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A typical diver inhales 0.450 liters of air per breath and carries a 25 L breathing tank containing air at a pressure of \(300 .\) bar. As she dives deeper, the pressure increases by 1 bar for every \(10.08 \mathrm{m}\). How many breaths can the diver take from this tank at a depth of \(35 \mathrm{m} ?\) Assume that the temperature remains constant.

Short Answer

Expert verified
At a depth of 35 meters, the diver can take approximately 16891 breaths from the 25 L tank.

Step by step solution

01

Determine the pressure at 35 meters depth

First, we need to calculate the pressure increase due to the depth. The pressure increases by 1 bar for every 10.08 meters. To find the total pressure at 35 meters, we'll multiply the increase per meter by the depth. Pressure increase at 35 meters = (35 meters) * (1 bar / 10.08 meters) ≈ 3.47 bar Next, we add this pressure increase to the initial pressure of the tank (300 bar) to get the total pressure at 35 meters: Total pressure at 35 meters = 300 bar + 3.47 bar ≈ 303.47 bar
02

Determine the volume of a single breath at 35 meters

Now, let's use the ideal gas law (PV=nRT) to determine the volume of a single breath at 35 meters. Since the temperature and amount of gas (number of moles) are constant, we can use the relationship: Initial pressure * Initial volume = Final pressure * Final volume We know the initial pressure (1 bar at the surface) and initial volume (0.450 L per breath), and we just calculated the final pressure (303.47 bar). Let V be the final volume: \(1 \cdot 0.450 = 303.47 \cdot V_LSB\) Now, solve for V: \(V_LSB = \frac{0.450}{303.47} ≈ 0.00148 \: L\)
03

Calculate the number of breaths using the adjusted volume at 35 meters

Finally, to find the number of breaths the diver can take, we'll divide the total volume of the tank by the volume of a single breath at 35 meters: Number of breaths = Total volume of the tank / Volume of a single breath at 35 meters Number of breaths = \( \frac{25L}{0.00148 L} \) Number of breaths ≈ 16891.89 Since the diver cannot take a fraction of a breath, we'll round down to the nearest whole number: Number of breaths ≈ 16891 At a depth of 35 meters, the diver can take approximately 16891 breaths from the 25 L tank.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pressure Increase with Depth
As a diver descends into the water, the pressure around them increases. This increase in pressure with depth is a fundamental concept in diving and physics. This happens because water has weight, so the deeper you go, the more water is above you, pressing downwards. In the given problem, it's established that pressure increases by 1 bar for every 10.08 meters of depth.

For instance, at 35 meters, the pressure increases by about 3.47 bar. This additional pressure must be added to the initial pressure of the air in the tank. Understanding this concept is crucial for divers, as increased pressure affects both the body and any gases they carry. This can have important implications for breathing and buoyancy control.
  • 1 bar increase per 10.08 meters of depth
  • Total pressure at depth includes base pressure
Breathing Volume Calculation
As a diver goes deeper, the pressure increases and affects the volume of each breath they take. According to the ideal gas law, pressure and volume are inversely related at constant temperature. This means that as pressure increases with depth, the volume for each breath decreases.

In our example, at the surface, a diver breathes 0.450 liters. However, at a depth of 35 meters, with increased pressure, this volume decreases significantly to about 0.00148 liters per breath.

The Role of Gas Laws

This smaller volume results from how the ideal gas law dictates adjustments in pressure and volume. Efficiently calculating these volume changes is essential for ensuring that divers have enough air supply for their dive duration.
Gas Laws in Diving
The behavior of gases under pressure is critical in diving. The Ideal Gas Law, represented as \(PV = nRT\), is a tool to understand this. In diving, usually, temperature (T) and the amount of gas (n) are constant, often simplifying it to \(P_1V_1 = P_2V_2\).

This relationship implies that pressure and volume will change inversely if other factors remain constant.
  • Pressure increases as depth increases
  • Volume adjusts inversely to pressure
This is why, as a diver goes deeper, they must adapt their breathing to the new environmental pressure changes. Correctly applying these gas laws helps divers plan appropriately for the air they will need at various depths.
Constant Temperature Assumption
In many gas-related calculations, especially those involving diving, a constant temperature assumption simplifies the math. This is because if the temperature remains unchanged, it excludes one variable from the ideal gas equation, making calculations more straightforward.

For our diver, the constant temperature means that any change in pressure directly affects the volume without temperature-induced complications. Effects like the compressibility of air under varying temperatures are excluded, focusing solely on depth-related pressure changes.

This assumption, while simplifying, doesn't always reflect real-world conditions perfectly, as divers might experience temperature variations at different depths. However, it provides a useful approximation for many scenarios.

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Most popular questions from this chapter

A rigid vessel of volume 0.400 \(\mathrm{m}^{3}\) containing \(\mathrm{H}_{2}\) at \(21.25^{\circ} \mathrm{C}\) and a pressure of \(715 \times 10^{3} \mathrm{Pa}\) is connected to a second rigid vessel of volume \(0.750 \mathrm{m}^{3}\) containing Ar at \(30.15^{\circ} \mathrm{C}\) at a pressure of \(203 \times 10^{3} \mathrm{Pa}\). A valve separating the two vessels is opened and both are cooled to a temperature of \(12.2^{\circ} \mathrm{C} .\) What is the final pressure in the vessels?

When Julius Caesar expired, his last exhalation had a volume of \(450 . \mathrm{cm}^{3}\) and contained 1.00 mole percent argon. Assume that \(T=300 . \mathrm{K}\) and \(P=1.00\) atm at the location of his demise. Assume further that \(T\) has the same value throughout Earth's atmosphere. If all of his exhaled Ar atoms are now uniformly distributed throughout the atmosphere, how many inhalations of \(450, \mathrm{cm}^{3}\) must we make to inhale one of the Ar atoms exhaled in Caesar's last breath? Assume the radius of Earth to be \(6.37 \times 10^{6} \mathrm{m}\). [Hint: Calculate the number of Ar atoms in the atmosphere in the simplified geometry of a plane of area equal to that of Earth's surface. See Problem P1.20 for the dependence of the barometric pressure and the composition of air on the height above Earth's surface.

Use the ideal gas and van der Waals equations to calculate the pressure when \(2.25 \mathrm{mol} \mathrm{H}_{2}\) are confined to a volume of \(1.65 \mathrm{L}\) at \(298 \mathrm{K}\). Is the gas in the repulsive or attractive region of the molecule-molecule potential?

Many processes such as the fabrication of integrated circuits are carried out in a vacuum chamber to avoid reaction of the material with oxygen in the atmosphere. It is difficult to routinely lower the pressure in a vacuum chamber below \(1.0 \times 10^{-10}\) Torr. Calculate the molar density at this pressure at \(300 .\) K. What fraction of the gas phase molecules initially present for \(1.0 \mathrm{atm}\) in the chamber are present at \(1.0 \times 10^{-10}\) Torr?

A balloon filled with 11.50 L of Ar at \(18.7^{\circ} \mathrm{C}\) and 1 atm rises to a height in the atmosphere where the pressure is 207 Torr and the temperature is \(-32.4^{\circ} \mathrm{C}\). What is the final volume of the balloon? Assume that the pressure inside and outside the balloon have the same value.

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