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Chapter 3: Problem 24

A process consists of two steps: (1) One \(\mathrm{kmol}\) of air at \(\mathrm{T}=800 \mathrm{K}\) and \(P=4\) bar is cooled at constant volume to \(\mathrm{T}=350 \mathrm{K}\). ( 2 ) The air is then heated at constant pressure until its temperature reaches \(800 \mathrm{K}\). If this two-step process is replaced by a single isothermal expansion of the air from \(800 \mathrm{K}\) and 4 bar to some final pressure \(\mathrm{P}\), what is the value of \(P\) that makes the work of the two processes the same? Assume mechanical reversibility and treat air as an ideal gas with \(C_{P}=(7 / 2) R\) and \(C_{V}=(5 / 2) R\).

Short Answer

Expert verified
The final pressure is approximately 2.28 bar.

Step by step solution

01

Understand the Problem

We need to find a final pressure after an isothermal expansion which results in the same work output as a two-step process of a gas being cooled and then reheated. We focus on equating the work done during each type of process.
02

Calculate Work for Two-Step Process

We first calculate the work for each step. For the first step, which is cooling at constant volume, no work is done because volume doesn't change: \ W = 0. For the second step, which is constant pressure heating, the work done is given by: \ \( W = P_2 (V_2 - V_3) \). We find the volumes using the ideal gas law: \( P_1V_1 = nRT_1 \), and \( P_2V_3 = nRT_3 \). \ Therefore, \( V_1 = \frac{nRT_1}{P_1} \), \( V_3 = \frac{nRT_3}{P_2} \). The work done: \ \( W = P_2(\frac{nRT_2}{P_2} - \frac{nRT_3}{P_2}) = nR(T_2 - T_3) \).
03

Calculate Isothermal Work

For an isothermal process, the work done is: \ \( W = nRT_1 \ln\left(\frac{V_f}{V_1}\right) \) but \( V_f/V_1 \) can also be expressed as \( P_1/P_f \) because the final pressure after expansion can be derived from the ideal gas law \( V_f = V_1 \times (P_1/P_f) \). So, \ \( W = nRT_1 \ln\left(\frac{P_1}{P_f}\right) \).
04

Equate the Two Works

Now, equate the work from the two-step process to the work from the isothermal process: \ \( nR(T_2 - T_3) = nRT_1 \ln\left(\frac{P_1}{P_f}\right) \) Simplifying, \ \( T_2 - T_3 = T_1 \ln\left(\frac{P_1}{P_f}\right) \). Now, solve for \( P_f \): \ \( T_1 \ln\left(\frac{4}{P_f}\right) = 450 \).
05

Solve for Final Pressure

Given that \( T_1 = 800 \) K, solve for \( P_f \): \ \( \ln\left(\frac{4}{P_f}\right) = \frac{450}{800} \) \ \( = 0.5625 \). Therefore, \ \( \frac{4}{P_f} = e^{0.5625} \) or \( P_f = 4/e^{0.5625} \). \ Finally, calculate \( P_f \): \ \( e^{0.5625} \approx 1.755 \), \ \( P_f = \frac{4}{1.755} \approx 2.28 \) bar.

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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 equation in thermodynamics that relates the pressure, volume, and temperature of an ideal gas. It is expressed as \( PV = nRT \), where:
  • \( P \) is the pressure of the gas
  • \( V \) is the volume of the gas
  • \( n \) is the amount of substance in moles
  • \( R \) is the universal gas constant
  • \( T \) is the temperature in Kelvin
This relationship allows us to determine one of these variables if the others are known. In the context of the exercise, the Ideal Gas Law was used to calculate the volumes \( V_1 \) and \( V_3 \) during the constant pressure process, enabling us to compute the work done in that process. This foundational equation helps in understanding the behavior of gases under various conditions, making it a critical component in thermodynamic calculations.
Isothermal Process
An isothermal process is one where the temperature of the system remains constant. For an ideal gas, this implies that any change in volume will be inversely proportional to the pressure change, as expressed by the Ideal Gas Law. In the context of this problem, we considered an isothermal expansion, where the gas expands and the pressure drops while maintaining a stable temperature of 800 K. The main feature of an isothermal process is that the internal energy change is zero because the temperature doesn't change. Therefore, any work done by or on the system results from heat transfer across the system boundary. Understanding isothermal processes aids in analyzing energy exchanges during the thermal operations of gases.
Work Calculation
Calculating work in thermodynamic processes depends heavily on the specific conditions under which the process occurs, such as constant volume, constant pressure, or constant temperature. In our exercise, both the two-step process and the isothermal expansion involve calculating work.For the two-step process:
  • Step one, cooling at constant volume, results in no work as volume does not change, reflecting \( W = 0 \).
  • Step two, heating at constant pressure, involves work calculation using \( W = nR(T_2 - T_3) \), which involves computing the volume change using the Ideal Gas Law.
For the isothermal process, work is determined using the formula \( W = nRT \ln\left(\frac{V_f}{V_1}\right) \), simplified to \( W = nRT \ln\left(\frac{P_1}{P_f}\right) \). This step-by-step breakdown clarifies the importance of different conditions and their impact on work in thermodynamic systems.
Mechanical Reversibility
Mechanical reversibility is a theoretical concept in thermodynamics where a process occurs in such a manner that it can be reversed by an infinitesimal change in external conditions. Reversible processes are idealizations that allow us to explore the maximum possible efficiency of thermodynamic transformations. In reversible processes, no energy is dissipated as friction or wasted, and the system remains constantly in equilibrium with its surroundings. This means the expansion or compression of a gas happens gradually, with pressure inside the system always equaling the pressure exerted externally. In the exercise, assuming mechanical reversibility enables us to calculate the work done precisely without accounting for real-world inefficiencies. This results in the most effective transfer of energy with minimal losses, a principle that's crucial to the design of efficient engines and machinery.

