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# Bulk Modulus

A moonshiner produces pure ethanol (ethyl alcohol) and stores it in a stainless steel tank in the form of a cylinder 0.300m in diameter with a tight-fitting piston at the top. The total volume of the tank is (250L or 0.250m^3). In an attempt to squeeze a little more into the tank, the moonshiner piles 1420kg of lead bricks on top of the piston. What additional volume of ethanol can the moonshiner squeeze into the tank? (Assume that the wall of the tank is perfectly rigid.)

Well the compressability of ethyl alcohol is (1.10 * 10^-13)

And I know that k (compressibility)...

k = - 1/v_0 * (v - v_0)/(p - p_0)

My target variable is v (finial volume), but how do I figure out p & p_0? Or what other equations can I use to figure out all the unknowns? Am I even taking the right approach?

Brian
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BrianGEFF719
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2 Solutions

Commented:
You are on the right track. In your formula I thingk p_0 is the ambient presure and p is the increased presure provided by the lead bricks.
Presure is force per unit area. The added force is weight of bricks times g.
That ought to get you started.
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Commented:
(p - p_0) is the increased presure provided by the lead bricks.
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Commented:
Compressibility should have units.  I think ethanol would be 1.1 * 10^-9 meters^2/newtons
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Author Commented:
How do I calculate the increased pressure?

Brian
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Commented:
1420kg * g / area of the piston
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Author Commented:
Ozo, my book defines pressure to be:

p = F (perpendicular)/SURFACE AREA

I have no idea how to apply this concept to this problem.
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Author Commented:

My answer was Change in Volume = -5.4 x 10^(-6) m^3

Is this correct?
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Author Commented:
Hey ozo, you just broke the 500,000 mark ;0
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