Question #89091
1 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
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 depthP_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)