#$&*
phy 232
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Problem Number 4
A tube 4.5 mm in diameter is run through the stopper of a sealed 2.5-liter container. The tube outside the container forms a U, then runs in a straight line with slope .021 with respect to horizontal. Alcohol is introduced into the tube, and fills the U, extending into the linear section of the tube. Both ends of the tube are open. The container is slightly heated, and the alcohol column is observed to move along the linear section of the tube. The material of which the container is constructed has coefficient of linear expansion `alpha = 86 * 10^-6 / C. If the temperature of the air in the container was originally 21 Celsius, then if the temperature increases by .59 Celsius how far will the alcohol column move?
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ok so I understand that you have a pressure a volume and a temperature on the left side of the alcohol, and you have a pressure and a volume on the right side of the alcohol in the sloped tube, but if the alcohol moves then that means that both the pressure and volume change in both sides of the tubes which cause the problem to be insolvable. If you could help to point me in the right direction that would be great.
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how do you set up this problem? i feel like there is some information missing.
Certainly under the given conditions the alcohol will move. The one thing that isn't given and I can't reasonably expect you to know is the density of alcohol, which is somewhat less than the density of water. For that reason I wouldn't count this specific problem against a student on the actual test. However there are other problem where the liquid in the tube is water, and I would expect the density of water to be common knowledge. So it's worth seeing how to solve this.
Strategy: The change in the position of the alcohol column tells you both the change in pressure and the change in volume, so it can be used to represent both of the two quantities (pressure and volume) you mention.
Let `dL stand for the change in the position of the alcohol column in the tube.
The level of the alcohol changes by `dy = .021 * `dL, and the pressure therefore changes by rho g `dy = .021 rho g `dL.
The volume of alcohol in the tube changes by A_cs * `dL, where A_cs is the cross-sectional area of the tube.
The volume of the container itself changes by 3 alpha * V0 * `dT, due to the expansion of the material.
The change in volume of the air is therefore A_cs * `dL + 3 alpha * V0 * `dT.
P V / T remains constant.
Thus P0 V0 / T0 = (P0 + .021 rho g `dL) * (V0 + A_cs * `dL + 3 alpha V0 `dT) / (T0 + `dT).
Assuming P0 = atmospheric pressure, T0 = 294 K, V0 = 2.5 liters, `dT = .59 Celsius, rho = 700 kg / m^3 (roughly the density of alcohol), g = 9.8 m/s^2, the only unknown in the equation is `dL.