A sample of an ideal gas is expanded \(1 \mathrm{~m}^{3}\) to \(3 \mathrm{~m}^{3}\) in a reversible process for which \(P=K V^{2}\), with \(K=6 \mathrm{bar} / \mathrm{m}^{6}\). Work done by the gas is : (a) \(5200 \mathrm{~kJ}\) (b) \(15600 \mathrm{~kJ}\) (c) \(52 \mathrm{~kJ}\) (d) \(5267.6 \mathrm{~kJ}\)

Short Answer

Expert verified
5267.6 \(\mathrm{kJ}\)

Step by step solution

01

Title - Understand the Given Data

The problem provides the initial and final volumes of an ideal gas expanding in a reversible process where the pressure is a function of volume, given by the equation \(P = KV^2\). The constant \(K\) is given as \(6 \mathrm{bar/m}^6\). We need to find the work done by the gas when it expands from \(1 \mathrm{m}^3\) to \(3 \mathrm{m}^3\).
02

Title - Apply the Work Formula for Expansion

The work done by an ideal gas during a reversible expansion can be calculated by integrating the pressure with respect to the volume: \[W = \int_{V_1}^{V_2} P\,dV\]. Since we know the pressure is a function of volume, we can substitute \(P = KV^2\) into the integral to get the work done.
03

Title - Integrate with Limits

Substitute \(P = KV^2\) and the values for \(K\), \(V_1\), and \(V_2\) into the equation and integrate to find the work done: \[W = \int_{1 \mathrm{m}^3}^{3 \mathrm{m}^3} 6 \mathrm{bar} \cdot V^2 \,dV\]. The integration will yield the work done in bar*m^3. You will then need to convert this to kJ.
04

Title - Calculate the Integral

Execute the integration: \[W = \int_{1 \mathrm{m}^3}^{3 \mathrm{m}^3} 6 \cdot V^2 \,dV = 6 \times \left. \frac{V^3}{3} \right|_{1 \mathrm{m}^3}^{3 \mathrm{m}^3} = 6 \times \left( \frac{3^3}{3} - \frac{1^3}{3} \right)\]. Solve the expression to get the result in bar*m^3.
05

Title - Convert Units to kJ

Convert the result from bar*m^3 to kJ using the conversion \(1 \mathrm{bar} \cdot \mathrm{m}^3 = 100 \mathrm{kJ}\). Multiply the integral result by this conversion factor to obtain the work done in kJ.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ideal Gas Law
The ideal gas law is a fundamental relationship in thermodynamics that relates the pressure (P), volume (V), temperature (T), and amount of substance (n) in moles of an ideal gas through the equation PV=nRT. Here, R is the universal gas constant. This law assumes the ideal behavior of a gas, meaning the individual gas particles do not attract or repel each other and occupy no volume themselves. While actual gases do not perfectly adhere to these principles, the ideal gas law provides a close approximation for many conditions.

The ideal gas law is crucial in understanding how gases respond to changes in temperature, pressure, and volume, and serves as the basis for deriving other important thermodynamic equations and processes, such as the one involved in our textbook problem.
Reversible Expansion
Reversible expansion refers to a process by which a gas expands in such a way that the system always stays in thermodynamic equilibrium with its surroundings. In other words, the process is carried out infinitely slowly and can be reversed at any point without any changes in the total entropy of the system plus environment.

In a reversible expansion, the external pressure applied to the gas matches the internal pressure of the gas at every moment, allowing for the most efficient work extraction. This concept is theoretical but provides a useful model for understanding the limits of efficiency in physical processes and helps to establish a foundation for calculating work in thermodynamic systems.
Integration in Thermodynamics
Integration in thermodynamics is used to calculate properties that vary with thermodynamic processes, such as work and heat. When we have variables that change continuously, such as volume or pressure, integration allows us to quantify the total change, considering every infinitesimal step of the process.

For example, when dealing with the work done by a gas under variable pressure, we use integration to sum up the work done at every infinitesimal increase in volume. Integration can be visualized graphically as the area under the curve on a pressure-volume (P-V) diagram. The correct setup of the integral, with accurate limits and integrated function, is crucial for determining the work or heat associated with a given thermodynamic process.
Pressure-Volume Work
Pressure-volume (P-V) work is the work done by or against a gas when it is compressed or expanded. It is a central concept in thermodynamics that relates to the energy transfer within a system undergoing a change in volume. The work done by a gas during an expansion is given by the integral of the pressure exerted by the gas with respect to its changing volume.

The general formula for P-V work in a reversible process is expressed as \(W = \[ \int_{V_1}^{V_2} P\text{d}V\] \) where W is the work done by or on the system, P is the pressure, and V is the volume. The limits of integration, \(V_1\) and \(V_2\), are the initial and final volumes of the gas, respectively. If the process is described by a specific P-V relationship, as was the case in our exercise where \(P = KV^2\), we can substitute this into the integral to calculate the total work done.

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Most popular questions from this chapter

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