Question
A spherical soap bubble inside an air chamber at pressure $P_0=10^5 Pa$ has a certain radius so that the excess pressure inside the bubble is $\Delta P=144 Pa$. Now, the chamber pressure is reduced to $8 P_0 / 27$ so that the bubble radius and its excess pressure change. In this process, all the temperatures remain unchanged. Assume air to be an ideal gas and the excess pressure $\Delta P$ in both the cases to be much smaller than the chamber pressure. The new excess pressure $\Delta P$ in $Pa$ is

Answer

(image)

$P - P _0=\Delta P =\frac{4 T }{ R }$

$P =\left( P _0+\frac{4 T }{ R }\right)$

Case-$2$

(image)

$P _1-\frac{8 P _0}{27}=\Delta P _1=\frac{4 T }{ R _1}$

$P _1=\frac{4 T }{ R _1}+\frac{8 P _0}{27}$

Constant temperature process

$PV = P _1 V_1$

$\left( P _0+\frac{4 T }{ R }\right) \frac{4}{3} \pi R ^3=\left(\frac{4 T }{ R _1}+\frac{8 P _0}{27}\right) \frac{4}{3} \pi R _1^3 ;\left(\frac{4 T }{ R }\right),\left(\frac{4 T }{ R _1}\right) \rightarrow \text { (Neglected) }$

$R =\frac{2}{3} R _1 \Rightarrow R _1=\frac{3}{2} R$

$\Delta P _1=\frac{4 T }{ R _1}=\frac{4 T }{3 R } \times 2=\frac{2}{3} \times(144)=96 Pa$

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