Initially area of soap bubble A1=4πr2 Under isothermal condition radius becomes 2r, Then, ‌‌ area A2=4π(2r)2 ‌=4π⋅4r2 ‌=16πr2 Increase in surface area ∆A‌=2(A2−A1) ‌=2(16πr2−4πr2)=24πr2 Energy spent W‌=T×∆A ‌=T⋅24πr2 or W=24πTr2J