Question

A quantity of a substance in a closed system is made to undergo a reversible process from an initial volume of $ 3 \,m^3 $ and initial pressure $ 10^5\, N/m^2 $ to a final volume of $ 5\, m^3 $ . If the pressure is proportional to the square of the volume (i.e. $ p = AV^2 $ ), the work done by the substance will be

• A. $ 3.6 \times 10^2\,J $
• B. $ 7.4 \times 10^3\,J $
• C. $ 2.2 \times 10^4\, J $
• D. $ 3.6 \times 10^5\,J $

Answer 1

Answer: (D)

Given, $p =AV^2$
For a closed system, $\Delta Q = 0$
$\therefore \Delta W = \Delta U$
Also, work done by the substance
$W = \int\limits_{v_i}^{v_{t}} p \cdot dV$
$ = \int\limits_{v_i}^{v_{t}} AV^{2} dV $
$ = A \left(\frac{V^{3}}{3}\right)_{v_i}^{v_{t}} \,\,\, \left(\because A = \frac{p_{i}}{V_{i}^{2}}\right)$
where, $A$ is constant.
$= \frac{A}{3} \left[V_{f}^{3} -V_{i}^{3}\right] $
$ = \frac{p_{i}}{3V_{i}} \left(V_{f}^{3} - V_{i}^{3}\right) $
$= \frac{10^{5}}{3\times 9} \left(125 - 27\right) $
$= \frac{10^{5}}{3\times 9} \times 98 $
$ = 3.6 \times 10^{5}\,J$