/*! 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 15 Suppose that a woman weighing \(... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose that a woman weighing \(130 \mathrm{lb}\) and wearing highheeled shoes momentarily places all her weight on the heel of one foot. If the area of the heel is 0.50 in. \(^{2}\), calculate the pressure exerted on the underlying surface in (a) kilopascals, (b) atmospheres, and (c) pounds per square inch.

Short Answer

Expert verified
The pressure exerted on the underlying surface by the woman's heel is approximately 260 psi, 1790.6 kPa, and 17.69 atm.

Step by step solution

01

Finding Pressure in Pounds per Square Inch

First, we will find the pressure exerted by the heel in pounds per square inch (psi). All our given values are already in the right units (lbs for weight and in² for area). Given: Weight (force) F = 130 lb Area A = 0.50 in² Calculate pressure P = F / A P (in psi) = (130 lb) / (0.50 in²) P (in psi) ≈ 260 psi
02

Conversion Factors

We need to convert the pressure from psi to kilopascals (kPa) and atmospheres (atm). The conversion factors are as follows: 1 psi = 6.89476 kPa 1 atm = 101.325 kPa 1 psi = 0.068046 atm (calculated by dividing 6.89476 kPa by 101.325 kPa)
03

Finding Pressure in Kilopascals

Now we will convert the pressure we found in psi to kilopascals (kPa) using the conversion factor: P (in kPa) = P (in psi) × 6.89476 kPa/psi P (in kPa) = 260 psi × 6.89476 kPa/psi P (in kPa) ≈ 1790.6 kPa
04

Finding Pressure in Atmospheres

Finally, we will convert the pressure we found in psi to atmospheres (atm) using the conversion factor: P (in atm) = P (in psi) × 0.068046 atm/psi P (in atm) = 260 psi × 0.068046 atm/psi P (in atm) ≈ 17.69 atm So, the pressure exerted on the underlying surface by the woman's heel is approximately 260 psi, 1790.6 kPa, and 17.69 atm.

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Key Concepts

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

Conversion Factors
Understanding conversion factors is crucial when dealing with different units of measurement. A conversion factor is a number used to change one set of units to another, by multiplying or dividing. In the context of pressure, we often need to convert between units like pounds per square inch (psi), kilopascals (kPa), and atmospheres (atm) to communicate measurements consistently and accurately.

For example, to convert psi to kPa, you would use the conversion factor of 1 psi equals approximately 6.89476 kPa. These factors are derived from the definitions of the units in terms of the base SI or metric units. Thus, knowing the correct conversion factor allows for quick and easy calculations across different measurement systems.
Kilopascals
The kilopascal (kPa) is a unit of pressure in the International System of Units (SI). One kilopascal equals 1,000 pascals and is based on the pascal, which is defined as one newton per square meter. The prefix 'kilo-' simply means 'thousand', so a kilopascal is one thousand pascals.

Common Uses

Kilopascals are commonly used in scientific contexts, as well as in industries like automotive to measure tire pressure, or meteorology for atmospheric pressure readings. When converting from the more commonly used psi, remember that it is a straightforward multiplication using the conversion factor. This importance in various fields emphasizes why understanding how to calculate and convert to kilopascals is a valuable skill.
Atmospheres
An atmosphere (atm) is another unit for measuring pressure. It is based on the average atmospheric pressure at sea level on Earth. One atmosphere is defined as the pressure that can support a column of mercury 760 millimeters high in a mercury barometer at sea level and at a temperature of 0 degrees Celsius.

Relation to Other Units

One atmosphere equals 101.325 kPa, which is derived from precise scientific measurement of air pressure. In practical calculations, we may use the conversion factor of 1 psi being equal to 0.068046 atm to change psi into atmospheres or vice versa. These figures are immensely important when working with systems where the atmospheric pressure reference is the standard, such as in scuba diving and aviation.
Pounds per Square Inch
Pounds per square inch, or psi, is a unit of pressure widely used in the United States and some other parts of the world. It reflects the force in pounds exerted on one square inch of area.

