How Much Ethanol Can Be Added to a Pressurized Tank?

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Homework Help Overview

The discussion revolves around a problem involving the calculation of additional volume of ethanol that can be added to a pressurized tank. The tank is cylindrical with specific dimensions and is subjected to additional pressure from lead weights placed on a piston. The subject area includes fluid mechanics and pressure-volume relationships in gases.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of pressure equations and the effects of added weight on the volume of liquid that can be accommodated in the tank. There are attempts to convert units and clarify the relationship between pressure and volume changes.

Discussion Status

Some participants have provided guidance on unit conversion, which has helped one individual arrive at the correct answer. Others are still seeking clarification on specific calculations and conversions related to the problem.

Contextual Notes

There is a noted confusion regarding the conversion between liters and cubic meters, as well as the use of different pressure units (atmospheres vs. Pascals). This indicates a potential gap in understanding the relationships between these measurements.

anubis01
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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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