Question #89091

Dec 20, 2017

The answer is (c) descending from a depth of $\text{0 m}$ to $\text{30 m}$ below sea level.

Explanation:

The idea here is that pressure and volume have an ((inverse relationship when temperature and the number of moles of gas are being kept constant $\to$ think Boyle's Law** here.

So right from the start, the fact that the pressure increases when descending below sea level tells you that in order for the volume of the balloon to decrease, it must descend.

Ascending will cause the pressure to decrease, which, in turn, will cause the volume of the balloon to increase, so options (a) and (b) are not valid solutions here.

Now, mathematically, you can express this inverse relationship as

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{P}_{1} {V}_{1} = {P}_{2} {V}_{2}}}}$

Here

• ${P}_{1}$, ${V}_{1}$ are the pressure and the volume of the gas inside the balloon at an initial depth
• ${P}_{2}$, ${V}_{2}$ are the pressure and the volume of the gas inside the balloon at a final depth

So, in order for the volume of the balloon to decrease by a factor of

${V}_{1} / {V}_{2} = \left(4.0 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{L"))))/(1.0color(red)(cancel(color(black)("L}}}}\right) = \textcolor{b l u e}{4}$

the pressure must increase by a factor of $\textcolor{b l u e}{4}$.

${P}_{2} / {P}_{1} = \textcolor{b l u e}{4}$

So now all you have to do is to look at which descent brings the pressure down by a factor of $\textcolor{b l u e}{4}$.

You know that you have--these are all depths

• $\text{0 m " -> " 1 atm}$
• $\text{10 m " -> " 2 atm}$
• $\text{20 m " -> " 3 atm}$
• $\text{30 m " -> " 4 atm}$
• $\text{40 m " -> " 5 atm}$
• $\text{50 m " -> " 6 atm}$
• $\text{60 m " -> " 7 atm}$

As you can see, the only descent that works here

$\text{0 m " -> " 30 m}$

because you have

${P}_{2} / {P}_{1} = \left(4 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{atm"))))/(1color(red)(cancel(color(black)("atm}}}}\right) = \textcolor{b l u e}{4}$