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Most popular questions from this chapter

Natural gas (assume pure methane) is delivered to a city via pipeline at a volumetric rate of 4 normal \(\mathrm{Mm}^{3}\) per day. Average delivery conditions are \(283.15 \mathrm{K}\left(10^{\circ} \mathrm{C}\right)\) and 20.7 bar. Determine: (a) The volumetric delivery rate in actual \(m^{3}\) per day. (b) The molar delivery rate in kmol per hour. (c) The gas velocity at delivery conditions in \(\mathrm{m} \mathrm{s}^{-1}\) The pipe is \(600 \mathrm{mm}\) heavy duty steel with an inside diameter of \(575 \mathrm{mm}\). Normal conditions are \(273.15 \mathrm{K}\left(0^{\circ} \mathrm{C}\right)\) and \(1 \mathrm{atm}\).

Calculate the reversible work done in compressing \(0.0283 \mathrm{m}^{3}\) of mercury at a constant temperature of \(273.15 \mathrm{K}\left(\mathrm{O}^{\circ} \mathrm{C}\right)\) from 1 atm to 3000 atm. The isothermal compressibility of mercury at \(273.15 \mathrm{K}\left(0^{\circ} \mathrm{C}\right)\) is $$\kappa=3.9 \times 10^{-6}-0.1 \times 10^{-9} P$$ where \(P\) is in atm and \(\mathrm{K}\) is in \(\mathrm{atm}^{-1}\).

To a good approximation, what is the molar volume of ethanol vapor at \(753.15 \mathrm{K}\left(480^{\circ} \mathrm{C}\right)\) and \(6000 \mathrm{kPa} ?\) How does this result compare with the ideal-gas value?

For methyl chloride at \(373.15 \mathrm{K}\left(100^{\circ} \mathrm{C}\right)\) the second and third virial coefficients are: $$B=-242.5 \mathrm{cm}^{3} \mathrm{mol}^{-1} \quad C=25200 \mathrm{cm}^{6} \mathrm{mol}^{-2}$$ Calculate the work of mechanically reversible, isothermal compression of 1 mol of methyl chloride from 1 bar to 55 bar at \(373.15 \mathrm{K}\left(100^{\circ} \mathrm{C}\right)\). Base calculations on the following forms of the virial equation: \((a)\) $$Z=1+\frac{B}{V}+\frac{C}{V^{2}}$$ \((b)\) $$Z=1+B^{\prime} P+C^{\prime} P^{2}$$ where $$B^{\prime}=\frac{B}{R T} \quad \text { and } \quad C^{\prime}=\frac{C-B^{2}}{(R T)^{2}}$$ Why don't both equations give exactly the same result?

An ideal gas flows through a horizontal tube at steady state. No heat is added and no shaft work is done. The cross-sectional area of the tube changes with length, and this causes the velocity to change. Derive an equation relating the temperatureto the velocity of the gas. If nitrogen at \(423.15 \mathrm{K}\left(150^{\circ} \mathrm{C}\right)\) flows past one section of the tube at a velocity of \(2.5 \mathrm{m} \mathrm{s}^{-1},\) what is its temperature at another section where its velocity is \(50 \mathrm{m} \mathrm{s}^{-1} ?\) Let \(C_{P}=(7 / 2) R\).
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