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)

$3.6\times {10}^5$ J

Given,

$V_1=3m^3$

$V_2=5m^3$

$P_1={10}^{5}N/m^2$

$P_1=A{V}_1^2$

$A=\frac{P_1}{V_1^2}=\frac{{10}^5}{3\times 3}=0.11\times {10}^5$

Work done, $W=\int PdV$

$W={\int }_3^{5}A{V}^{2}dV$

$W=A{\left[\frac{V^3}{3}\right]}_3^5$

$W=0.11\times {10}^5\times \frac{[5^3-3^3]}{3}$

$W=\frac{0.11\times {10}^5\times (125-27)}{3}$

$W=3.6\times {10}^{5}J$

The correct option is D.