Calculating Pressure in PSI

In the given problem, we calculated the pressure exerted by the heel by dividing the weight, in pounds, by the area of the heel in square inches. This calculation gives us a direct reading in psi, a unit that is very intuitive for day-to-day use in various applications like monitoring car tire pressure or the pressure in hydraulic systems.

Being able to switch between psi and metric units ensures that pressures can be compared and understood in multiple contexts, expanding the understanding and communication of pressure across different industries and countries.

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Most popular questions from this chapter

Propane, \(\mathrm{C}_{3} \mathrm{H}_{8}\), liquefies under modest pressure, allowing a large amount to be stored in a container. (a) Calculate the number of moles of propane gas in a 110 -L container at 3.00 atm and \(27^{\circ} \mathrm{C} .\) (b) Calculate the number of moles of liquid propane that can be stored in the same volume if the density of the liquid is \(0.590 \mathrm{~g} / \mathrm{mL} .\) (c) Calculate the ratio of the number of moles of liquid to moles of gas. Discuss this ratio in light of the kinetic-molecular theory of gases.

Consider a mixture of two gases, \(A\) and \(B\), confined in a closed vessel. A quantity of a third gas, \(C,\) is added to the same vessel at the same temperature. How does the addition of gas \(\mathrm{C}\) affect the following: (a) the partial pressure of gas \(A,\) (b) the total pressure in the vessel, \((\mathbf{c})\) the mole fraction of gas \(\mathrm{B} ?\)

(a) Calculate the number of molecules in a deep breath of air whose volume is \(2.25 \mathrm{~L}\) at body temperature, \(37{ }^{\circ} \mathrm{C},\) and a pressure of 735 torr. (b) The adult blue whale has a lung capacity of \(5.0 \times 10^{3} \mathrm{~L} .\) Calculate the mass of air (assume an average molar mass \(28.98 \mathrm{~g} / \mathrm{mol}\) ) contained in an adult blue whale's lungs at \(0.0{ }^{\circ} \mathrm{C}\) and 1.00 atm, assuming the air behaves ideally.

Suppose you are given two 1 -L flasks and told that one contains a gas of molar mass \(30,\) the other a gas of molar mass 60 , both at the same temperature. The pressure in flask \(\mathrm{A}\) is \(\mathrm{X}\) atm, and the mass of gas in the flask is \(1.2 \mathrm{~g} .\) The pressure in flask \(\mathrm{B}\) is \(0.5 \mathrm{X}\) atm, and the mass of gas in that flask is \(1.2 \mathrm{~g} .\) Which flask contains the gas of molar mass \(30,\) and which contains the gas of molar mass \(60 ?\)

Natural gas is very abundant in many Middle Eastern oil fields. However, the costs of shipping the gas to markets in other parts of the world are high because it is necessary to liquefy the gas, which is mainly methane and has a boiling point at atmospheric pressure of \(-164{ }^{\circ} \mathrm{C}\). One possible strategy is to oxidize the methane to methanol, \(\mathrm{CH}_{3} \mathrm{OH},\) which has a boiling point of \(65^{\circ} \mathrm{C}\) and can therefore be shipped more readily. Suppose that \(10.7 \times 10^{9} \mathrm{ft}^{3}\) of methane at atmospheric pressure and \(25^{\circ} \mathrm{C}\) is oxidized to methanol. (a) What volume of methanol is formed if the density of \(\mathrm{CH}_{3} \mathrm{OH}\) is \(0.791 \mathrm{~g} / \mathrm{mL} ?\) (b) Write balanced chemical equations for the oxidations of methane and methanol to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\). Calculate the total enthalpy change for complete combustion of the \(10.7 \times 10^{9} \mathrm{ft}^{3}\) of methane just described and for complete combustion of the equivalent amount of methanol, as calculated in part (a). (c) Methane, when liquefied, has a density of \(0.466 \mathrm{~g} / \mathrm{mL} ;\) the density of methanol at \(25^{\circ} \mathrm{C}\) is \(0.791 \mathrm{~g} / \mathrm{mL}\). Compare the enthalpy change upon combustion of a unit volume of liquid methane and liquid methanol. From the standpoint of energy production, which substance has the higher enthalpy of combustion per unit volume?

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