Solve Pressure Problem for Scuba Diver: Find Duration Underwater

[myogi]

Ok, I got this scuba diver and he has a tank that is 0.015 meters cubed and filled with compressed air at an absolute pressure of 2.02e7 Pa. Assuming air is consumed at a rate of 0.03 meters cubed per minute and the temperature is constant, how long can this diver stay under water at a depth of 30.0 meters? Assuming the density of the water is 1025 kilograms per meter cubed

I found the pressure at 30 ft. below, but I didn't really know what to do with it. Help??

 

[Andrew Mason]

Ok, I got this scuba diver and he has a tank that is 0.015 meters cubed and filled with compressed air at an absolute pressure of 2.02e7 Pa. Assuming air is consumed at a rate of 0.03 meters cubed per minute and the temperature is constant, how long can this diver stay under water at a depth of 30.0 meters? Assuming the density of the water is 1025 kilograms per meter cubed.
I found the pressure at 30 ft. below, but I didn't really know what to do with it.

You will need to find the pressure at a depth of 30 metres.

I assume that the .03 m^3/min. is the volume rate of air at a depth of 30 m. So work out how much volume the tank contains when let out with the ambient pressure at 30 m. ($V \propto 1/P$). With that total volume, work out the consumption time from the consumption rate.

Keep in mind, if he stays that that long at 30 m. he will die. He needs some air to get down to 30m and some air to get back. He has to come up gradually.

AM

 

[wais]

To solve this problem, we can use the ideal gas law which states that pressure, volume, and temperature are related by the equation PV=nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature.

First, we need to convert the tank volume from 0.015 meters cubed to liters by multiplying it by 1000. This gives us a volume of 15 liters.

Next, we need to find the number of moles of gas in the tank. We can do this by using the ideal gas law and solving for n. Since we know the pressure, volume, and temperature are constant, we can set up the following equation:

(2.02e7 Pa)(15 L) = n(8.31 J/mol*K)(298 K)

Solving for n gives us a value of approximately 98.4 moles of gas in the tank.

Now, we can use this information to find the amount of time the diver can stay underwater. We know that the diver is consuming 0.03 meters cubed of air per minute, which is equal to 30 liters.

To find the duration, we can use the formula:

Duration = (Number of moles of gas in tank) / (Rate of consumption of gas)

Plugging in our values, we get:

Duration = (98.4 moles) / (30 L/min) = 3.28 minutes

Therefore, the diver can stay underwater for approximately 3.28 minutes at a depth of 30.0 meters using the given tank and assuming a constant temperature.

However, we also need to take into account the density of the water. This means that the pressure at a depth of 30.0 meters will not be the same as the pressure at the surface. Using the equation P = ρgh, where P is pressure, ρ is density, g is the acceleration due to gravity, and h is the depth, we can find the pressure at a depth of 30.0 meters.

Plugging in the values, we get:

P = (1025 kg/m^3)(9.8 m/s^2)(30.0 m) = 299400 Pa

This is the actual pressure at a depth of 30.0 meters, which is significantly lower than the absolute