Problem 51 - Calculus: Early Transcendental Functions

Question

For a particular ideal gas at constant temperature, pressure \(P\) and volume \(V\) are related by \(P V=10 .\) The work required to increase the volume from \(V=2\) to \(V=4\) is given by the integral \(\int_{2}^{4} P(V) d V .\) Estimate the value of this integral.

Step-by-Step Solution

Answer

The work required to increase the volume from \(V=2\) to \(V=4\) is \(10 ln 2\)

Step 1: Identify the Given Expressions

The given equation for pressure and volume is \(P V = 10\). It's also been specified that the work required comes from the integral \(\int_{2}^{4} P(V) d V\), where V ranges from 2 to 4.

Step 2: Substitute the Pressure Expression

Use the equation \(P V = 10\) to express pressure as a function of volume, as \(P = \frac{10}{V}\). Substitute this expression for P in the integral, which becomes \(\int_{2}^{4} \frac{10}{V} d V\).

Step 3: Compute the Integral

Compute the integral using standard integration rules. The antiderivative of \( \frac{1}{V} \) is \( ln |V|\), so simplifying the integral we get \( 10 \left[ ln |V| \right]_{2}^{4}\).

Step 4: Evaluate the Integral

Evaluate the resulting expression at the limits V=4 and V=2, therefore: \( 10 \left[ ln |4| - ln |2| \right] = 10 ln 2\).

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in thermodynamics that describes the relationship between pressure, volume, and temperature for an ideal gas. The formula is generally given as \(PV = nRT\), where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin.

In the context of this problem, we are looking at a specific scenario where temperature is constant. Under such conditions, the equation simplifies because the term \(nRT\) becomes a constant. As given in our problem statement, this constant is 10, leading to the simpler equation \(PV = 10\).

• When volume increases, pressure decreases to maintain the same constant product.
• This relationship is a direct application of Boyle's Law, a particular case of the Ideal Gas Law when temperature is constant.

Understanding this link between pressure and volume is crucial for determining how changes in one affect the other when working with gases.

Integration in Calculus

Integration is a fundamental concept in calculus used to calculate areas under curves, among other applications. In this exercise, we use integration to compute the work done by a gas.

Our task is to evaluate the integral \(\int_{2}^{4} \frac{10}{V} \, dV\). Here's how you can approach it:

• First, recognize that the function \(\frac{1}{V}\) resembles a common integral form.
• The antiderivative of \(\frac{1}{V}\) is well-known to be the natural logarithm \(\ln |V|\).
• Thus, the integral becomes \(10 \int_{2}^{4} \frac{1}{V} \, dV = 10 \left[ \ln |V| \right]_{2}^{4}\).

By applying the fundamental theorem of calculus, we compute the definite integral by evaluating it at the upper and lower limits, which simplifies the process of calculating the work involved in this thermodynamic system.

Work in Thermodynamics

Work is a key concept in thermodynamics, representing the energy transfer that occurs when a force is applied over a distance. Studying work in thermodynamics involves understanding how changing the state variables of a gas, such as volume and pressure, affects the energy transformations.

In this exercise, the work done by the gas is calculated by the integral \(\int P(V) \, dV\).

• The computed work represents the area under the curve of \(P(V)\) as a function of volume.
• For an ideal gas at constant temperature, the work done is given by \(W = nRT \ln(\frac{V_f}{V_i})\), relating it to our final integrated solution \(10 \ln 2\).

Understanding work in thermodynamics helps us understand how energy is converted within a system, which is essential for understanding engines, refrigerators, and various other systems that rely on thermodynamic principles.