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During a normal breath, our lungs expand about 0.50 L against an external pressure of 1.0 atm. How much work is involved in this process (in J)?

Short Answer

Expert verified
The work involved in this process is approximately -50.66 J (in Joules).

Step by step solution

01

Convert volume to SI units

We are given the volume expansion in liters (L) and need to convert it to cubic meters (m³) which is the SI unit for volume. To do so, we will use the conversion factor: 1 L = 0.001 m³ \newline \(\Delta V\) = 0.50 L \newline \(0.50 L \times \frac{0.001 m³}{1 L} \) \newline \(\Delta V = 0.0005 m³\)
02

Convert pressure to SI units

We are given the pressure in atmospheres (atm) and need to convert it to pascals (Pa) which is the SI unit for pressure. To do so, we will use the conversion factor: 1 atm = 101325 Pa \newline \(P =\) 1.0 atm \newline \(1.0 atm \times \frac{101325 Pa}{1 atm}\) \newline \(P = 101325 Pa\)
03

Calculate the work done

Now that we have the volume and pressure in SI units, we can apply the formula for calculating the work done against a constant external pressure: \newline \(W = -P \Delta V\) \newline Substituting the values, we get: \newline \(W = -(101325 Pa)(0.0005 m³)\) \newline \(W = -50.6625 J\) Since the work done is negative, it means that the work is done by the body to expand the lungs. Therefore, the work involved in this process is approximately -50.66 J (in Joules).

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Key Concepts

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

Work Done
When discussing work done, especially in thermodynamics, it refers to the amount of energy transferred when a force causes an object to move or a system to change volume. In the context of gas expansion, like our lung example, work is done when the gas within the lungs expands against an external pressure. This concept of work is crucial because it describes energy changes in thermodynamic processes.

The formula to calculate work done (\( W \)) in a system where the volume changes at constant pressure is:
  • \( W = -P \, \Delta V \)
Let's break this down:
  • The negative sign indicates that when a system expands, it does work against the external force, which decreases the system's energy.
  • \( P \) is the external pressure that the substance is expanding against.
  • \( \Delta V \) represents the change in volume of the gas. It's the final volume minus the initial volume.
When performing calculations, ensure that units are correct; this often means converting to SI units. For example, pressure in pascals (Pa) and volume in cubic meters (m³). Remember, positive work is done on the system, while negative work indicates work done by the system.
Expansion of Gases
Expansion of gases is a fundamental concept in thermodynamics and occurs when the gas increases its volume. This can happen due to heating or, as in our lung example, through a pressure difference.

In a controlled expansion, like breathing, the gas (air) inside the lungs expands because the muscles around the lungs contract and create a space for expansion. This expansion has significant implications:
  • Thermodynamics describes this as an increase in volume with corresponding pressure changes. The pressure usually decreases as the volume increases for ideal gases if the temperature stays constant (Boyle's Law).
  • Energy is required for this expansion, demonstrated by the work done, as the gas pushes against external pressure.
Understanding gas expansion helps in fields such as meteorology, engineering, and health sciences, where scientists often predict how gases will behave under different conditions. Moreover, expansion illustrates practical applications of the conservation of energy principles.
SI Units Conversion
In science, especially physics and engineering, the International System of Units (SI units) provides consistency for measurements, ensuring that everyone speaks the same language worldwide. Converting measurements to SI units is crucial for correct and universally understood calculations.

For the problem of gas expansion in the lungs:
  • Volume is converted from liters (\( L \)) to cubic meters (\( m^3 \)):\( 1 \, \text{L} = 0.001 \, m^3 \)
  • Pressure is converted from atmospheres (\( atm \)) to pascals (\( Pa \)):\( 1 \, atm = 101325 \, Pa \)
Converting these units correctly ensures reliable input into equations like the work done formula:
  • Accurate conversions prevent miscalculations that could lead to incorrect interpretations of physical phenomena.
  • Inconsistent units can derail engineering projects or scientific experiments since the results hinge on precise measurements.
Always take the time to verify your units. It can be the difference between success and error in your mathematical conclusions.

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Most popular questions from this chapter

(a) According to the first law of thermodynamics, what quantity is conserved? (b) What is meant by the internal energy of a system? (c) By what means can the internal energy of a closed system increase?

A coffee-cup calorimeter of the type shown in Figure 5.18 contains 150.0 g of water at \(25.1^{\circ} \mathrm{C} .\) A \(121.0-\mathrm{g}\) block of copper metal is heated to \(100.4^{\circ} \mathrm{C}\) by putting it in a beaker of boiling water. The specific heat of \(\mathrm{Cu}(s)\) is \(0.385 \mathrm{J} / \mathrm{g}-\mathrm{K}\) . The Cu is added to the calorimeter, and after a time the contents of the cup reach a constant temperature of \(30.1^{\circ} \mathrm{C}\) (a) Determine the amount of heat, in J, lost by the copper block. (b) Determine the amount of heat gained by the water. The specific heat of water is \(4.18 \mathrm{J} / \mathrm{g}-\mathrm{K}\) . (c) The difference between your answers for (a) and (b) is due to heat loss through the Styrofoam cups and the heat necessary to raise the temperature of the inner wall of the apparatus. The heat capacity of the calorimeter is the amount of heat necessary to raise the temperature of the apparatus (the cups and the stopper) by 1 K. Calculate the heat capacity of the calorimeter in J/K. (d)What would be the final temperature of the system if all the heat lost by the copper block were absorbed by the water in the calorimeter?

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Assume that the following reaction occurs at constant pressure: $$2 \mathrm{Al}(s)+3 \mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{AlCl}_{3}(s)$$ (a) If you are given \(\Delta H\) for the reaction, what additional information do you need to determine \(\Delta E\) for the process? (b) Which quantity is larger for this reaction? (c) Explain your answer to part (b).

Write balanced equations that describe the formation of the following compounds from elements in their standard states, and then look up the standard enthalpy of formation for each substance in Appendix C: (a) \(\mathrm{H}_{2} \mathrm{O}_{2}(g),(\mathbf{b}) \mathrm{CaCO}_{3}(s)\) (c) \(\mathrm{POCl}_{3}(l),(\mathbf{d}) \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l) .\)

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