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Compressed air in aerosol cans is used to free electronic equipment of dust. Does the air do any work as it escapes from the can?

Short Answer

Expert verified
Yes, the air does work when it escapes from the can because it is displaced and moves into an area of lower pressure.

Step by step solution

01

Concept understanding

One must first understand the definition of work in physics. Work is done when a force applied on an object is moved. Hence, it involves two key things force and displacement.
02

Applying the concept

Secondly, when compressed air is released from an aerosol can, it quickly disperses or moves to an area of lower pressure. Looking back at our definition, we clearly have displacement or movement.
03

Deriving the answer

Given the displacement, we can therefore conclude that work has been done.

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Key Concepts

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

Work in Physics
In physics, the concept of work is a fundamental aspect that describes how energy is transferred when a force is applied to move an object. If you can picture pushing a box across the floor or lifting a book off a table, you're already familiar with the basic idea. But, there's a bit more to it:

  • Definition: Work is done when a force causes an object to move in the direction of the force.
  • Formula: It can be mathematically defined as: \( W = F \cdot d \cdot \cos(\theta) \), where \( W \) is the work done, \( F \) is the force applied, \( d \) is the displacement, and \( \theta \) is the angle between the force and direction of motion.
  • Units of Measurement: The unit of work in the SI system is the Joule (J), where 1 Joule is equivalent to 1 Newton-meter.

When air escapes from an aerosol can, it moves because of the force created by pressure difference. This movement signifies that work is being done, even if it's not always visible. The compressed air uses its internal pressure to perform work as it travels to lower pressure regions. This concept is crucial in understanding how energy and forces interact in various systems.
Pressure and Displacement
Pressure and displacement are closely linked concepts that explain a lot about how gases and liquids behave.

When air rushes out of an aerosol can, it's all about pressure:
  • Pressure: Inside the aerosol can, air is under high pressure. Pressure itself is defined as force per unit area, described mathematically as \( P = \frac{F}{A} \), with \( P \) being pressure, \( F \) the force, and \( A \) the area.
  • Displacement: The movement or dispersion of the air towards a region of lower pressure is referred to as displacement. In this context, displacement is crucial because it's what allows work to happen.

As air displaces, it demonstrates how these two elements of pressure and displacement work together. The air naturally seeks equilibrium, moving from an area of higher pressure inside the can to an area of lower pressure outside. This natural movement of air particles results in displacement. Thus, the principles of pressure and displacement not only explain the behavior of the escaping air but also validate the work that is done in this process. Understanding this helps us see the invisible forces at play in common situations.
Aerosol Cans
Aerosol cans are an interesting application of thermodynamics at work in everyday life. They encapsulate the concept of stored energy and controlled release perfectly:
  • Functionality: An aerosol can is essentially a sealed container with a pressurized liquid or gas. The pressure's primary role is to propel the substance out when the can's valve is opened.
  • Principles: They operate on the principle that when the nozzle is pressed, the pressure difference causes the propellant to push the liquid or gas out, transforming potential energy into kinetic energy.
  • Applications: Aerosol cans are used widely, from cleaning electronics to dispensing deodorants, because they distribute products evenly and efficiently.

Understanding the principles behind how aerosol cans function, including the interplay of pressure and displacement, enhances our appreciation of these everyday objects. Additionally, recognizing that work is involved when using an aerosol can underscores the idea that even simple actions engage complex physical principles. This insight aligns our everyday experiences with scientific theory, illustrating thermodynamics in action.

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Most popular questions from this chapter

The heat of solution of \(\mathrm{KI}(\mathrm{s})\) in water is \(+20.3 \mathrm{kJ} / \mathrm{mol}\) KI. If a quantity of KI is added to sufficient water at \(23.5^{\circ} \mathrm{C}\) in a Styrofoam cup to produce \(150.0 \mathrm{mL}\) of 2.50 M KI, what will be the final temperature? (Assume a density of \(1.30 \mathrm{g} / \mathrm{mL}\) and a specific heat of \(2.7 \mathrm{Jg}^{-1}\) \(\left.^{\circ} \mathrm{C}^{-1} \text {for } 2.50 \mathrm{M} \mathrm{KI} .\right)\)

There are other forms of work besides \(\mathrm{P}-\mathrm{V}\) work. For example, electrical work is defined as the potential \(x\) change in charge, \(w=\phi d q\). If a charge in a system is changed from \(10 \mathrm{C}\) to \(5 \mathrm{C}\) in a potential of \(100 \mathrm{V}\) and \(45 \mathrm{J}\) of heat is liberated, what is the change in the internal energy? (Note: \(1 \mathrm{V}=1 \mathrm{J} / \mathrm{C})\).

