Pressure p, volume V and temperature T for a certain gas are related by
`p=(alphaT-betaT^2)/(V)`
where, `alpha` and `beta` are constants. Find the work done by the gas if the temperature changes from `T_1` to `T_2` while the pressure remains the constant.

Explanation

To solve the problem, we need to find the work done by the gas when the temperature changes from \( T_1 \) to \( T_2 \) while keeping the pressure constant. The relationship between pressure \( p \), volume \( V \), and temperature \( T \) is given by:

\[ p = \frac{\alpha T - \beta T^2}{V} \]

### Step 1: Rearranging the Equation
We can rearrange the equation to express volume \( V \) in terms of pressure \( p \) and temperature \( T \):

\[ V = \frac{\alpha T - \beta T^2}{p} \]

### Step 2: Differentiating Volume with Respect to Temperature
Next, we differentiate \( V \) with respect to \( T \):

\[ dV = \frac{d}{dT} \left( \frac{\alpha T - \beta T^2}{p} \right) \]

Using the quotient rule, we find:

\[ dV = \frac{1}{p} \left( \alpha - 2\beta T \right) dT \]

### Step 3: Work Done by the Gas
The work done \( W \) by the gas during this process, when pressure is constant, is given by the integral of \( p \) with respect to \( V \):

\[ W = \int p \, dV \]

Substituting the expression for \( dV \):

\[ W = \int p \left( \frac{\alpha - 2\beta T}{p} \right) dT \]

The \( p \) cancels out:

\[ W = \int (\alpha - 2\beta T) \, dT \]

### Step 4: Integrating
Now we integrate from \( T_1 \) to \( T_2 \):

\[ W = \left[ \alpha T - \beta T^2 \right]_{T_1}^{T_2} \]

### Step 5: Evaluating the Integral
Evaluating the integral at the limits \( T_1 \) and \( T_2 \):

\[ W = \left( \alpha T_2 - \beta T_2^2 \right) - \left( \alpha T_1 - \beta T_1^2 \right) \]

This simplifies to:

\[ W = \alpha (T_2 - T_1) - \beta (T_2^2 - T_1^2) \]

### Final Answer
Thus, the work done by the gas is:

\[ W = \alpha (T_2 - T_1) - \beta (T_2^2 - T_1^2) \]

---