A balloon containing air at and 1 bar initially is filled with air further, isothermally and reversi — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

Given $P \propto D \implies P = kD$. At state 1, $P_1 = 10^5\text{ Pa}$, $D_1 = 1\text{ m} \implies k = 10^5\text{ Pa/m}$. At state 2, $P_2 = 4\text{ bar} \implies D_2 = 4\text{ m}$. Volume $V = \frac{\pi}{6} D^3 \implies dV = \frac{\pi}{2} D^2 dD$. Work $W = -\int P dV = -\int_1^4 (kD)(\frac{\pi}{2} D^2) dD = -\frac{k\pi}{2} \int_1^4 D^3 dD = -\frac{k\pi}{2} [\frac{D^4}{4}]_1^4 = -\frac{10^5 \pi}{8} (256 - 1) = -\frac{255 \pi}{8} \times 10^5 \approx -100 \times 10^5\text{ J} = -1.0 \times 10^7\text{ J}$. Magnitude is $1 \times 10^7\text{ J}$, so the value is 1.