Pressure `P`, Volume `V` and temperature `T` of a certain material are related by the `P=(alphaT^(2))/(V)`. Here `alpha` is constant. Work done by the material when temparature changes from `T_(0)` to `2T_(0)` while pressure remains constant is :

Options

A. `6alphaT_(0)^(3)`
B. `(3)/(2)alphaT_(0)^(2)`
C. `2alphaT_(0)^(2)`
D. `3alphaT_(0)^(2)`

Answer

To solve the problem, we need to find the work done by the material when the temperature changes from \( T_0 \) to \( 2T_0 \) while the pressure remains constant. The relationship between pressure \( P \), volume \( V \), and temperature \( T \) is given by:

\[ P = \frac{\alpha T^2}{V} \]

### Step 1: Rearranging the Equation
From the given equation, we can rearrange it to express \( PV \):

\[ PV = \alpha T^2 \]

### Step 2: Differentiate the Equation
We differentiate the equation \( PV = \alpha T^2 \):

\[ P dV + V dP = 2\alpha T dT \]

Let this be Equation (1).

### Step 3: Considering Constant Pressure
Since the pressure \( P \) remains constant during the process, we have \( dP = 0 \). Thus, Equation (1) simplifies to:

\[ P dV = 2\alpha T dT \]

### Step 4: Express Work Done
The work done \( W \) by the material is given by:

\[ W = P dV \]

From the previous step, we can substitute \( P dV \):

\[ W = 2\alpha T dT \]

### Step 5: Integrate Both Sides
Now, we integrate both sides. The limits for \( T \) will change from \( T_0 \) to \( 2T_0 \):

\[ W = 2\alpha \int_{T_0}^{2T_0} T dT \]

### Step 6: Solve the Integral
The integral of \( T \) is:

\[ \int T dT = \frac{T^2}{2} \]

Thus, we have:

\[ W = 2\alpha \left[ \frac{T^2}{2} \right]_{T_0}^{2T_0} \]

### Step 7: Evaluate the Limits
Now we evaluate the limits:

\[ W = 2\alpha \left( \frac{(2T_0)^2}{2} - \frac{(T_0)^2}{2} \right) \]

Calculating this gives:

\[ W = 2\alpha \left( \frac{4T_0^2}{2} - \frac{T_0^2}{2} \right) = 2\alpha \left( 2T_0^2 - \frac{T_0^2}{2} \right) \]

\[ W = 2\alpha \left( \frac{4T_0^2 - T_0^2}{2} \right) = 2\alpha \left( \frac{3T_0^2}{2} \right) \]

\[ W = 3\alpha T_0^2 \]

### Conclusion
Thus, the work done by the material when the temperature changes from \( T_0 \) to \( 2T_0 \) while the pressure remains constant is:

\[ W = 3\alpha T_0^2 \]