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Chapter 5: Problem 1

Define pressure. From the definition, obtain the SI unit of pressure in terms of SI base units.

Short Answer

Expert verified
The SI unit of pressure is Pascal (Pa), defined as 1 kg路m鈦宦孤穝鈦宦�.

Step by step solution

01

Understand Pressure

Pressure is defined as the amount of force exerted per unit area. Mathematically, pressure \( P \) is given by the formula:\[P = \frac{F}{A}\]where \( F \) is the force applied and \( A \) is the area over which the force is applied.
02

Identify SI Units for Force and Area

Force \( F \) is measured in Newtons (N), and area \( A \) is measured in square meters (m虏). The SI unit of force, Newton, is derived from the base units as follows:\[ \text{Newton} = \text{kilogram} \times \text{meter/s}^2 = ext{kg} \cdot ext{m/s}^2 \] means \( \text{N} = \text{kg} \cdot \text{m/s}^2 \)
03

Calculate the SI Unit of Pressure

Using the formula for pressure and the SI units for force and area, we can deduce the SI unit of pressure. Since pressure \( P \) is defined as force \( F \) over area \( A \), substituting the units gives:\[P = \frac{F}{A} = \frac{ ext{N}}{ ext{m}^2} = \frac{ ext{kg} \cdot ext{m/s}^2}{ ext{m}^2}\]Simplifying this expression results in:\[P = ext{kg} imes ext{m}^{-1} imes ext{s}^{-2}\]
04

Conclude the SI Unit for Pressure

The derived SI unit for pressure is Pascal (Pa), which equates to:\[1 ext{ Pa} = 1 ext{ kg} imes ext{m}^{-1} imes ext{s}^{-2}\]

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Key Concepts

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

Definition of Pressure
Pressure is a fundamental concept in physics that describes how a force is distributed over an area. Imagine you are pushing on a door; the force you apply is spread over the surface area of the part of the door you are touching. This relationship is what we call pressure.
Mathematically, pressure (\( P \)) is defined as force divided by area:\[ P = \frac{F}{A} \]where \( F \) is the force applied and \( A \) is the area over which the force is distributed.
The practical importance of understanding pressure comes into play in numerous situations, such as inflating tires, measuring blood pressure, and in engineering applications.
Pascal
The Pascal (Pa) is the SI unit of pressure. Named after the French mathematician and physicist Blaise Pascal, it was established to measure how much force is applied over a certain area.
In terms of its definition:\[ 1 \text{ Pa} = 1 \text{ N/m}^2 \]This means that a Pascal is equivalent to a force of one Newton applied uniformly across an area of one square meter.
The Pascal is a relatively small unit of pressure, so for practical purposes, pressure is often measured in kilopascals (kPa) or megapascals (MPa). For example, atmospheric pressure at sea level is about 101.3 kPa.
Force and Area Relationship
The concept of pressure hinges on the relationship between force and area. To understand this, consider the equation:\[ P = \frac{F}{A} \]Here, increasing the force on a given area results in higher pressure, whereas increasing the area over which a force is applied results in lower pressure.
For example:
  • When you use a sharp knife (small area of application), it requires less force to cut than a blunt knife (larger area).
  • Tires are an example where distributing the weight of a vehicle over a larger area helps in reducing the pressure and prevents damage to the surfaces.
SI Base Units
The International System of Units (SI) provides a coherent set of units that are universally used to measure physical quantities. Pressure, as measured in Pascals, is based on the products and ratios of these base units.
For pressure, the key SI base units involved are:
  • Kilogram (kg) for mass
  • Meter (m) for length
  • Second (s) for time
Considering these, we can express pressure in base units:\[ \text{Pascal (Pa)} = \frac{\text{kg} \cdot \text{m/s}^2}{\text{m}^2} = \text{kg} \cdot \text{m}^{-1} \cdot \text{s}^{-2} \]Understanding these relationships helps in comprehending complex problems in physics and engineering, ensuring consistent measurement practices.

