# 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?

Oct 31, 2016

The volume at the surface is 10 L.

#### Explanation:

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

${P}_{\text{water" = "1000 kg"·stackrelcolor(blue)("m"^"-1")(color(red)(cancel(color(black)("m"^"-3"))))× 9.81 color(red)(cancel(color(black)("m")))·"s"^"-2" × 40 color(red)(cancel(color(black)("m"))) = 3.92 × 10^5color(white)(l) "kg·m"^"-1""s"^"-2" = 3.92 × 10^5 color(white)(l)"Pa}}$

${P}_{\text{water" = 3.92 × 10^5 color(red)(cancel(color(black)("Pa"))) × "1 atm"/(101.325 × 10^3 color(red)(cancel(color(black)("Pa")))) = "3.87 atm}}$

${P}_{\text{atm" = 14.7 "psi" × "1 atm"/(14.70 "psi") = "1.00 atm}}$

∴ The total pressure at 40 m is

${P}_{\text{tot" = "3.87 atm + 1.00 atm" = "4.87 atm}}$

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

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} {P}_{1} {V}_{1} = {P}_{2} {V}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

V_2 = V_1 × P_1/P_2

${P}_{1} = \text{4.87 atm"; V_1 = "2 L}$
${P}_{2} = \text{1.00 atm;"color(white)(l) V_2 = "?}$
${V}_{2} = \text{2 L" × (4.87 color(red)(cancel(color(black)("atm"))))/(1.00 color(red)(cancel(color(black)("atm")))) = "10 L}$ (1 significant figure)