For a thermodynamic system the pressure, volume and temperature are related as new gas law given as
`P=(alphaT^2)/V`
Here `alpha` is a constant. Find the work done by the system in this process when pressure remains constant and its temperature changes from `T_0` to `2 T_0`.

Answer

To find the work done by the system in the given process where the pressure remains constant and the temperature changes from \( T_0 \) to \( 2T_0 \), we can follow these steps:

### Step 1: Understand the relationship between pressure, volume, and temperature
The gas law provided is:
\[ P = \frac{\alpha T^2}{V} \]
Here, \( P \) is the pressure, \( V \) is the volume, \( T \) is the temperature, and \( \alpha \) is a constant.

### Step 2: Identify the work done formula
The work done \( W \) by the system during a thermodynamic process is given by:
\[ W = \int P \, dV \]
Since the pressure \( P \) is constant in this case, we can express the work done as:
\[ W = P \Delta V = P (V_f - V_i) \]
where \( V_f \) is the final volume and \( V_i \) is the initial volume.

### Step 3: Relate pressure to temperature and volume
From the gas law, we can express volume in terms of pressure and temperature:
\[ V = \frac{\alpha T^2}{P} \]
Since pressure is constant, we can write:
\[ V_i = \frac{\alpha T_0^2}{P} \]
and
\[ V_f = \frac{\alpha (2T_0)^2}{P} = \frac{\alpha \cdot 4T_0^2}{P} \]

### Step 4: Calculate the change in volume
Now, calculate the change in volume:
\[ \Delta V = V_f - V_i = \frac{\alpha \cdot 4T_0^2}{P} - \frac{\alpha T_0^2}{P} \]
\[ \Delta V = \frac{\alpha (4T_0^2 - T_0^2)}{P} = \frac{\alpha \cdot 3T_0^2}{P} \]

### Step 5: Substitute into the work done formula
Substituting \( \Delta V \) back into the work done formula:
\[ W = P \Delta V = P \left(\frac{\alpha \cdot 3T_0^2}{P}\right) \]
The pressure \( P \) cancels out:
\[ W = 3\alpha T_0^2 \]

### Final Answer
Thus, the work done by the system when the temperature changes from \( T_0 \) to \( 2T_0 \) is:
\[ W = 3\alpha T_0^2 \]

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