Two moles of a diatomic ideal gas is taken through `pT=` constant. Its temperature is increased from T to 2T. Find the work done by the system?

Explanation

To solve the problem of finding the work done by the system when two moles of a diatomic ideal gas are taken through a constant \( pT \) path while its temperature is increased from \( T \) to \( 2T \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information**:
- We have 2 moles of a diatomic ideal gas.
- The process is along a path where \( pT = \text{constant} \).
- The initial temperature \( T_i = T \) and the final temperature \( T_f = 2T \).

2. **Express the Constant**:
- Let \( k = pT \) be the constant. Therefore, we can write:
\[ p = \frac{k}{T} \]

3. **Use the Ideal Gas Law**:
- According to the ideal gas law, we have:
\[ pV = nRT \]
- Substituting \( p \) from the previous step:
\[ \frac{k}{T} V = nRT \]
- Rearranging gives:
\[ V = \frac{nRT^2}{k} \]

4. **Differentiate Volume with Respect to Temperature**:
- Differentiate both sides with respect to \( T \):
\[ dV = \frac{d}{dT} \left( \frac{nRT^2}{k} \right) \]
- This results in:
\[ dV = \frac{2nRT}{k} dT \]

5. **Calculate Work Done**:
- The work done \( W \) in a process can be expressed as:
\[ W = \int P \, dV \]
- Substitute \( P = \frac{k}{T} \) and \( dV = \frac{2nRT}{k} dT \):
\[ W = \int \frac{k}{T} \left( \frac{2nRT}{k} dT \right) \]
- Simplifying gives:
\[ W = 2nR \int dT \]

6. **Evaluate the Integral**:
- The integral of \( dT \) from \( T \) to \( 2T \) is:
\[ W = 2nR \left[ T \right]_{T}^{2T} = 2nR (2T - T) = 2nRT \]

7. **Substitute the Number of Moles**:
- Since \( n = 2 \) (given in the problem):
\[ W = 2 \cdot 2RT = 4RT \]

### Final Answer:
The work done by the system is:
\[ \boxed{4RT} \]