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Thermodynamics help

S

Suspicious-Teach8788

Lifer
Ahhhh this sucks...

Ok, so I'm given:

Flask w/ air connected via a tube to another beaker of oil which has a column of oil sticking out initially at h = 0.

The flask w/ air is sitting in a beaker of ice, but we remove it and put it in room temperature and the oil rises. We put it in hot water and it rises..

I have the volume of the flask of air + tubing which gives me the total air present initially...

Is there a way to find out how high the oil column goes @ T = 100C? I am given density of oil , g = 9.8 m/s^2 and the cross sectional area of the oil column A (circular)....

 
Are you saying that the column of oil is a fixed height? i.e. does the oil exert a constant pressure no matter what? if so, then just use the ideal gas law.
 
How exactly does the change in pressure in the air flask affect the beaker of oil, again? You need a drawing or something to show how they're connected. Oil is basically an incompressible fluid, I'm not seeing how the oil will rise if it's sitting in a beaker. Sounds like Bernoulli's gets used somewhere, but not the way you describe it.
 
So if I understand it right, the beaker of oil has TWO spouts on top? One spout is connected on top to the beaker of air and the other spout is connected to a column which is open to the atmosphere. Or am I reading the problem incorrectly?

 
I'll give some hints.

Change in Air pressure = Density * g * Change in Oil Height

The cross section of the oil is irrelevant.

Edit: I may have over-simplified. Working it out now.
 
If I have the problem correct, then this is what you need to do (using the ideal gas law like Umberger suggested):

Assumptions:
1) Assume the beaker of ice is partially melted.
2) Assume the ice/water is pure.
3) Thus, the beaker of ice/water is at 0°C.
4) Assume the entire volume of air is at the same temperature of the water bath. This assumption is probably never going to be correct in real life, but if the volume of the air beaker is large and the volume of the air line to the oil beaker is small, and if the tubing is well insulated, it is a decent assumption. Also, a long period of time is needed for the whole contraption to reach equilibrium.
5) Assume the oil density is constant with temperature and pressure.
6) Assume the ideal gas law applies (the error is quite small with the temperatures we are using.
7) Assume the tubing from the air beaker to the oil beaker is the same inner diameter as the oil column height.
8) Assume the oil column is open to the atmosphere and that you know that pressure.

Here is what you know:
1) V1 = initial volume of air.
2) T1 = 0°C= initial temperature of air.
3) h1 = 0*m = initial oil column height difference in the two tubes.
4) D = density of oil.
5) g = 9.8*m/s^2 = gravity.
6) Patm = atmospheric pressure.
7) A = inner cross-sectional area of both tubes.

Calculations:
1) Find the pressure of air in the beaker. P1 = Patm + h1*D*g. Since h1 = 0*m, P1 = Patm.
2) Find out how much air you have. n = (P1*V1) / (R*T1) = (Patm*V1) / (R*T1). Note: you don?t need to solve this, I?ll just use this later.
3) Find out the new pressure of air in the beaker at T2. P2 = Patm + h2*D*g. Note: you can?t solve this yet because you don't know h2.
3) Find out the new volume of air due to displacement of oil. V2 = V1 + h2*A. Note: you can?t solve this yet because you don't know h2.
4) Find out the new volume of air due to increasing temperature. V2 = (n*R*T2) / P2 = (n*R*T2) / (Patm + h2*D*g) = (Patm*V1*T2) / (T1*(Patm + h2*D*g)). Note: you can?t solve this yet because you don't know h2.
5) Combine equations (3) and (4). V1 + h2*A = (Patm*V1*T2) / (T1*(Patm + h2*D*g)). This is one equation with one unknown, it can now be solved for h2. I?ll leave that up to you. Do your own homework.
 
Was the air initially at environmental pressure at Temp = 0? We need to know the temperature at which the flask was sealed, I think.

Geez, so easy to forget this stuff.
 
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