For a mercury drop of radius $3.00 ext{ mm}$ (surface tension $4.65 imes 10^{-1} ext{ N/m}$), wh

Question

For a mercury drop of radius $3.00	ext{ mm}$ (surface tension $4.65 	imes 10^{-1}	ext{ N/m}$), what is the total absolute pressure inside the drop? (Assume standard atmospheric pressure is $1.01 	imes 10^5	ext{ Pa}$)

Professor PC Thomas Classes · Answer

Total absolute pressure $P = P_{atm} + \Delta P$. The excess pressure $\Delta P = 310	ext{ Pa} = 0.0031 	imes 10^5	ext{ Pa}$. Thus, $P = 1.01 	imes 10^5 + 0.0031 	imes 10^5 = 1.0131 	imes 10^5	ext{ Pa}$.