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Ch 18: A Macroscopic Description of Matter
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 18: A Macroscopic Description of MatterProblem 50a
Chapter 18, Problem 50a

A diving bell is a 3.0-m-tall cylinder closed at the upper end but open at the lower end. The temperature of the air in the bell is 20°C. The bell is lowered into the ocean until its lower end is 100 m deep. The temperature at that depth is 10°C. How high does the water rise in the bell after enough time has passed for the air inside to reach thermal equilibrium?

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Step 1: Understand the problem. The diving bell is a cylinder with air trapped inside, and as it is lowered into the ocean, the pressure increases due to the depth. The air inside the bell compresses, causing water to rise into the bell. We need to calculate how high the water rises after the air reaches thermal equilibrium at the new temperature and pressure.
Step 2: Apply the ideal gas law. The relationship between pressure, volume, and temperature for the air inside the bell can be expressed as \( P_1 V_1 / T_1 = P_2 V_2 / T_2 \), where \( P_1 \), \( V_1 \), and \( T_1 \) are the initial pressure, volume, and temperature, and \( P_2 \), \( V_2 \), and \( T_2 \) are the final pressure, volume, and temperature.
Step 3: Determine the initial and final pressures. The initial pressure \( P_1 \) is the atmospheric pressure at the surface (approximately \( 1.0 \, \text{atm} \)). The final pressure \( P_2 \) is the sum of atmospheric pressure and the pressure due to the water column at 100 m depth, calculated using \( P_2 = P_1 + \rho g h \), where \( \rho \) is the density of water (\( 1000 \, \text{kg/m}^3 \)), \( g \) is the acceleration due to gravity (\( 9.8 \, \text{m/s}^2 \)), and \( h \) is the depth (\( 100 \, \text{m} \)).
Step 4: Relate the volume change to the water rise. The initial volume \( V_1 \) of the air is the full volume of the cylinder, \( V_1 = \text{cross-sectional area} \times \text{height} \). The final volume \( V_2 \) is reduced due to the water rising into the bell. The height of the water rise \( h_w \) can be calculated using \( V_2 = \text{cross-sectional area} \times (3.0 \, \text{m} - h_w) \).
Step 5: Solve for \( h_w \). Substitute the values for \( P_1 \), \( P_2 \), \( T_1 \), \( T_2 \), \( V_1 \), and \( V_2 \) into the ideal gas law equation. Rearrange the equation to isolate \( h_w \), and solve for the height of the water rise. Ensure that temperatures are converted to Kelvin (\( T = \text{°C} + 273.15 \)) before substitution.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Thermal Equilibrium

Thermal equilibrium occurs when two objects or systems reach the same temperature and no heat flows between them. In this scenario, the air inside the diving bell will eventually cool down to match the temperature of the surrounding water at 10°C. This process is crucial for determining the pressure changes inside the bell, which directly affects how high the water will rise.
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Ideal Gas Law

The Ideal Gas Law, expressed as PV = nRT, relates the pressure (P), volume (V), and temperature (T) of an ideal gas. In this problem, as the temperature of the air in the diving bell decreases, the pressure will also change, influencing the volume of air and the height of the water column. Understanding this relationship is essential for calculating the new equilibrium state of the gas inside the bell.
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Hydrostatic Pressure

Hydrostatic pressure is the pressure exerted by a fluid at equilibrium due to the force of gravity. It increases with depth in a fluid and is given by the formula P = ρgh, where ρ is the fluid density, g is the acceleration due to gravity, and h is the depth. This concept is vital for determining the pressure exerted by the water at 100 m depth, which will help in calculating how high the water rises in the diving bell.
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Related Practice
Textbook Question

A 6.0-cm-diameter, 10-cm-long cylinder contains 100 mg of oxygen (O₂) at a pressure less than 1 atm. The cap on one end of the cylinder is held in place only by the pressure of the air. One day when the atmospheric pressure is 100 kPa, it takes a 184 N force to pull the cap off. What is the temperature of the gas?

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Textbook Question

The 3.0-m-long pipe in FIGURE P18.49 is closed at the top end. It is slowly pushed straight down into the water until the top end of the pipe is level with the water's surface. What is the length L of the trapped volume of air?

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On a cool morning, when the temperature is 15°C, you measure the pressure in your car tires to be 30 psi. After driving 20 mi on the freeway, the temperature of your tires is 45°C . What pressure will your tire gauge now show?

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