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, we need to find the work done by the system when 2 moles of a diatomic ideal gas undergo a process at constant \( P \cdot T \) while the temperature is increased from \( T \) to \( 2T \).

### Step-by-Step Solution:

1. **Understanding the Process**:
The process is defined as \( P \cdot T = \text{constant} \). We can denote this constant as \( K \).
\[ P \cdot T = K \]

2. **Using the Ideal Gas Law**:
The ideal gas equation is given by:
\[ PV = nRT \]
From the relation \( P = \frac{nRT}{V} \), we can substitute \( P \) in the equation \( P \cdot T = K \):
\[ \frac{nRT}{V} \cdot T = K \]
This simplifies to:
\[ \frac{nRT^2}{V} = K \]

3. **Rearranging for Volume**:
Rearranging gives us:
\[ V = \frac{nRT^2}{K} \]

4. **Differentiating Volume with Respect to Temperature**:
To find the change in volume with respect to temperature, we differentiate:
\[ V = \frac{nR}{K} T^2 \]
Differentiating both sides with respect to \( T \):
\[ dV = \frac{nR}{K} \cdot 2T \, dT \]
Thus,
\[ dV = \frac{2nR}{K} T \, dT \]

5. **Finding Work Done**:
The work done \( W \) by the gas is given by the integral of pressure with respect to volume:
\[ W = \int P \, dV \]
Substituting \( P \) from the relation \( P = \frac{K}{T} \):
\[ W = \int \frac{K}{T} \, dV \]
Now substituting \( dV \):
\[ W = \int \frac{K}{T} \cdot \frac{2nR}{K} T \, dT \]
Simplifying gives:
\[ W = \int 2nR \, dT \]

6. **Evaluating the Integral**:
The limits of integration are from \( T \) to \( 2T \):
\[ W = 2nR \int_{T}^{2T} dT = 2nR [T]_{T}^{2T} = 2nR (2T - T) = 2nR T \]

7. **Substituting the Number of Moles**:
Since we have 2 moles of gas:
\[ W = 2 \cdot 2RT = 4RT \]

### Final Answer:
The work done by the system is:
\[ W = 4RT \]