV \]
Substitute \( p = \frac{\alpha}{T} \) and \( dV = \frac{2nRT}{\alpha} dT \):
\[ W = \int_{T_0}^{2T_0} \frac{\alpha}{T} \cdot \frac{2nRT}{\alpha} \, dT \]

5. **Simplify the Integral**:
The \( \alpha \) terms cancel out:
\[ W = \int_{T_0}^{2T_0} 2nR \, dT \]

6. **Evaluate the Integral**:
The integral simplifies to:
\[ W = 2nR \left[ T \right]_{T_0}^{2T_0} = 2nR (2T_0 - T_0) = 2nR T_0 \]

7. **Final Result**:
Thus, the work done in this process is:
\[ W = 2nR T_0 \]