Solving the Fluids Straw Problem: Compute Water Fraction in Ideal Gas Model

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The discussion focuses on solving a physics problem involving a drinking straw partially filled with water and air. The key points include the need to calculate the pressure in the air trapped in the straw before and after it is removed from the water. The problem emphasizes that it is a hydrostatics issue rather than a buoyancy problem, as 95% of the straw is submerged initially. Participants highlight the importance of understanding pressure differentials and the behavior of the water surface inside the straw when it is drawn out. The solution requires applying hydrostatic principles to determine the new pressure and the movement of the water surface.
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Homework Statement


This problem pertains to the drinking straw trick. You decide to place your finger over a 20.0cm straw. 95.0% of the straw is filled with water, while the top is full of air. Treating the straw as an ideal gas, compute fraction of the straw that is filled with water when the straw is drawn out of the glass.

Homework Equations


B = \rho gV
P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2

The Attempt at a Solution


I am not really sure how to start this, but I am pretty sure I have to do something with the pressure of the straw. The atmospheric pressure is equal to 101325 Pa, and the pressure at the bottom of the straw should be Patm + rho*g*h, so there is a pressure differential. If 95% of the straw is filled with water, does that mean that the water contributes to 95% of the buoyant force of the straw while air contributes 5%? I have these ideas but I am not too sure how to apply them to proceed with this problem
 
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This is not a buoyancy problem. It is a hyrostatics problem. Initially, 95% of the straw is below the water surface, i.e., 19 cm. There is 1 cm of straw sticking out above the water surface. You put your finger over the top of the straw, and take the straw out of the water.

What is the pressure in the air trapped in the straw before you take the straw out of the water? After you take the straw out of the water, the water surface inside the straw moves down a little, and a small amount of water leaks out the bottom. You need to determine the new pressure in the air head space, and how much the water surface inside the straw moves down.

Chet
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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