Pressure $P$, volume $V$ and temperature $T$ for a certain gas are related by $P=$ $\dfrac{\alpha \mathrm{T}-\beta \mathrm{T}^{2}}{\mathrm{V}}$ where $\alpha$ and $\beta$ constants. Find the work done by the gas if the temperature changes from $\mathrm{T}_{1}$ to $\mathrm{T}_{2}$ while the pressure remains constant.

Hint:
The state of a gas is determined by the values, such as the pressure, temperature, and volume occupied by the gas, of certain measurable properties. It's possible to change the values of these variables and the state of the gas. Determine the amount of work and the amount of heat required to alter the gas state. Calculate Volume, differentiate it with respect to time and then take out the work done by the gas.

Formula used: $\quad \mathrm{W}=\int \mathrm{P} \mathrm{dV}$

Complete answer:

For a gas, during a change of volume, work is the product of the pressure p and volume V. The work is the area under the curve on a graph of pressure versus volume that describes how the state is altered from state 1 to state 2.
The given relation is $\quad \mathrm{P}=\dfrac{\alpha \mathrm{T}-\beta \mathrm{T}^{2}}{\mathrm{V}}$
$\text{V}=\dfrac{\alpha \text{T}-\beta {{\text{T}}^{2}}}{\text{P}}$
Differentiating we get,
$\text{dV}=\dfrac{\alpha -2\beta \text{T}}{\text{P}}\text{dT}$
Work done by the gas $\text{W}=\int{\text{P}}\text{dV}$
$\text{W}=\int_{{{\text{T}}_{1}}}^{{{\text{T}}_{2}}}{(\alpha -2\beta \text{T})}\text{dT}$
$\Rightarrow \text{W}=\left. \alpha \text{T} \right|_{{{\text{T}}_{1}}}^{{{\text{T}}_{2}}}-2\beta \times \left. \dfrac{{{\text{T}}^{2}}}{2} \right|_{{{\text{T}}_{1}}}^{{{\text{T}}_{2}}}$
$=\alpha \left( {{\text{T}}_{2}}-{{\text{T}}_{1}} \right)-\beta \left( \text{T}_{2}^{2}-\text{T}_{1}^{2} \right)$
\[\therefore \] The work done by the gas if the temperature change from $\mathrm{T}_{1}$ to $\mathrm{T}_{2}$ while the pressure remains the constant is $\alpha \left( {{\text{T}}_{2}}-{{\text{T}}_{1}} \right)-\beta \left( \text{T}_{2}^{2}-\text{T}_{1}^{2} \right)$

Note:
We focus on the effects that the system (e.g. an engine) has on its surroundings when defining work. Thus, when the system operates in the environment (energy leaves the system), we define work as positive. If work (energy added to the system) is done on the system, the work is negative. F the pressure has Pascals units and the volume units of m3 will have Joules units in the work. Work done by a system has a negative value (as the system expands). Work done on a system (as the system shrinks) has a positive value.