Hi,
just need a little help with this physics question please.
A cylindrical tank of water is located 68m deep in water.
An air lock at the bottome of the tank allows access. The sailor must wait while the air pressure in the lock is increased until it equals the water pressure in the tank. The bottom half of the air lock is filled with water and the door to the tank then can be opened.
A sailor inflates a balloon to a volume of 6.0L with high pressure air in the air lock. He takes the ballon with him as he ascends to the surface. what is the approximate volume of the balloon when it reaches the surface, assuming standard temperature and pressure?
So, what I've worked out so far is that maybe I could use Boyle's Law
P1V1=P2V2
using V1=6Land P2=1atm=10^5Pa
I thought the value of P1, the pressure in the airlock could be obtained by using the formula P1=r * g * h=1000*9.8*68=666400Pa*6
so if I solve for V2=666400*6/10^5=39.98L.
But the answer is 45.5L. Where did I go wrong??
Please help!
sorry as in P1=666400 Pa not 666400pa*6
To solve this problem, you correctly identified that Boyle's Law can be used. However, it seems there is a mistake in the calculation of P1, the pressure in the airlock.
The formula to calculate the pressure at a certain depth in a fluid is P = ρ * g * h, where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth. In this case, the fluid is water, so the density ρ of water is 1000 kg/m^3.
Therefore, the correct calculation for P1 would be:
P1 = ρ * g * h = 1000 kg/m^3 * 9.8 m/s^2 * 68 m = 666,400 Pa
Now, using Boyle's Law, we have:
P1 * V1 = P2 * V2
Substituting the known values:
666,400 Pa * 6 L = 1 atm * V2
To convert 1 atm to Pascals, we multiply by 10^5:
666,400 Pa * 6 L = 10^5 Pa * V2
Solving for V2:
V2 = (666,400 * 6) / 10^5 ≈ 39.98 L
It seems you made a calculation error in your previous calculation, resulting in a slightly lower value for V2. However, the correct answer should be approximately 39.98 L, not exactly 45.5 L as you mentioned.
So, the approximate volume of the balloon when it reaches the surface, assuming standard temperature and pressure, is indeed 39.98 L.