The combustion of methane gas, the principal constituent of natural gas, is represented by the equation $$\begin{aligned} \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+& 2 \mathrm{H}_{2} \mathrm{O}(1) \\ \Delta H^{\circ} &=-890.3 \mathrm{kJ} \end{aligned}$$ (a) What mass of methane, in kilograms, must be burned to liberate \(2.80 \times 10^{7} \mathrm{kJ}\) of heat? (b) What quantity of heat, in kilojoules, is liberated in the complete combustion of \(1.65 \times 10^{4} \mathrm{L}\) of \(\mathrm{CH}_{4}(\mathrm{g})\) measured at \(18.6^{\circ} \mathrm{C}\) and \(768 \mathrm{mmHg} ?\) (c) If the quantity of heat calculated in part (b) could be transferred with \(100 \%\) efficiency to water, what volume of water, in liters, could be heated from 8.8 to \(60.0^{\circ} \mathrm{C}\) as a result?

You are planning a lecture demonstration to illustrate an endothermic process. You want to lower the temperature of \(1400 \mathrm{mL}\) water in an insulated container from 25 to \(10^{\circ} \mathrm{C} .\) Approximately what mass of \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s})\) should you dissolve in the water to achieve this result? The heat of solution of \(\mathrm{NH}_{4} \mathrm{Cl}\) is \(+14.7 \mathrm{kJ} / \mathrm{mol} \mathrm{NH}_{4} \mathrm{Cl}\).

In the Are You Wondering \(7-1\) box, the temperature variation of enthalpy is discussed, and the equation \(q_{P}=\) heat capacity \(\times\) temperature change \(=C_{P} \times \Delta T\) was introduced to show how enthalpy changes with temperature for a constant-pressure process. Strictly speaking, the heat capacity of a substance at constant pressure is the slope of the line representing the variation of enthalpy (H) with temperature, that is $$C_{P}=\frac{d H}{d T} \quad(\text { at constant pressure })$$ where \(C_{P}\) is the heat capacity of the substance in question. Heat capacity is an extensive quantity and heat capacities are usually quoted as molar heat capacities \(C_{P, \mathrm{m}},\) the heat capacity of one mole of substance; an intensive property. The heat capacity at constant pressure is used to estimate the change in enthalpy due to a change in temperature. For infinitesimal changes in temperature, $$d H=C_{p} d T \quad(\text { at constant pressure })$$ To evaluate the change in enthalpy for a particular temperature change, from \(T_{1}\) to \(T_{2}\), we write $$\int_{H\left(T_{1}\right)}^{H\left(T_{2}\right)} d H=H\left(T_{2}\right)-H\left(T_{1}\right)=\int_{T_{1}}^{T_{2}} C_{P} d T$$ If we assume that \(C_{P}\) is independent of temperature, then we recover equation (7.5) $$\Delta H=C_{P} \times \Delta T$$ On the other hand, we often find that the heat capacity is a function of temperature; a convenient empirical expression is $$C_{P, \mathrm{m}}=a+b T+\frac{c}{T^{2}}$$ What is the change in molar enthalpy of \(\mathrm{N}_{2}\) when it is heated from \(25.0^{\circ} \mathrm{C}\) to \(100.0^{\circ} \mathrm{C} ?\) The molar heat capacity of nitrogen is given by$$C_{P, \mathrm{m}}=28.58+3.77 \times 10^{-3} T-\frac{0.5 \times 10^{5}}{T^{2}} \mathrm{JK}^{-1} \mathrm{mol}^{-1}$$

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