Question #46e82

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Question #46e82

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Answer 1 · 2015-12-17T17:28:13

Here's what I got.

Explanation:

As it is written, your problem doesn't provide enough information to allow for a solution to be formulated exclusively in terms of the original volume of the balloon.

More specifically, you would need to know the density of seawater at that depth and temperature.

With this being said, I will assume that the density of seawater at that depth is about $1.025 g/mL$ $"1.025 g/mL"$ .

Now, the idea here is that the number of moles of gas and its temperature will remain constant as the balloon makes its way to $130 m$ $"130 m"$ . Although the temperature of the water decreases with depth, $130 m$ $"130 m"$ is still a relatively low depth, so you can safely assume that it remains constant.

Even if that were not the case, the insufficient data would force you to assume temperature constant anyway.

Now, pressure and volume have an inverse relationship when number of moles and temperature are kept constant - this is known as Boyle's Law.

Mathematically, this is written like this

$P_{1} V_{1} = P_{2} V_{2}$ $color(blue)(P_1V_1 = P_2V_2)" "$ , where

$P_{1}$ $P_1$ , $V_{1}$ $V_1$ - the pressure and volume of the gas at an initial state
$P_{2}$ $P_2$ , $V_{2}$ $V_2$ - the pressure and volume of the gas at a final state

At the surface of the Earth, atmospheric pressure is equal to $1 atm$ $"1 atm"$ , or $101,325 Pa$ $"101,325 Pa"$ .

The pressure exerted by a fluid at a depth $h$ $h$ is given by the equation

$P_{h} = ρ \cdot g \cdot h$ $color(blue)(P_h = rho * g * h)" "$ , where

$ρ$ $rho$ - the density of the liquid at that depth
$g$ $g$ - the gravitational acceleration, equal to ${9.81 m s}^{- 2}$ $"9.81 m s"^(-2)$
$h$ $h$ - the given depth

Before plugging in your values, convert the density from grams per milliliter to kilograms per cubic meter

$1.025 color(red)(cancel(color(black)("g")))/(color(red)(cancel(color(black)("mL")))) * "1 kg"/(1000color(red)(cancel(color(black)("g")))) * (10^6color(red)(cancel(color(black)("mL"))))/"1 m"^3 = "1025 kg/m"^3$

The pressure exerted by the seawater at $"130 m"$ is equal to

$P_"water" = 1025 "kg"/"m"^3 * 9.81"m"/"s"^2 * "130 m"$

$P_"water" = "1,307,182.5" overbrace("kg"/("m s"^(2)))^(color(blue) (="Pa"))$

$P_"water" = "1,307,182.5 Pa"$

The total pressure exerted on the balloon will include the atmospheric pressure

$P_"total" = P_"atm" + P_"water"$

$P_"total" = "101,325 Pa" + "1,307,182.5 Pa" = "1,408,507.5 Pa"$

Rearrange the Boyle Law equation and solve for $V_2$ to get

$P_1V_1 = P_2V_2 implies V_2 = P_1/P_2 * V_2$

Here $P_2$ will be $P_"total"$ , $P_1$ is the atmospheric pressure at the surface, and $V_1$ is the initial volume of the balloon

$V_2 = ("101,325" color(red)(cancel(color(black)("Pa"))))/("1,408,507.5" color(red)(cancel(color(black)("Pa")))) * V_1$

$V_2 = 0.07194 * V_1$

Rounded to two sig figs, the answer will be

$V_2 = color(green)(0.072 * V_1)$

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Question #46e82

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