A scuba diver has an air tank with a volume of 0.020 m^3. the air in the tank is

Question

A scuba diver has an air tank with a volume of 0.020 m^3. the air in the tank is initially at a pressure of 1.0 times 10^7 pa. Assume that the diver breathes 0.500 L/s of air. Find how long the tank will last at a depth of each of the following. Incandescent lightbulbs are filled with an inert gas to lengthen the filament life. With the current 9at T = 20.0 degree C) the gas inside a lightbulb has a pressure of 117 kPa. When the bulb is burning temperature rises to 78.0 degree C. what is the pressure at the higher temperature?

Explanation / Answer

Hi,

In this case I assume you are looking for the answer of the question 2 as this is the one that can be seen clearly.

A way to solve this problem involves making the following suppositions:

1. The air in the tank can be treated as an ideal gas despite its high pressure.

2. The temperature of the air in the tank remains constant, therefore the product PV is also a constant.

3. The rate at which the diver consumes air from the tank is independent from the pressure or the volume of the tank.

4. The water density is constant and its value is 1000 kg/m3

Intuitively, we know that at a higher depth, the pressure over the diver will be higher and the air it is breathed will be compressed by this increasing pressure. Therefore, the more the diver goes down, the more volume or pressure inside the tank is needed. If the pressure inside the tank or the volume of this one is a constant, then the time the diver can remain will be lower as the depth increases.

(a) We can use the following procedure to find the compressed volume of air the diver has:

Vc = Vo (Po / P) ; the pressure P can be found if we express said pressure as a colum of fluid (water in this case)

P = pgh = (1000 kg/m3) * (9.8 m/s2) * (1.7 m) = 16660 Pa

Vc = 0.020 m3 (1*107 Pa / 16660 Pa) = 0.020 m3 (600.2) = 12 m3 = 12000 L

The time can be calculated using the rate given:

t = Vc / r = 12000 L / (0.5 L/s) = 24000 s = 400 min

(b) This question can be answered following the previous procedure:

P = (1000 kg/m3)(9.8 m/s2)(17 m) = 166600 Pa

Vc = 0.020 m3 (1*107 Pa/166000 Pa) = 1.2 m3 = 1200 L

t = 1200 L / (0.5 L/s) = 40 min

Note: if any of the suppositions is not valid, then the results would be different.

I hope it helps.

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