/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 64 On a day when the pressure of th... [FREE SOLUTION] | 91Ó°ÊÓ

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On a day when the pressure of the atmosphere is \(1.000 \times 10^{5} \mathrm{~Pa}\), a chemist distills a liquid under slightly reduced pressure. The pressure within the distillation chamber is read by an oil-filled manometer (density of oil \(\left.=0.78 \mathrm{~g} / \mathrm{cm}^{3}\right)\). The difference in heights on the two sides of the manometer is \(27 \mathrm{~cm}\). What is the pressure in the distillation chamber?

Short Answer

Expert verified
The pressure in the distillation chamber is 97947.814 Pa.

Step by step solution

01

Understand the Problem

We need to find the pressure inside the distillation chamber. The pressure difference is indicated by the manometer reading and the known atmospheric pressure. We will calculate the pressure change due to the height of the oil column.
02

Convert Units

Convert the density of oil from g/cm³ to kg/m³ for consistency in units: \(0.78 \text{ g/cm}^3 = 780 \text{ kg/m}^3\). Similarly, convert the height from cm to m: \(27 \text{ cm} = 0.27 \text{ m}\).
03

Calculate Pressure Change

Use the formula for pressure change due to a liquid column, \( \Delta P = \rho g h \), where \( \rho = 780 \text{ kg/m}^3\), \( g = 9.81 \text{ m/s}^2\), and \( h = 0.27 \text{ m}\). Calculate \( \Delta P \): \[\Delta P = 780 \times 9.81 \times 0.27 = 2052.186 \text{ Pa}.\]
04

Calculate Pressure in the Chamber

The pressure in the distillation chamber \(P_{chamber}\) is the atmospheric pressure minus the pressure change: \[P_{chamber} = 1.000 \times 10^5 - 2052.186 \text{ Pa} = 97947.814 \text{ Pa}.\]

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Key Concepts

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

Pressure Calculation
In fluid mechanics, understanding how to calculate pressure is crucial because it allows us to determine how forces are distributed along surfaces in a fluid environment. Pressure is essentially the force exerted per unit area, and it is measured in pascals (Pa) in the metric system.
The foundation of pressure calculation in fluids involves using the principle of communicating vessels. This principle tells us that in a fluid at rest, pressure changes linearly with depth due to gravity. This is encapsulated in the formula: \[ \Delta P = \rho g h \] where
  • \( \Delta P \) is the change in pressure,
  • \( \rho \) is the fluid density (mass per unit volume),
  • \( g \) is the acceleration due to gravity, typically 9.81 \( \text{m/s}^2 \),
  • \( h \) is the height difference of the fluid column.
By calculating \( \Delta P \), we can determine the impact of a fluid's height on pressure, which is key when working with liquid columns in devices like manometers.
Manometer
A manometer is a device used for measuring the pressure in a fluid. It is a U-shaped tube filled with a liquid, usually mercury or oil, that can measure the pressure of a gas or a liquid. By observing the height difference of the liquid in the two arms of the tube, one can determine the pressure difference between the fluid whose pressure is being measured and a reference point, often the atmospheric pressure.
The working principle of a manometer is straightforward. In the case of a fluid-filled manometer, the pressure difference can be directly related to the height difference by the pressure formula: \[ \Delta P = \rho g h \] The density \( \rho \), the gravitational constant \( g \), and the height \( h \) all factor into creating this pressure difference. Thus, manometers are critical in applications where precise pressure measurements are necessary, such as in laboratory settings or industrial plant operations.
Unit Conversion
Unit conversion is an essential skill in fluid mechanics, as it ensures consistency and accuracy in calculations, especially when dealing with different measurement systems. In many scientific problems, values need to be converted between units to facilitate correct calculations—a common requirement when applying pressure formulas.
For instance, the density of a fluid might be given in grams per cubic centimeter (g/cm³), but the pressure formula \( \Delta P = \rho g h \) requires density in kilograms per cubic meter (kg/m³). To convert, you multiply by 1000, since there are 1000 grams in a kilogram and \(\left(1 \text{ cm} = 0.01 \text{ m}\right)\) leads to \(\left(1\text{ cm}^3 = 1 \times 10^{-6} \text{ m}^3\right)\).
Another common conversion task involves changing measurement of distance from centimeters to meters. There are 100 centimeters in a meter, so divide the centimeter measure by 100.
Successful unit conversion paves the way for accurate computations and results, aiding in exercises like determining the pressure in a distillation chamber using a manometer.

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Most popular questions from this chapter

A cubic covered tank \(5.00 \mathrm{~m}\) by \(5.00 \mathrm{~m}\) by \(5.00 \mathrm{~m}\) is completely filled with water through an \(8.00-\mathrm{cm}^{2}\) hole in its lid. A plug is then forced into the hole with a vertical force of \(200 \mathrm{~N}\). What change in pressure, if any, will be read by a gauge in the side of the tank as a result of inserting the plug?

A large open rectangular tank \(2.00 \mathrm{~m}\) by \(2.00 \mathrm{~m}\) by \(11.0 \mathrm{~m}\) deep is filled with ethyl alcohol to a depth of \(10.0 \mathrm{~m}\). What is the value of the net force exerted by the liquid on the bottom of the tank?

A spring whose composition is not completely known might be either bronze (sp gr \(8.8\) ) or brass (sp gr \(8.4\) ). It has a mass of \(1.26\) g when measured in air and \(1.11\) g in water. Which is it made of?

At a height of \(10 \mathrm{~km}\) (33 \(000 \mathrm{ft}\) ) above sea level, atmospheric pressure is about \(210 \mathrm{~mm}\) of mercury. What is the net resultant normal force on a \(600 \mathrm{~cm}^{2}\) window of an airplane flying at this height? Assume the pressure inside the plane is \(760 \mathrm{~mm}\) of mercury. The density of mercury is \(13600 \mathrm{~kg}\).

A barrel will rupture when the gauge pressure within it reaches 350 \(\mathrm{kPa}\). It is attached to the lower end of a vertical pipe, with the pipe and barrel filled with oil \(\left(\rho=890 \mathrm{~kg} / \mathrm{m}^{3}\right)\). How long can the pipe be if the barrel is not to rupture? From \(P=\rho g h\) we have $$ h=\frac{P}{\rho g}=\frac{350 \times 10^{3} \mathrm{~N} / \mathrm{m}^{2}}{\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)\left(890 \mathrm{~kg} / \mathrm{m}^{3}\right)}=40.1 \mathrm{~m} $$

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