How Much Ethanol Can Be Added to a Pressurized Tank?

AI Thread Summary
The discussion focuses on calculating the additional volume of ethanol that can be added to a pressurized tank by applying pressure from lead weights on a piston. The moonshiner's tank has a total volume of 275 liters and a diameter of 0.290 m. The calculation involves using the pressure formula and the compressibility factor for ethanol, resulting in an additional volume of approximately 58.3 liters. A key point raised is the importance of using consistent units, as pressure should be converted to atmospheres for accurate results. The conversation highlights the need for clarity in unit conversions and understanding the physics behind the problem.
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Homework Statement


A moonshiner produces pure ethanol (ethyl alcohol) late at night and stores it in a stainless steel tank in the form of a cylinder 0.290 m in diameter with a tight-fitting piston at the top. The total volume of the tank is 275 L(0.250 m3) . In an attempt to squeeze a little more into the tank, the moonshiner piles lead breaks of total mass 1420kg on top of the piston.

a)What additional volume of ethanol can the moonshiner squeeze into the tank? (Assume that the wall of the tank is perfectly rigid.)?

Homework Equations


v'=-kv0p' where ' means change
p'=F/A


The Attempt at a Solution


okay the value of k for ethyl is 111x10-6 atm-1
p'=F/A=mg/pir2=1420*9.8/pi(0.145)2=210211.48
v'=-(111x10-6)(0.250m3)(210211.48)=-5.83x10-2m3=58.3L (we can ommit negative because question asks what additional amount of volume can be added.)

Now using this method I still have the wrong answer, Can anyone please point in the right direction. Any help is appreciated.
 
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anubis01 said:

Homework Statement


A moonshiner produces pure ethanol (ethyl alcohol) late at night and stores it in a stainless steel tank in the form of a cylinder 0.290 m in diameter with a tight-fitting piston at the top. The total volume of the tank is 275 L(0.250 m3) . In an attempt to squeeze a little more into the tank, the moonshiner piles lead breaks of total mass 1420kg on top of the piston.

a)What additional volume of ethanol can the moonshiner squeeze into the tank? (Assume that the wall of the tank is perfectly rigid.)?

Homework Equations


v'=-kv0p' where ' means change
p'=F/A


The Attempt at a Solution


okay the value of k for ethyl is 111x10-6 atm-1
p'=F/A=mg/pir2=1420*9.8/pi(0.145)2=210211.48
v'=-(111x10-6)(0.250m3)(210211.48)=-5.83x10-2m3=58.3L (we can ommit negative because question asks what additional amount of volume can be added.)

Now using this method I still have the wrong answer, Can anyone please point in the right direction. Any help is appreciated.

SI uses Pascals as a measure of pressure; your k for ethyl uses atmospheres.
 
Okay I changed my p' into atm and I got the right answer thanks for the tip.
 
I have the same question and I am stuck. How did he get 275L = 0.250m^3?
 

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