How Much Ethanol Can Be Added to a Pressurized Tank?

[anubis01]

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.

 

[asleight]

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.

 

[anubis01]

Okay I changed my p' into atm and I got the right answer thanks for the tip.

 

[Hockeystar]

I have the same question and I am stuck. How did he get 275L = 0.250m^3?

 

[rwisz]

I would approach this problem by first considering the concept of bulk stress and strain. Bulk stress is the force per unit area acting on a material, while bulk strain is the change in volume of a material due to the applied stress.

In this scenario, the moonshiner is applying a force (the weight of the lead bricks) on the piston, which is causing a stress on the ethanol in the tank. This stress will result in a change in volume, or strain, of the ethanol.

To calculate the additional volume of ethanol that can be squeezed into the tank, we can use the bulk modulus equation: ΔV = -(V0ΔP)/B, where ΔV is the change in volume, V0 is the initial volume, ΔP is the change in pressure, and B is the bulk modulus of the material.

In this case, the initial volume of the tank is 0.250 m^3 and the initial pressure is atmospheric pressure, which we can assume to be 1 atm. The additional pressure applied by the lead bricks can be calculated as follows: P = F/A = (1420 kg * 9.8 m/s^2)/(π*0.145 m^2) = 210211.48 Pa.

Plugging these values into the bulk modulus equation, we get: ΔV = -[(0.250 m^3)(210211.48 Pa)]/(111x10^-6 atm^-1) = -4.76 L.

Therefore, the additional volume of ethanol that can be squeezed into the tank is 4.76 L.