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The pressure $p$, volume $V$ and temperature $T$ for a certain gas are related by $p=\frac{A T-B T^{2}}{V}$ where $A$ and $B$ are constants. The work done by the gas when the temperature changes from $T_{1}$ to $T_{2}$ while the pressure remains constant, is given by
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The correct answer is:
$ A\left(\tau_{2}-T_{1}\right)-\frac{B}{2}\left(T_{2}^{2}-T_{1}^{2}\right)$
Given $P=\frac{A T-B T^{2}}{V}$
$\begin{array}{lc}\Rightarrow & P V=A T-B T^{2} \\ \Rightarrow & P \Delta V=A \Delta T-B T \Delta T\end{array}$
On integrating, we get Work $=\int P d V=A \int_{T_{1}}^{T_{2}} d T-B \int_{T_{1}}^{T_{2}} T d T$
$=A\left(T_{2}-T_{1}\right)-\frac{B}{2}\left[\left(T_{2}\right)^{2}-\left(T_{1}\right)^{2}\right]$
$\begin{array}{lc}\Rightarrow & P V=A T-B T^{2} \\ \Rightarrow & P \Delta V=A \Delta T-B T \Delta T\end{array}$
On integrating, we get Work $=\int P d V=A \int_{T_{1}}^{T_{2}} d T-B \int_{T_{1}}^{T_{2}} T d T$
$=A\left(T_{2}-T_{1}\right)-\frac{B}{2}\left[\left(T_{2}\right)^{2}-\left(T_{1}\right)^{2}\right]$
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