Question

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

Short answer

The work involved in this process is approximately \(W = -50.66 \mathrm{~J}\). This indicates that the lungs do work on the surroundings as they expand against the external pressure during a normal breath.

Step-by-step solution

Convert given values to SI units

To convert the given volume change and external pressure to SI units, we use the following conversion factors: 1 L = 0.001 m³ 1 atm = 101325 Pa Converting 0.50 L to cubic meters: \[0.50 \mathrm{~L} \times \frac{1 \mathrm{~m}^3}{1000 \mathrm{~L}} = 0.0005 \mathrm{~m}^3\] Converting 1.0 atm to pascals: \[1.0 \mathrm{~atm} \times 101325 \frac{\mathrm{Pa}}{1 \mathrm{~atm}} = 101325 \mathrm{~Pa}\]

Calculate the work done using the formula

Now that we have the volume change and external pressure in their proper units, we can use the formula to calculate the work done: \[W = -P \times \Delta V \] Substitute the given values: \[W = - (101325 \mathrm{~Pa})\times (0.0005 \mathrm{~m}^3)\]

Calculate the result

Now, we calculate the work done: \begin{align*} W &= - (101325 \mathrm{~Pa}) \times (0.0005 \mathrm{~m}^3) \\ &= -50.6625 \mathrm{~J} \end{align*} Since the work done is negative, it means that the lungs do work on the surroundings, as they expand against the external pressure. So, there are approximately \(50.66 \mathrm{~J}\) of work involved in this process.

Key concepts

Work Done in Thermodynamics

In thermodynamics, work done refers to the energy transferred when a system changes its volume under an external pressure. Work can be either positive or negative, depending on the direction of the transfer.

• If a system does work on its surroundings (like a gas expanding), the work done is considered negative.
• If work is done on the system (like compressing a gas), the work is positive.

In the context of the exercise, when the lungs expand, they exert a force against the atmospheric pressure. This expansion involves doing work against that pressure. The formula used to calculate this is:\[ W = -P \times \Delta V \]where \(W\) is the work done, \(P\) is the pressure, and \(\Delta V\) is the change in volume. The negative sign indicates work done by the system on the surroundings. This concept is crucial for understanding various processes in science and engineering.

Pressure-Volume Work

Pressure-volume work is a key concept in thermodynamics, particularly in processes where gases are involved. It focuses on how a force exerted by pressure results in a volume change.For this type of work, it's important to remember:

• The pressure must be constant for the calculation to be straightforward. Otherwise, integration is needed for varying pressures.
• Volume change \(\Delta V\) is crucial as it determines the extent of the work.

In our exercise scenario, a constant pressure (1 atm converted to Pa) is applied to the lungs as they expand, performing work on the surroundings. The equation used is simple yet foundational, reflecting how systems in thermodynamics interact with their environment through volume changes under pressure.

SI Units Conversion

Understanding and converting to International System of Units (SI) is essential for solving physics problems effectively. Often, measurements are initially given in non-SI units and require conversion.In the given exercise:

• The volume change was provided in liters. The conversion to cubic meters involves the factor \(1 \text{ L} = 0.001 \text{ m}^3\).
• The pressure was given in atmospheres, converted to pascals, using the conversion factor \(1 \text{ atm} = 101325 \text{ Pa}\).

Conversion to SI units ensures consistency and accuracy when applying formulas. This practice is a standard approach in science, enabling universally understandable calculations that are necessary for engaging with the global scientific community. Mastering these conversions helps avoid errors and simplifies complex problem-solving efforts.