A scuba diver has an air tank with a volume of 0.010 m^3. The air in the tank is initially at a pressure of 1.0x10^7 Pa. Assume that the diver breathes 0.400 L/s of air. Find how long the tank will last at a depth of each of the following.
(a) 1.0 m
min
(b) 10.0 m
Please someone help me on this I don't know how to do this. thank you.
min
To solve this problem, we need to use the formula for Boyle's Law, which states that the pressure and volume of a gas are inversely proportional at a constant temperature. The formula is:
P1V1 = P2V2
Where P1 is the initial pressure, V1 is the initial volume, P2 is the final pressure, and V2 is the final volume.
First, let's convert the given values to SI units.
The initial volume V1 is given as 0.010 m^3. No conversion is needed.
The initial pressure P1 is given as 1.0x10^7 Pa. No conversion is needed.
The diver breathes 0.400 L/s of air. To find the change in volume per second, we need to convert liters to cubic meters:
0.400 L/s = 0.400 x (1/1000) m^3/s = 0.0004 m^3/s
Now let's find the final volume V2 at each depth.
(a) Depth of 1.0 m:
According to Boyle's Law, the pressure increases by 1 atm for every 10 m of depth. Therefore, at a depth of 1.0 m, the pressure will increase by 1/10 atm.
The final pressure P2 at a depth of 1.0 m is:
P2 = P1 + ΔP
= 1.0 x 10^7 Pa + (1/10 x 1.013 x 10^5 Pa) [1 atm = 1.013 x 10^5 Pa]
= 1.0 x 10^7 Pa + 1.013 x 10^6 Pa
= 1.1013 x 10^7 Pa
Now we can find the final volume V2 at a depth of 1.0 m using Boyle's Law:
P1V1 = P2V2
(1.0 x 10^7 Pa)(0.010 m^3) = (1.1013 x 10^7 Pa)V2
V2 = (1.0 x 10^7 Pa)(0.010 m^3) / (1.1013 x 10^7 Pa)
= 0.00908 m^3
Now, let's calculate how long the tank will last at a depth of 1.0 m:
The change in volume per second is 0.0004 m^3/s. Therefore, the tank will last:
Time = V2 / (change in volume per second)
= 0.00908 m^3 / 0.0004 m^3/s
= 22.7 seconds
(b) Depth of 10.0 m:
Using the same steps as above, we can find the final volume V2 and calculate the time:
Final pressure P2 at a depth of 10.0 m is:
P2 = P1 + ΔP
= 1.0 x 10^7 Pa + (10/10 x 1.013 x 10^5 Pa) [1 atm = 1.013 x 10^5 Pa]
= 1.0 x 10^7 Pa + 1.013 x 10^6 Pa
= 1.1013 x 10^7 Pa
Using Boyle's Law:
P1V1 = P2V2
(1.0 x 10^7 Pa)(0.010 m^3) = (1.1013 x 10^7 Pa)V2
V2 = (1.0 x 10^7 Pa)(0.010 m^3) / (1.1013 x 10^7 Pa)
= 0.00908 m^3
The tank will last:
Time = V2 / (change in volume per second)
= 0.00908 m^3 / 0.0004 m^3/s
= 22.7 seconds
Therefore, regardless of the depth, the tank will last approximately 22.7 seconds.