Question

Pressure $P$ , volume $V$ and temperature $T$ of a certain material are related by $P = \frac{\alpha T^{2}}{V} ,$ where $\alpha $ is constant. Work done by the material when temperature changes from $T_{0}$ to $2T_{0}$ and pressure remains constant is

• A. $3\alpha T_{0}^{2}$
• B. $5\alpha T_{0}^{2}$
• C. $\frac{3}{2}\alpha T_{0}^{2}$
• D. $7\alpha T_{0}^{2}$

Answer 1

Answer: (A)

The given relation is $p=\frac{\alpha T^{2}}{V} \, \, \, \therefore \, \, V=\frac{\alpha T^{2}}{P}$
As pressure is kept constant, $dV=\left(\frac{2 \alpha T}{P}\right)dT$
$W=\displaystyle \int p \, d V = \displaystyle \int _{T_{0}}^{2 T_{0}} p \left(\frac{2 \alpha T}{p}\right) d T$
$=2\alpha \left[\frac{T^{2}}{2}\right]_{T_{0}}^{2 T_{0}}=3\alpha T_{0}^{2}$