Thermodynamics height of air Help

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SUMMARY

The discussion focuses on a physics problem involving a steel can submerged in water, specifically analyzing the behavior of air pressure and volume changes as the can is submerged. The relevant equation, \( P_1 \times V_1 = P_2 \times V_2 \), is utilized to determine the height of the air inside the can at different water levels. As the can is submerged deeper, the increasing water pressure compresses the air, leading to a decrease in its volume. The problem is framed under constant temperature conditions, emphasizing the principles of thermodynamics.

PREREQUISITES
  • Understanding of the ideal gas law and pressure-volume relationships
  • Familiarity with basic thermodynamics concepts
  • Knowledge of hydrostatic pressure and its effects on submerged objects
  • Ability to manipulate algebraic equations for problem-solving
NEXT STEPS
  • Study the ideal gas law and its applications in thermodynamics
  • Learn about hydrostatic pressure calculations in fluid mechanics
  • Explore the concept of buoyancy and its effects on submerged objects
  • Investigate real-world applications of pressure-volume relationships in engineering
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the principles of thermodynamics and fluid mechanics, particularly in relation to pressure changes in gases.

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Homework Statement



A steel can, open at one end, closed at the other, 8inches high, with a diameter of 3inches, is submerged in water vertically with the open end facing down. What is the height of the air inside the can when:
a) the cans closed end is in level with water surface?
b) the can is 7inches below water surface?

Temperature is constant.


Homework Equations



pressure1*volume1 = presure2*volume2

The Attempt at a Solution




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any help would be helpful. thanx.
 
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What can you say about the air in the can as you lower the can in the water? As you lower the can, the water pressure increases. Since the air trapped in the can has no where to go, the air pressure will equalize with the water pressure (which increases with depth), but the volume will have to change because the air compresses.

You have the equation you need. You simply need to get the physical situation straight.
 

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