Question

If pressure is expressed in atmospheres and volume is expressed in liters, \(P \Delta V\) has units of \(L\) atm (liters \(x\) atmospheres). In Chapter 10 you learned that 1 atm \(=\) \(101,325 \mathrm{~Pa}\), and in Chapter 1 you learned that \(1 \mathrm{~L}=\) \(1 \mathrm{dm}^{3}\). Use this information to determine the number of joules corresponding to \(1 \mathrm{~L}\) atm.

Short answer

1 L atm is equivalent to 101.325 Joules.

Step-by-step solution

Convert Liters to Cubic Meters

Since 1 liter is equal to 1 cubic decimeter (dm^3), and there are 1000 dm^3 in 1 cubic meter (m^3), you convert liters to cubic meters by using the conversion factor: 1 L = 1 dm^3 = 0.001 m^3.

Convert Atmospheres to Pascals

The pressure is given in atmospheres, and it is known that 1 atmosphere is equivalent to 101,325 Pa (Pascals). Therefore, to convert atmospheres to Pascals, you multiply the atmospheric pressure by 101,325 Pa/atm.

Calculate Work Done in Joules

Work done can be calculated as the product of pressure and change in volume. With the conversions made, the product of 1 L atm will be equivalent to the product of 0.001 m^3 (Volume in cubic meters) and 101,325 Pa (Pressure in Pascals). The product gives us the work done in Joules since 1 Joule is equivalent to 1 Pa * m^3.

Perform the Multiplication

Now, multiply the volume in cubic meters with the pressure in Pascals: Work (Joules) = 0.001 m^3 * 101,325 Pa = 101.325 Joules.

Key concepts

Pressure Volume Work

Understanding the concept of pressure volume work is key to grasping various scientific principles, particularly within thermodynamics, where it is frequently encountered. Pressure volume work, often symbolized as \( P\text{\textDelta}V \), involves the work done by or on a system when the volume of a system changes in the presence of an external pressure.

When a gas expands in a chamber, it does work on the surroundings, pushing the piston outwards against the ambient pressure. Conversely, if the gas is compressed, the surroundings do work on the gas. The calculation of this work is straightforward when dealing with ideal cases: work done is the product of the external pressure and the change in volume, \( W = P\text{\textDelta}V \). However, it is essential to ensure that pressure and volume are in consistent units when performing this calculation - in this case, Pascals for pressure and cubic meters for volume.

Units Conversion

Units conversion is an essential skill in science and everyday life, as it allows for the standardization of measurements and the comparison of values. In our exercise, liters (L), typically used to measure volume, and atmospheres (atm), a unit of pressure, are converted to cubic meters (m³) and Pascals (Pa), respectively.

This conversion is crucial because the Joule, the SI unit of work, is defined as one Pascal times one cubic meter (\(1 J = 1 Pa \times 1 m^3\)). Thus, to work with Joules, we must express our measurements in the correct units. It's beneficial to remember common conversion factors, like \(1 L = 0.001 m^3\) and \(1 atm = 101,325 Pa\). Keeping track of units and ensuring correct conversions can prevent errors and confusion in scientific calculations.

Gas Laws

Gas laws are fundamental principles that describe the behavior of gases and how they respond to changes in temperature, volume, and pressure. Understanding these laws is vital for handling exercises involving work done by gases.

There are several basic gas laws, with Boyle's Law stating that pressure and volume are inversely proportional at a constant temperature, and Charles's Law describing how gases expand when heated. The combined gas law puts these relationships together, making it possible to predict the behavior of a gas when multiple properties change. For energy calculations and problems involving pressure volume work, these laws help us understand how a gas might behave under different conditions. In our exercise, understanding the relationship between pressure and volume would help illuminate why work can be calculated with the simple equation \( W = P\text{\textDelta}V \).

Energy Calculations

Energy calculations are crucial in various scientific and engineering fields to understand how systems operate and use energy. In the context of our exercise, these calculations are used to determine the work done during the expansion or compression of a gas.

The concept of work in a physics sense involves a force causing displacement. When a gas expands against a force (like the atmospheric pressure), it does work on its surroundings, converting its own internal energy into work. The fundamental formula \( W = P\text{\textDelta}V \) provides a means to calculate the energy transferred in the process of this work. Remembering that work and energy are interchangeable terms in physics (both measured in Joules), it helps to emphasize the role of energy conservation in these calculations. The ability to calculate energy transfer accurately is vital for applications ranging from engine efficiency to the thermodynamics of the atmosphere.