$(a)\;4 \normalsize PV+3 \normalsize ST=0\qquad(b)\;3PV+4ST=0\qquad(c)\;2PV+3ST=0\qquad(d)\;3PV+2ST=0$

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Surface tension $$\displaystyle T=P\frac{(R^{3} - 2r^{3})}{(4(2r^{2} - R^{2})}$$

Or, $$\displaystyle T= \frac{3P\delta V}{-4\delta A}$$

where V & VA are the change in volume and surface area. Therefore, .

$$3P\delta V+4T\delta A=0$$

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