Pressure P, volume V and temperature T for a certain gas are related by `P=(AT-BT^(2))/(V)`, where A and B are constatns .The work done by the gas as its temperature change from `T_(1)` to `T_(2)` while pressure remaining constatn is

Options

A. `A-(B)/(2)(R_(2)-T_(1))`
B. `A(T_(2)-T_(1))-B(T_(2)^(2)-T_(1)^(2))`
C. `(A)/(2)(T_(2)^(2)-T_(1)^(2))-(B)/(3)(T_(2)^(3)-T_(1)^(3))`
D. `A(T_(2)-T_(1))-(B)/(3)(T_(2)-T_(1))^(3)`

Explanation

To find the work done by the gas as its temperature changes from \( T_1 \) to \( T_2 \) while the pressure remains constant, we will follow these steps:

### Step 1: Understand the relationship between P, V, and T
Given the equation:
\[ P = \frac{AT - BT^2}{V} \]
we can rearrange this to express \( PV \):
\[ PV = AT - BT^2 \]

### Step 2: Set up the work done equation
The work done \( dW \) by the gas during an infinitesimal change in volume \( dV \) at constant pressure \( P \) is given by:
\[ dW = P \, dV \]

### Step 3: Differentiate the equation with respect to T
To find \( dV \), we differentiate the equation \( PV = AT - BT^2 \) with respect to \( T \):
\[ \frac{d(PV)}{dT} = \frac{d(AT - BT^2)}{dT} \]
Since \( P \) is constant, we have:
\[ P \frac{dV}{dT} = A - 2BT \]
Thus, we can express \( dV \) in terms of \( dT \):
\[ dV = \frac{A - 2BT}{P} \, dT \]

### Step 4: Substitute \( dV \) into the work equation
Substituting \( dV \) into the work done equation:
\[ dW = P \left(\frac{A - 2BT}{P} \, dT\right) = (A - 2BT) \, dT \]

### Step 5: Integrate to find the total work done
Now, we integrate \( dW \) from \( T_1 \) to \( T_2 \):
\[ W = \int_{T_1}^{T_2} (A - 2BT) \, dT \]
Calculating the integral:
\[ W = \left[ AT - BT^2 \right]_{T_1}^{T_2} \]
This gives:
\[ W = \left( AT_2 - BT_2^2 \right) - \left( AT_1 - BT_1^2 \right) \]
Simplifying this, we get:
\[ W = A(T_2 - T_1) - B(T_2^2 - T_1^2) \]

### Final Result
Thus, the work done by the gas as its temperature changes from \( T_1 \) to \( T_2 \) while pressure remains constant is:
\[ W = A(T_2 - T_1) - B(T_2^2 - T_1^2) \]

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