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?

Answer

To find the work done by the system when two moles of a diatomic ideal gas are taken through a process where \( P \cdot T = \text{constant} \) and the temperature is increased from \( T \) to \( 2T \), we can follow these steps:

### Step 1: Understand the process
We are given that \( P \cdot T = k \) (a constant). This means that the product of pressure \( P \) and temperature \( T \) remains constant throughout the process.

### Step 2: Relate pressure, volume, and temperature
Using the ideal gas law, we know:
\[ PV = nRT \]
From this, we can express pressure \( P \) as:
\[ P = \frac{nRT}{V} \]

### Step 3: Substitute for \( P \)
Since \( P \cdot T = k \), we can write:
\[ P = \frac{k}{T} \]

### Step 4: Differentiate the equation
We know that:
\[ P = \frac{k}{T} \implies V = \frac{nRT}{P} \]
Substituting for \( P \):
\[ V = \frac{nRT^2}{k} \]
Differentiating this equation with respect to \( T \):
\[ dV = \frac{d(nRT^2/k)}{dT} = \frac{2nRT}{k} dT \]

### Step 5: Set up the work integral
The work done \( W \) by the gas during the process is given by:
\[ W = \int P \, dV \]
Substituting \( P \) and \( dV \):
\[ W = \int \frac{k}{T} \cdot \left(\frac{2nRT}{k} dT\right) \]
This simplifies to:
\[ W = 2nR \int dT \]

### Step 6: Evaluate the integral
The limits of integration are from \( T \) to \( 2T \):
\[ W = 2nR \left[ T \right]_{T}^{2T} = 2nR(2T - T) = 2nRT \]

### Step 7: Substitute the number of moles
Given \( n = 2 \) (two moles of gas):
\[ W = 2 \cdot 2RT = 4RT \]

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

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