If the balloon had a volume of 2 L at a depth of 40 m, what was the original volume of the balloon if we assume the pressure at the surface of the water is 14.7 psi?

The pressure at a depth of 40 m is the hydrostatic pressure of the water plus the atmospheric pressure.

The hydrostatic pressure #P# of a liquid is given by the formula

#color(blue)(bar(ul(|color(white)(a/a)P = ρghcolor(white)(a/a)|)))" "#

where

#ρ# = the density of the liquid
#g# = the acceleration due to gravity
#h# = the depth of the liquid

#P_"water" = "1000 kg"·stackrelcolor(blue)("m"^"-1")(color(red)(cancel(color(black)("m"^"-3"))))× 9.81 color(red)(cancel(color(black)("m")))·"s"^"-2" × 40 color(red)(cancel(color(black)("m"))) = 3.92 × 10^5color(white)(l) "kg·m"^"-1""s"^"-2" = 3.92 × 10^5 color(white)(l)"Pa"#

#P_"water" = 3.92 × 10^5 color(red)(cancel(color(black)("Pa"))) × "1 atm"/(101.325 × 10^3 color(red)(cancel(color(black)("Pa")))) = "3.87 atm"#

#P_"atm" = 14.7 "psi" × "1 atm"/(14.70 "psi") = "1.00 atm"#

∴ The total pressure at 40 m is

#P_"tot" = "3.87 atm + 1.00 atm" = "4.87 atm"#

Now we can use Boyle's Law to calculate the volume of the balloon at the surface.

#color(blue)(bar(ul(|color(white)(a/a) P_1V_1 = P_2V_2color(white)(a/a)|)))" "#

#V_2 = V_1 × P_1/P_2#

In your problem,

#P_1 = "4.87 atm"; V_1 = "2 L"#
#P_2 = "1.00 atm;"color(white)(l) V_2 = "?"#

∴ #V_2 = "2 L" × (4.87 color(red)(cancel(color(black)("atm"))))/(1.00 color(red)(cancel(color(black)("atm")))) = "10 L"# (1 significant figure)