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Most popular questions from this chapter

5.161 Power plants driven by fossil-fuel combustion generate substantial greenhouse gases (e.g., \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) ) as well as gases that contribute to poor air quality (e.g., \(\mathrm{SO}_{2}\) ). To evaluate exhaust emissions for regulation purposes, the generally inert nitrogen supplied with air must be included in the balanced reactions; it is further assumed that air is composed of only \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) with a volume/volume ratio of \(3.76: 1.00,\) respectively. In addition to the water produced, the fuel's \(\mathrm{C}\) and \(\mathrm{S}\) are converted to \(\mathrm{CO}_{2}\) and \(\mathrm{SO}_{2}\) at the stack. Given these stipulations, answer the following for a power plant running on a fossil fuel having the formula \(\mathrm{C}_{18} \mathrm{H}_{10} \mathrm{~S}\). Including the \(\mathrm{N}_{2}\) supplied in air, write a balanced combustion equation for the complex fuel assuming \(120 \%\) stoichiometric combustion (i.e., when excess oxygen is present in the product gases). Except for \(\mathrm{N}_{2},\) use only integer coefficients. (See problem \(3.141 .\) ) What gas law is used to effectively convert the given \(\mathrm{N}_{2} / \mathrm{O}_{2}\) volume ratio to a molar ratio when deriving the balanced combustion equation in (a)? (i) Boyle's law (ii) Charles's law (iii) Avogadro's law (iv) ideal gas law c) Assuming the product water condenses, use the result from (b) to determine the stack gas composition on a "dry" basis by calculating the volume/volume percentages for \(\mathrm{CO}_{2}, \mathrm{~N}_{2},\) and \(\mathrm{SO}_{2}\) Assuming the product water remains as vapor, repeat (c) on a "wet" basis.Assuming the product water remains as vapor, repeat (c) on a "wet" basis. Some fuels are cheaper than others, particularly those with higher sulfur contents that lead to poorer air quality. In port, when faced with ever-increasing air quality regulations, large ships operate power plants on more costly, higher-grade fuels; with regulations nonexistent at sea, ships' officers make the cost-saving switch to lower-grade fuels. Suppose two fuels are available, one having the general formula \(\mathrm{C}_{18} \mathrm{H}_{10} \mathrm{~S}\) and the other \(\mathrm{C}_{32} \mathrm{H}_{18} \mathrm{~S} .\) Based on air quality concerns alone, which fuel is more likely to be used when idling in port? Support your answer to (d) by calculating the mass ratio of \(\mathrm{SO}_{2}\) produced to fuel burned for \(\mathrm{C}_{18} \mathrm{H}_{10} \mathrm{~S}\) and \(\mathrm{C}_{32} \mathrm{H}_{18} \mathrm{~S},\) assuming \(120 \%\) stoichiometric combustion.

Sodium hydrogen carbonate is also known as baking soda. When this compound is heated, it decomposes to sodium carbonate, carbon dioxide, and water vapor. Write the balanced equation for this reaction. What volume (in liters) of carbon dioxide gas at \(77^{\circ} \mathrm{C}\) and \(756 \mathrm{mmHg}\) will be produced from \(26.8 \mathrm{~g}\) of sodium hydrogen carbonate?

Hydrogen has two stable isotopes, \({ }^{1} \mathrm{H}\) and \({ }^{2} \mathrm{H}\), with atomic weights of 1.0078 amu and 2.0141 amu, respectively. Ordinary hydrogen gas, \(\mathrm{H}_{2}\), is a mixture consisting mostly of \({ }^{1} \mathrm{H}_{2}\) and \({ }^{1} \mathrm{H}^{2} \mathrm{H}\). Calculate the ratio of rates of effusion of \({ }^{1} \mathrm{H}_{2}\) and \({ }^{1} \mathrm{H}^{2} \mathrm{H}\) under the same conditions.

A McLeod gauge measures low gas pressures by compressing a known volume of the gas at constant temperature. If \(315 \mathrm{~cm}^{3}\) of gas is compressed to a volume of 0.0457 \(\mathrm{cm}^{3}\) under a pressure of \(2.51 \mathrm{kPa}\), what was the original gas pressure?

Calculate the ratio of rates of effusion of \({ }^{235} \mathrm{UF}_{6}\) and \({ }^{238} \mathrm{UF}_{6}\), where \({ }^{235} \mathrm{U}\) and \({ }^{238} \mathrm{U}\) are isotopes of uranium. The atomic weights are \({ }^{235} \mathrm{U}, 235.04 \mathrm{amu} ;{ }^{238} \mathrm{U}, 238.05 \mathrm{amu} ;{ }^{19} \mathrm{~F}\) (the only naturally occurring isotope), 18.998 amu. Carry five significant figures in the calculation.
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