Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

A volume \(v\) of gas is confined in a cylinder, one end of which is closed by a movable piston. If \(A\) is the area in square inches of the face of the piston and \(x\) is the distance in inches from the cylinder head to the piston, then \(v=A x .\) The pressure of the confined gas is a continuous function \(p\) of the volume, and \(p(v)=p(A x)\) will be denoted by \(f(x)\). Show that the work done by the piston in compressing the gas from a volume \(v_{1}=A x_{1}\) to a volume \(v_{2}=A x_{2}\) is $$ W=A \int_{x_{2}}^{x_{1}} f(x) d x $$

Step-by-Step Solution

Answer

The work done is given by \( W = A \int_{x_2}^{x_1} f(x) \, dx \).

Step 1: Define the Concept of Work

Work done by a force is calculated as the integral of the force applied over the distance through which the force is applied. In this case, the force is exerted by the pressure of the gas on the piston, and the work done is due to the movement of the piston.

Step 2: Express Pressure as a Function of Distance

Consider the pressure of the gas as a function of volume, which can be written as a function of distance, i.e., \( p = f(x) \). This is given by the problem as \( f(x) = p(Ax) \).

Step 3: Relate Force to Pressure and Area

The force exerted by the gas on the piston is the product of the pressure and the area of the piston, \( F = p imes A \). Therefore, \( F = f(x) imes A \).

Step 4: Set Up the Work Integral

The work done, \( W \), can be calculated by integrating the force over the distance that the force is applied. Here, the distance is the path covered by the piston, from \( x_1 \) to \( x_2 \). Thus, the integral setup becomes: \[ W = \int_{x_2}^{x_1} F \, dx = \int_{x_2}^{x_1} A f(x) \, dx \]

Step 5: Simplify the Integral Expression

Since the area \( A \) is constant, it can be factored out of the integral:\[ W = A \int_{x_2}^{x_1} f(x) \, dx \] This expression gives us the work done by the piston when compressing the gas from volume \( v_1 = Ax_1 \) to volume \( v_2 = Ax_2 \).

Key Concepts

Volume and Pressure Relationship

In the context of a gas confined within a cylinder with a movable piston, understanding the relationship between volume and pressure is key. When the piston moves, it changes the volume inside the cylinder. This change in volume directly impacts the pressure of the gas. According to Boyle's Law, for a given amount of gas at constant temperature, the pressure of the gas is inversely proportional to its volume. This means that as the volume decreases, the pressure increases, and vice versa.

This relationship is crucial for understanding how a piston works. The gas's pressure causes it to exert force on the piston, leading to compression or expansion, depending on the direction of movement. Thus, understanding this relationship helps us predict how changes in one variable (volume) can affect the other (pressure). As the gas is compressed by the piston, this changing dynamic plays a pivotal role in the work done by the system.

Force Exerted by Gas

The force exerted by the gas on the piston is determined by its pressure and the area over which this pressure is applied. The formula for calculating this force is straightforward: the force (\( F \) ) is the product of pressure (\( p \)) and the piston's area (\( A \)). So, \( F = p \cdot A \).

This concept is rooted in the basic understanding that pressure is defined as force per unit area. Therefore, rearranging the relation gives us the force when the pressure and area of a surface are known. This is extremely important because this force is responsible for the movement of the piston.

• Pressure \( p \) is obtained from the behavior of the gas as a function of volume and subsequently as \( f(x) \) when considering distance.
• The piston's face area \( A \) remains constant throughout the process.

Understanding this helps in effectively setting up the integral to find the work done by the piston.

Pressure as a Function of Volume

In this scenario, pressure is not simply a constant but a function of volume, iterated as \( f(x) \). When the volume changes, we see that pressure changes in correspondence. This relationship can be expressed as \( p = f(v) \), converting it to \( f(x) \) by relating volume directly to the piston's position in terms of distance.

To understand how pressure changes, we look at how it varies as a function of the piston's position along the cylinder. This dependency is crucial because it provides a way to calculate work done when considering variable pressure over the movement path of the piston.

• Challenges arise when dealing with real-life gases, as they may not always behave ideally, leading to more complex functions.
• However, in many practical applications and problems, assumptions of ideal behavior are made to simplify calculations.

Understanding how pressure functions over distance allow us to create meaningful calculations of work done by the piston during its movement.

Integration of Force Over Distance

To find the work done by a piston compressing gas, we utilize the integral of force over the distance the piston travels. It's important to remember that work is the product of force and the distance over which that force is applied. When force varies because of changing pressure, the situation requires integration.

The work done, denoted as \( W \), is calculated by integrating the force exerted by the piston from one position to another. Mathematically, this is represented as:\[ W = A \int_{x_2}^{x_1} f(x) \, dx\]In this integral:

• \( A \), the area of the piston, remains constant.
• \( f(x) \) is the variable force exerted by the gas depending on distance \( x \).
• This integration captures the cumulated effect of force applied across the range of motion from \( x_2 \) to \( x_1 \), accounting for all changes in pressure.

By understanding how to set up this integral, one can accurately determine how much work is done by the piston during compression, providing critical insights into thermodynamic processes.