Divers know that the pressure exerted buy the water increases about 100kpa with every 10.2 m of depth. This means that at 10.2 m below the surface, the pressure is 201 kpa; at 20.4 m , the pressure is 301 Kpa; and so forth. Given that the volume of a balloon is 3.5L at STP and that the temperature of the water remains the same, what is the volume 51 m below the water's surface?
So the new pressure is
101 + (100 x 51/10.2) = ?
Then P1V1 = P2V2
To determine the volume of the balloon 51 m below the water's surface, we need to understand the relationship between pressure and volume using Boyle's Law. Boyle's Law states that at a constant temperature, the pressure and volume of a gas are inversely proportional.
Let's begin by calculating the change in pressure from the surface to 51 m below the water's surface. We know that the pressure increases by 100 kPa for every 10.2 m of depth. Thus, the pressure change can be calculated as follows:
Pressure change = (number of 10.2 m increments) × (pressure increase per increment)
Pressure change = (51 m / 10.2 m) × (100 kPa)
Pressure change = 5 × 100 kPa
Pressure change = 500 kPa
Since we are given that the initial pressure of the balloon is at standard temperature and pressure (STP), which is approximately 101.3 kPa, we can find the new pressure 51 m below the water's surface:
New pressure = Initial pressure + Pressure change
New pressure = 101.3 kPa + 500 kPa
New pressure = 601.3 kPa
Now that we have the new pressure, we can use the relationship between pressure and volume from Boyle's Law to determine the final volume of the balloon. Boyle's Law states that the pressure multiplied by the volume of a gas is constant, as long as the temperature remains constant.
Initial pressure × Initial volume = New pressure × New volume
We are given that the initial volume of the balloon at STP is 3.5 L. Substituting the values into the equation:
(101.3 kPa) × (3.5 L) = (601.3 kPa) × (New volume)
Solving for the new volume:
New volume = (101.3 kPa × 3.5 L) / 601.3 kPa
New volume = 589.55 L / 601.3 kPa
New volume ≈ 0.978 L
Therefore, the volume of the balloon 51 m below the water's surface would be approximately 0.978 L.