/*! 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 31 During a normal breath, our lung... [FREE SOLUTION] | 91Ó°ÊÓ

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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

At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat \(\mathrm{KClO}_{3} :\) $$2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g) \quad \Delta H=-89.4 \mathrm{kJ}$$ For this reaction, calculate \(\Delta H\) for the formation of (a) 1.36 \(\mathrm{mol}\) of \(\mathrm{O}_{2}\) and \((\mathbf{b}) 10.4 \mathrm{g}\) of \(\mathrm{KCl}\) (c) The decomposition of \(\mathrm{KClO}_{3}\) proceeds spontaneously when it is heated. Do you think that the reverse reaction, the formation of \(\mathrm{KClO}_{3}\) from \(\mathrm{KCl}\) and \(\mathrm{O}_{2},\) is likely to be feasible under ordinary conditions? Explain your answer.

Consider a system consisting of the following apparatus, in which gas is confined in one flask and there is a vacuum in the other flask. The flasks are separated by a valve. Assume that the flasks are perfectly insulated and will not allow the flow of heat into or out of the flasks to the surroundings. When the valve is opened, gas flows from the filled flask to the evacuated one. (a) Is work performed during the expansion of the gas? (b) Why or why not? (c) Can you determine the value of \(\Delta E\) for the process?

(a) What is meant by the term standard conditions with reference to enthalpy changes? (b) What is meant by the term enthalpy of formation? (c) What is meant by the term standard enthalpy of formation?

A 1.800 -g sample of phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right)\) was burned in a bomb calorimeter whose total heat capacity is 11.66 \(\mathrm{kJ} /^{\circ} \mathrm{C}\) The temperature of the calorimeter plus contents increased from 21.36 to \(26.37^{\circ} \mathrm{C}\) (a) Write a balanced chemical equation for the bomb calorimeter reaction. (b) What is the heat of combustion per gram of phenol? Per mole of phenol?

One of the best-selling light, or low-calorie, beers is 4.2\(\%\) alcohol by volume and a 12 -oz serving contains 110 Calories; remember: 1 Calorie \(=1000\) cal \(=1\) kcal. To estimate the percentage of Calories that comes from the alcohol, consider the following questions. (a) Write a balanced chemical equation for the reaction of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) , with oxygen to make carbon dioxide and water. (b) Use enthalpies of formation in Appendix \(\mathrm{C}\) to determine \(\Delta H\) for this reaction. \((\mathbf{c})\) If 4.2\(\%\) of the total volume is ethanol and the density of ethanol is \(0.789 \mathrm{g} / \mathrm{mL},\) what mass of ethanol does a 12 - oz serving of light beer contain? (\boldsymbol{d} ) How many Calories are released by the metabolism of ethanol, the reaction from part (a)? (e) What percentage of the 110 Calories comes from the ethanol?

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