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 $$ W=A \int_{x_{2}}^{x_{1}} f(x) d x $$ Hint: The total force on the face of the piston is \(p(v) \cdot A=\) \(p(A x) \cdot A=A \cdot f(x) .\)

Step-by-Step Solution

Answer

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

Step 1: Understand the problem

First, we need to understand that the work done by the piston in compressing the gas is the integral of the force applied by the piston as it moves from one position to another. The force is expressed as the pressure of the gas times the area of the piston.

Step 2: Identify relevant equations

We are given that the volume of the gas is defined as \( v = A x \), where \( A \) is the area of the piston and \( x \) is the position of the piston. The pressure as a function of volume is \( p(v) = f(x) \) and the total force on the piston is \( A \cdot f(x) \).

Step 3: Express work done in terms of force and displacement

The work done \( W \) by the piston can be expressed as the integral of force over displacement: \[ W = \int_{x_2}^{x_1} A \cdot f(x) \cdot dx \] where \( A \cdot f(x) \) represents the changing force as a function of \( x \), and \( dx \) is the infinitesimal displacement.

Step 4: Integrate the expression

Since \( A \) is constant with respect to \( x \), you can take \( A \) out of the integral: \[ W = A \int_{x_2}^{x_1} f(x) \; dx \]. This represents the work done due to the fluid compression from \( x_1 \) to \( x_2 \).

Step 5: Conclude the expression

The derived expression \( W = A \int_{x_2}^{x_1} f(x) \; dx \) reflects the total work done by the piston in compressing the gas as the force, which depends on the volume, is applied over the displacement between \( x_1 \) and \( x_2 \).

Key Concepts

Integration

Integration is a fundamental concept in calculus, crucial for calculating areas, volumes, and understanding changes over time. In our context, integration helps us calculate the work done by the piston compressing gas. When a variable force, such as pressure, acts over a distance, integration allows us to sum the infinitesimal bits of work done over that distance.In the exercise, integration is used to add up all the small amounts of work (\( A \cdot f(x) \cdot dx \)) done by the piston as it moves from position \( x_2 \) to \( x_1 \). By integrating, we can find the total work, hence the integral sign \( \int \) is crucial as it signifies the accumulation process.Think of integration as the calculus tool that turns a continuous variation into a total quantity.

Work Done

Work done in physics is defined as the force applied multiplied by the distance over which it acts. In this scenario, we consider the work to compress the gas inside the cylinder. By moving the piston, we apply force against the gas’s pressure, doing work against it.The expression \( W = A \int_{x_2}^{x_1} f(x) \; dx \) tells us this concept: \( W \) represents work. \( A \cdot f(x) \) stands for the force exerted by the piston, with \( A \) as area and \( f(x) \) representing dynamic pressure. \( dx \) denotes an infinitesimal distance traveled by the piston.This formula highlights how small changes add up to the total work done, where force can vary based on piston position.

Pressure

In this exercise, pressure refers to the force applied per unit area on the piston by the confined gas. Pressure is a pivotal concept here, as the force exerted by the gas on the piston is determined by its pressure. This force drives the piston as it compresses the gas, changing its volume.Pressure as a function, denoted \( f(x) = p(Ax) \), indicates how it dynamically alters with changes in the piston's position. The greater the pressure, the higher the force, and vice versa.Understanding how pressure varies with volume or position is key to solving problems involving gases and pistons, as this defines the required work to compress the gas from one state to another.

Continuous Function

A continuous function has no breaks, jumps, or undefined points over its domain. In our scenario, the pressure \( p(v) \) is a continuous function \( f(x) \), a crucial assumption as it implies that pressure smoothly varies with volume.The continuous nature of \( f(x) \) ensures that work calculations via integration are valid and interpretable. Without continuity, predictions about the total work done would lack precision, as the assumption is required to ensure that \( f(x) \) behaves predictably across all volumes.In calculus, continuous functions play a key role because they allow us to apply limits, derivatives, and integrals confidently, knowing that the behavior of the function remains steady throughout its defined range.