# Question #89091

##### 1 Answer

#### Answer:

The answer is **(c)** descending from a depth of

#### 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 #-># 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

#color(blue)(ul(color(black)(P_1V_1 = P_2V_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 = (4.0 color(red)(cancel(color(black)("L"))))/(1.0color(red)(cancel(color(black)("L")))) = color(blue)(4)#

the pressure must **increase** by a factor of

#P_2/P_1 = color(blue)(4)#

So now all you have to do is to look at which descent brings the pressure down by a factor of

You know that you have--these are all *depths*

#"0 m " -> " 1 atm"# #"10 m " -> " 2 atm"# #"20 m " -> " 3 atm"# #"30 m " -> " 4 atm"# #"40 m " -> " 5 atm"# #"50 m " -> " 6 atm"# #"60 m " -> " 7 atm"#

As you can see, the only descent that works here

#"0 m " -> " 30 m"#

because you have

#P_2/P_1 = (4 color(red)(cancel(color(black)("atm"))))/(1color(red)(cancel(color(black)("atm")))) = color(blue)(4)#