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During a deep breath, our lungs expand about \(2.0 \mathrm{~L}\) against an external pressure of \(101.3 \mathrm{kPa}\). How much work is involved in this process (in J)?

Short Answer

Expert verified
The work done during this process is \(-202.6 \mathrm{J}\).

Step by step solution

01

Write the formula for work done

We will use the formula of mechanical work done by a gas in this situation. Work done (W) by a gas during the expansion is given by: W = -P × ΔV where P = external pressure, ΔV = change in volume. We have a negative sign because the work done by the gas on the surroundings is negative as the gas is expanding.
02

Substitute the given values

Now we will substitute the given values in the formula: P = 101.3 kPa = 101.3 × 10^3 Pa (converting kPa to Pa) ΔV = 2.0 L = 2.0 × 10^-3 m³ (converting L to m³) W = -101.3 × 10^3 Pa × 2.0 × 10^-3 m³
03

Calculate the work done

Now, find the work done by multiplying the given values. W = -101.3 × 10^3 × 2.0 × 10^-3 W = -202.6 J The work done during this process is -202.6 J. The negative sign indicates that the work is done by the body, expanding the lungs against the external pressure.

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Key Concepts

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

Work Done by Gas
When gas within an enclosed space is made to expand, it performs work against the surrounding environment. This is known as "work done by gas." Essentially, any time a gas pushes against its container, such as when you inflate a balloon or take a deep breath, it does work. The amount of work done is determined by the formula:
\[ W = -P \times \Delta V \]
In this formula, \(W\) represents the work done, \(P\) is the external pressure exerted upon the gas, and \(\Delta V\) is the change in volume of the gas. Notably, there is a negative sign in the equation. This negative sign indicates that the work executed by the gas is done against the environment, meaning the surroundings resist the expansion.
Bullets to help:
  • Work is done by the gas when it expands against an opposing pressure.
  • The negative sign in the formula indicates the direction of work (outward from the system).
  • This type of work is important in fields like engineering and biology, such as measuring lung function or engine efficiency.
Pressure-Volume Work
Pressure-volume work refers to the work done as a result of the pressure acting over a volume change in an elastic container, like our lungs or a piston. Whenever there is a change in volume while a constant pressure is applied, pressure-volume work comes into play.
The formula that governs this is essentially the same:\[ W = -P \times \Delta V \]With pressure-volume work, understanding the relationship between pressure and volume, known as Boyle's Law, can be very helpful:
  • \(P_1 \times V_1 = P_2 \times V_2\)
  • This law underscores that pressure and volume are inversely proportional, meaning if volume goes up, pressure must decrease to maintain the same amount of energy (and vice-versa).
    For further comprehension:
    • Pressure-volume work is essential in many natural and engineered systems.
    • Understanding it assists in designing more efficient machines and devices, like motors or lung ventilators.
    • It illustrates the mechanical energy transferred in processes involving gas expansion or compression.
    Conversion of Units
    To accurately calculate and understand processes like work done by gas, you often need to convert units to a consistent system. For instance, pressures are often expressed in kilopascals (kPa) while volumes might be in liters (L). However, our mathematical formulas require uniform units, typically Pascals (Pa) for pressure and cubic meters (m³) for volume, in the SI unit system.
    Here's a guide on how to carry out these conversions:
    • To convert kPa to Pa: Multiply by \(10^3\). For example, \(101.3\, \text{kPa} = 101,300\, \text{Pa}\).
    • To convert liters to cubic meters: Multiply by \(10^{-3}\). For example, \(2.0\, \text{L} = 2.0 \times 10^{-3}\, \text{m}^3\).
    Using consistent units
    • Ensures accuracy in calculations and helps in understanding the relationship between different physical quantities.
    • Is crucial for scientific research and standardized industrial practices.
    This practice of unit consistency is vital to avoid errors and misinterpretations.

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    Most popular questions from this chapter

    One of the best-selling light, or low-calorie, beers is \(4.2 \%\) alcohol by volume and a 355 -mL serving contains 110 Calories; remember: 1 Calorie \(=1000 \mathrm{cal}=1 \mathrm{kcal} .\) 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 \(355-\mathrm{mL}\) serving of light beer contain? (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?

    (a) Why does the standard enthalpy of formation of both the very reactive fluorine \(\left(\mathrm{F}_{2}\right)\) and the almost inert gas nitrogen \(\left(\mathrm{N}_{2}\right)\) both read zero? (b) Write the chemical equation for the reaction whose enthalpy change is the standard enthalpy of formation of naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\).

    Assume that 2 moles of water are formed according to the following reaction at constant pressure \((101.3 \mathrm{kPa})\) and constant temnerature \((298 \mathrm{~K});\) $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) $$ (a) Calculate the pressure-volume work for this reaction. (b) Calculate \(\Delta E\) for the reaction using your answer to (a).

    The air bags that provide protection in automobiles in the event of an accident expand because of a rapid chemical reaction. From the viewpoint of the chemical reactants as the system, what do you expect for the signs of \(q\) and \(w\) in this process?

    (a) Derive an equation to convert the specific heat of a pure substance to its molar heat capacity. (b) The specific heat of aluminum is \(0.9 \mathrm{~J} /(\mathrm{g} \cdot \mathrm{K}) .\) Calculate its molar heat capacity. (c) If you know the specific heat of aluminum, what additional information do you need to calculate the heat capacity of a particular piece of an aluminum component?

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