Question

A quantity of a substance in a closed system is made to undergo a reversible process from an initial volume of $3m^3$ and initial pressure $10^{5}N/m^2$ to a final volume of $5m^3$ . If the pressure is proportional to the square of the volume (i.e. $p=A{V}^2$ ), 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, $p=A{V}^2$
For a closed system, $\Delta Q=0$
$\therefore \Delta W=\Delta U$
Also, work done by the substance
$W=\underset{v_i}{\overset{v_t}{\int }}p\cdot dV$
$=\underset{v_i}{\overset{v_t}{\int }}A{V}^{2}dV$
$=A{\left(\frac{V^3}{3}\right)}_{v_i}^{v_t}\left(\because A=\frac{p_i}{V_i^2}\right)$
where, $A$ is constant.
$=\frac{A}{3}\left[V_f^3-V_i^3\right]$
$=\frac{p_i}{3V_i}\left(V_f^3-V_i^3\right)$
$=\frac{10^5}{3\times 9}\left(125-27\right)$
$=\frac{10^5}{3\times 9}\times 98$
$=3.6\times 10^{5}J$