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Chapter 11: Problem 95

A moonshiner produces pure ethanol (ethyl alcohol) late at night and stores it in a stainless steel tank in the form of a cylinder 0.300 \(\mathrm{m}\) in diameter with a tight-fitting piston at the top. The total volume of the tank is 250 \(\mathrm{L}\left(0.250 \mathrm{m}^{3}\right) .\) In an attempt to squeeze a little more into the tank, the moonshiner piles 1420 \(\mathrm{kg}\) 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.)

Short Answer

Expert verified
About 5.4 cm鲁 more ethanol can be squeezed into the tank.

Step by step solution

01

Understand the Problem

We have a cylinder tank with a tight-fitting piston and 1420 kg of lead bricks on top. The question asks how much additional ethanol (\( \Delta V \) ) can be squeezed into the tank due to the pressure created by the bricks.
02

Calculate the Pressure Exerted by the Lead Bricks

First, find the force exerted by the lead bricks, which is the weight \( F = mg \) where \( m = 1420 \) kg and \( g = 9.8 \text{ m/s}^2 \), then divide by the area \( A \) of the piston to get the pressure: \( A = \pi \left(\frac{D}{2}\right)^2 = \pi \left(0.15\right)^2 \text{ m}^2.\)So,\( P = \frac{1420 \times 9.8}{\pi \times 0.15^2} = \frac{13916}{0.0707} = 196792 \text{ Pa}.\)
03

Use the Bulk Modulus Concept

Now, apply the concept of bulk modulus. The bulk modulus \( B \) for ethanol is a measure of its resistance to compression, typically around \( 9.1 \times 10^9 \text{ Pa} \).The formula relating pressure change \( \Delta P \) and change in volume \( \Delta V \) is: \( \Delta P = -B \frac{\Delta V}{V_0} \)where\( V_0 \) is the initial volume \( 0.250 \text{ m}^3 \).Rearranging gives:\( \Delta V = -\frac{\Delta P \cdot V_0}{B}.\)
04

Calculate the Additional Volume of Ethanol

Substitute into the rearranged bulk modulus formula:\( \Delta V = -\frac{196792 \times 0.250}{9.1 \times 10^9} \)Compute this:\(\Delta V = -\frac{49198}{9.1 \times 10^9} = -0.0000054 \text{ m}^3= -5.4 \text{ cm}^3.\)
05

Interpret the Result

The negative sign indicates a decrease in volume due to compression. However, since we are concerned with the additional amount of ethanol squeezed in, we take the positive value: The additional volume of ethanol that can be squeezed in is approximately 5.4 cm each to help the moonshiner fit more fluid into the tank.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pressure Calculation Basics
To understand how pressure affects volume, let's start with the basics of pressure calculation. In physics, pressure is defined as force applied per unit area. It tells us how the force is distributed over a surface.
  • Force: It is the weight of the object, calculated as mass times the acceleration due to gravity. For the moonshiner's lead bricks, this is the mass (1420 kg) times gravity (9.8 m/s虏).
  • Area: This is the surface area of the piston, which is determined by its diameter. Here, the diameter is 0.3 meters, and the area is \(\pi \left(\frac{D}{2}\right)^2\).
By dividing the force by the piston area, you find the pressure exerted by the bricks: \( P = \frac{Force}{Area} = \frac{1420 \times 9.8}{\pi \times 0.15^2} \). This equation allows us to convert the weight of an object, such as lead bricks, into a pressure value in Pascals, suitable for further analysis.
Understanding Volume Change
Next, let's focus on how pressure affects the volume of liquids through volume change. Volume change is a response to the applied external pressure. In this exercise, we're interested in how much extra ethanol can be squeezed into a tank.

Liquids are typically resistant to changes in their volume, which is why we use the concept of bulk modulus. The bulk modulus (\( B \)) is a property of materials that indicates their ability to withstand changes in volume under pressure. The formula used is: \( \Delta P = -B \frac{\Delta V}{V_0} \).
  • \( \Delta V \): The change in volume you are looking for.
  • \( V_0 \): The initial volume before pressure is applied.
  • \( \Delta P \): The change in pressure from any external source, like lead bricks.
By rearranging the formula, you can solve for the change in volume (\( \Delta V \)), helping you find how much more liquid can fit in by compressing the existing volume.
Ethanol Compression
Ethanol, like any liquid, has a bulk modulus that indicates its compression resistance. Here, the bulk modulus for ethanol is used to understand how its volume decreases when subjected to external pressure. In this scenario, the moonshiner compresses ethanol by stacking bricks on the tank.

Let's apply the bulk modulus concept to ethanol compression. The bulk modulus of ethanol is approximately \(9.1 \times 10^9 \) Pascals. By using the formula \( \Delta V = -\frac{\Delta P \cdot V_0}{B} \), we can determine how much the initial ethanol volume (0.250 \(m^3\)) decreases.
Given the pressure calculated in the previous section, substitute into the equation to find the change in volume: \[ \Delta V = -\frac{196792 \times 0.250}{9.1 \times 10^9} \.\] Calculate this to get about -5.4 cm鲁.
This negative sign reflects a 'compression', or volume reduction. In practical terms, it means that approximately 5.4 cm鲁 more ethanol can fit into the tank due to this compressive force. Understanding ethanol compression not only enlightens us about the moonshining process but also exemplifies fundamental physics principles.

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Most popular questions from this chapter

\(\mathrm{A} 1.05-\mathrm{m}-\operatorname{lon} \mathrm{g}\) rod of negligible weight is supported at its ends by wires \(A\) and \(B\) of equal length (Fig. P11.91). The cross-sectional area of \(A\) is 2.00 \(\mathrm{mm}^{2}\) and that of \(B\) is 4.00 \(\mathrm{mm}^{2} .\) Young's modulus for wire \(A\) is \(1.80 \times 10^{11} \mathrm{Pa}\) ; that for \(B\) is \(1.20 \times 10^{11} \mathrm{Pa}\) . At what poing the rod should a weight \(w\) be suspended to produce (a) equal stresses in \(A\) and \(B\) and (b) equal strains in \(A\) and \(B ?\)

A pickup truck has a wheelbase of 3.00 \(\mathrm{m} .\) Ordinarily, \(10,780 \mathrm{N}\) rests on the front wheels and 8820 \(\mathrm{N}\) on the rear wheels when the truck is parked on a level road. (a) A box weighing 3600 \(\mathrm{N}\) is now placed on the tailgate, 1.00 \(\mathrm{m}\) behind the rear axle. How much total weight now rests on the front wheels? On the rear wheels? (b) How much weight would need to be placed on the tailgate to make the front wheels come off the ground?

A metal rod that is 4.00 \(\mathrm{m}\) long and 0.50 \(\mathrm{cm}^{2}\) in cross sectional area is found to stretch 0.20 \(\mathrm{cm}\) under a tension of 5000 \(\mathrm{N.}\) What is Young's modulus for this metal?

A solid gold bar is pulled up from the hold of the sunken RMS Titanic. (a) What happens to its volume as it goes from the pressure at the ship to the lower pressure at the ocean's surface? (b) The pressure difference is proportional to the depth. How many times greater would the volume change have been had the ship been twice as deep? (c) The bulk modulus of lead is one- fourth that of gold. Find the ratio of the volume change of a solid lead bar to that of a gold bar of equal volume for the same pressure change.

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