Problem 37 - Calculus of a Single Variable

Question

Boyle's Law In Exercises 37 and 38 , find the work done by the gas for the given volume and pressure. Assume that the pressure is inversely proportional to the volume. (See Example \(6 . )\) 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.

Step-by-Step Solution

Answer

The work done by the gas is 1386.29 foot-pounds.

Step 1: Calculate the Final Pressure

Using Boyle's Law, which states that \(P1 \times V1 = P2 \times V2\), where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume. Here, the initial volume \(V1 = 2\) cubic feet and initial pressure \(P1 = 1000\) pounds per square foot, while the final volume \(V2 = 3\) cubic feet. Solving this equation for \(P2\), one will get \(P2 = P1 \times (V1/V2) = 1000 \times (2/3) = 666.67\) pounds per square foot.

Step 2: Compute the Work Done

The work done by the gas, W, is given by the integral, \(W = \int PdV\). Since P is inversely proportional to V, this becomes \(W = \int_{V1}^{V2} P1 \times (V1/V) dV = P1 \times V1 \times \int_{V1}^{V2} 1/V dV = P1 \times V1 \times [ln(V)]_{V1}^{V2} = 1000 \times 2 \times (ln(3) - ln(2)) = 1386.29\) foot-pounds.

Key Concepts

Physics of Gases

Understanding the behavior of gases is crucial in physics, particularly when studying thermodynamics, which deals with heat, work, and the energy of a system. Boyle's Law is a fundamental principle that relates the pressure of a gas to its volume at a constant temperature.

According to Boyle's Law, for a given amount of an ideal gas kept at a constant temperature, the pressure exerted by the gas is inversely proportional to its volume. Mathematically, this can be expressed as: \( P1 \times V1 = P2 \times V2 \) where \(P1\) and \(P2\) are the initial and final pressures, and \(V1\) and \(V2\) are the initial and final volumes of the gas, respectively.

This principle is essential for a variety of applications, including breathing mechanisms in biology, the functioning of syringes and pneumatic systems in engineering, and the understanding of atmospheric pressure in weather forecasting.

Integral Calculus

Integral calculus is a branch of mathematics that is concerned with the accumulation of quantities and the areas under and between curves. It allows us to solve problems related to volumes, areas, and the lengths of curves. One common application in physics, especially in the context of Boyle's Law, is the calculation of work done by a gas during expansion or compression.

In the provided exercise, the work done by a gas when expanding from volume \(V1\) to \(V2\) is found using the integral \(W = \int PdV\), which accounts for the variable pressure with respect to volume. By integrating the pressure function over the volume change, we obtain a quantitative measurement of the work involved in the process. Integral calculus helps bridge the gap between theory and practical quantification in physical scenarios.

Work Done by Gas

The concept of 'work done by gas' involves the energy transferred when a gas expands or compresses. In thermodynamics, work is done by a gas when there is a volume change against an external pressure. The work done by an expanding gas is positive, as the gas has to exert a force to push against the external pressure and increase its volume.

The equation \(W = \int PdV\) represents the work done on or by the gas. When dealing with a gas like in the given exercise, where the pressure is inversely proportional to the volume, this equation requires modification to accommodate this relationship, leading to the integration of a variable pressure function across volume. The result is the total work done throughout the gas's expansion or compression, and understanding this calculation is critical for comprehension of energy transfer within gaseous systems.