Problem 23 - Essential Calculus

Question

Boyle's Law A quantity of gas with an initial volume of 2 cubic feet and a pressure of 1000 pounds per square foot expands to a volume of 3 cubic feet. Find the work done by the gas using the integral \(W=\int_{V_{0}}^{v_{1}} p d V\) Assume that the pressure is inversely proportional to the volume.

Step-by-Step Solution

Answer

The work done by the gas is \(W = 2000 * (\ln(3) - \ln(2))\) pounds per foot

Step 1: Express pressure as a function of volume

Since the pressure \(p\) is inversely proportional to the volume, there exists some constant \(k\) such that \(p = k/V\). As it was given that initially when the volume \(V_{0} = 2\) cubic feet, the pressure \(p = 1000\) pounds per square foot, we can put these values into the equation to find that the constant \(k\) is \(k = p*V_{0} = 1000*2 = 2000\).

Step 2: Plug pressure expression into the integral

The formula to calculate the work done by the gas is given by \(W = \int_{V_{0}}^{V_{1}} p dV\). Thus we can substitute \(p = k/V = 2000/V\) into the formula to obtain \(W = \int_{2}^{3} 2000/V dV\).

Step 3: Solve the integral

Integrating the pressure function will give us the work done by the gas. The integral of \(2000/V\) with respect to \(V\) from 2 to 3 is \(2000 * [\ln(V)]_{2}^{3}\). This is equal to \(2000 * (\ln(3) - \ln(2))\).

Step 4: Calculate the numerical value

Evaluating the log values for 2 and 3, and solving the equation yields a numerical value for work done.

Key Concepts

Work done by gas

When discussing the work done by a gas, think of it as the energy transferred when a gas expands or contracts. In simpler terms, it's the area under the curve in a pressure-volume graph.

In our problem, the gas initially occupies 2 cubic feet and expands to 3 cubic feet. According to Boyle's Law, when the volume of a gas changes, it alters the pressure, and thus the total work done by the gas is calculated by integrating the pressure over the change in volume.

• The formula used is: \[ W = \int_{V_0}^{V_1} p \, dV \]This integral calculates the accumulated "pressure times change in volume" over the entire expansion process.
• By solving this integral, we can determine the work done in expanding the volume of the gas.

Understanding work done by gas is crucial in thermodynamic studies and illustrates how energy is transferred in processes involving gas.

Integral calculus

Integral calculus gives us the tools and methods to find areas, volumes, central points, and many useful things by summing "small quantities." In this exercise, we use integrals to determine the work done by applying it to the pressure-volume relationship.

Here are key steps in the process:

• We initially know that pressure is a function of volume in this context, specifically inversely proportional, described by \( p = \frac{k}{V} \). Knowing this helps us set up the integral for \( W = \int_{2}^{3} \frac{2000}{V} \, dV \).
• The integral \( \int \frac{1}{V} \, dV \) computes to a natural logarithmic function \( \ln(V) \). Integrals often transform complicated expressions into manageable solutions.
• Once integrated, we use the limits of the volume expansion (2 to 3 cubic feet) to compute the exact work done by evaluating this logarithmic expression.

With these steps, integral calculus enables us to capture the dynamic changes in pressure and volume and summarize the impact of these changes effectively.

Pressure-volume relationship

The relationship between pressure and volume is at the heart of the problem and draws from Boyle's Law, which states that for a given mass of gas at constant temperature, the volume is inversely proportional to the pressure.

In simpler terms:

• As the volume of the gas increases, the pressure decreases and vice-versa. This relationship is mathematically described as \( p = \frac{k}{V} \), where \( k \) is a constant based on the properties of the gas and initial conditions.
• In our problem, we determine the constant \( k \) by using the initial conditions: a pressure of 1000 pounds per square foot at a volume of 2 cubic feet. Thus, \( k = 1000 \times 2 = 2000 \).
• Understanding this relationship allows us to properly express pressure as a function of volume and use it in our integral calculation for work done by the expanding gas.

Grasping this concept is important not only for this exercise but also for understanding more complex physical phenomena involving gases and their behavior.