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Chapter 7: Problem 17

If a system expands in volume by an amount \(\Delta V\left(\mathrm{m}^{3}\right)\) against a constant restraining pressure \(P\left(\mathrm{N} / \mathrm{m}^{2}\right),\) a quantity \(P \Delta V(\mathrm{J})\) of energy is transferred as expansion work from the system to its surroundings. Suppose that the following four conditions are satisfied for a closed system: (a) the system expands against a constant pressure (so that \(\Delta P=0\) ); (b) \(\Delta E_{\mathrm{k}}=0 ;\) (c) \(\Delta E_{\mathrm{p}}=0 ;\) and (d) the only work done by or on the system is expansion work. Prove that under these conditions, the energy balance simplifies to \(Q=\Delta H\)

Short Answer

Expert verified
The energy balance simplifies to \(Q = \Delta H\) under these specific conditions given that the only work done by the system is expansion work and there are no changes in the kinetic and potential energy of the system.

Step by step solution

01

Understand the given conditions

According to the conditions, the system is expanding against a constant pressure (\(\Delta P=0\)), and there are no changes in kinetic and potential energy (\(\Delta E_k=0, \Delta E_p=0\)). The only work done by or on the system is expansion work.
02

Apply the first law of thermodynamics

The first law of thermodynamics states that the change in internal energy of a closed system is equivalent to the heat added to the system minus the work done by the system. Mathematically, it is represented as \(\Delta U = Q - W\). In this scenario, since the only work done is the expansion work, we can replace \(W\) with \(P \Delta V\). Therefore, \(\Delta U = Q - P\Delta V\).
03

Express the enthalpy change

The change in enthalpy \(\Delta H\) is defined as \(\Delta U + P \Delta V\). If we replace \(\Delta U\) from the previous step, we get \(\Delta H = (Q - P \Delta V) + P \Delta V\). Simplifying this, we find that \(Q = \Delta H\).

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Key Concepts

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

First Law of Thermodynamics
The First Law of Thermodynamics is a fundamental principle that serves as a cornerstone in the study of energy within systems. It is essentially a restatement of the concept of energy conservation, which indicates that energy cannot be created or destroyed; it can only be transformed from one form to another. In the context of a closed system, the First Law can be expressed in the equation \(\Delta U = Q - W\), where \(\Delta U\) represents the change in internal energy, \(Q\) is the heat added to the system, and \(W\) is the work done by the system.
This equation tells us that any increase in the internal energy of the system comes from either heat being transferred to it or from work being done on it. Conversely, if the internal energy decreases, it is due to heat being transferred out of the system or work being done by the system on its surroundings.
In situations where the system does expansion work, such as in this problem, the work \(W\) can be expressed as the product of pressure and change in volume, \(P \Delta V\). This alteration allows us to directly relate changes in the system's energy to physical processes that can be easily measured.
Enthalpy Change
Enthalpy is a concept in thermodynamics that extends the idea of internal energy to include energy used in doing external work, particularly pertinent during processes at constant pressure. It is represented by the symbol \(H\), and the change in enthalpy, \(\Delta H\), is given by the equation \(\Delta H = \Delta U + P \Delta V\).
This equation implies that the change in enthalpy is the sum of the change in internal energy and the work required to allow the system to expand or contract at constant pressure. It provides a useful way of considering heat and work in processes that are common in chemical reactions and other scenarios.
In the given exercise, after applying the First Law of Thermodynamics, we substitute \(\Delta U\) in the enthalpy formula resulting in \(\Delta H = (Q - P \Delta V) + P \Delta V\). The \(P \Delta V\) terms cancel out, leading to the simplified relationship \(Q = \Delta H\). This relationship often appears in chemical reactions where pressure remains constant, making it very handy for calculations.
Closed System
A closed system is one in which mass does not cross the system boundary, meaning no material is added or removed during processes. However, energy in the form of heat or work can still be exchanged between the system and its surroundings. Understanding this distinction is crucial, especially when applying the First Law of Thermodynamics.
In this problem, the system is closed with the specified conditions which lead to simplifications such as having no change in kinetic energy (\(\Delta E_k = 0\)) or potential energy (\(\Delta E_p = 0\)). These constraints simplify the energy balance equation, focusing solely on changes in internal energy and enthalpy.
This unique characteristic of closed systems often simplifies analysis, making it more straightforward to use concepts like enthalpy where only energy changes due to heat and work need consideration. This aligns perfectly with problems requiring calculation of heat or work in processes where pressure is constant, like in the given exercise, supporting the conclusion \(Q = \Delta H\).

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Most popular questions from this chapter

Steam at \(260^{\circ} \mathrm{C}\) and 7.00 bar absolute is expanded through a nozzle to \(200^{\circ} \mathrm{C}\) and 4.00 bar. Negligible heat is transferred from the nozzle to its surroundings. The approach velocity of the steam is negligible. The specific enthalpy of steam is \(2974 \mathrm{kJ} / \mathrm{kg}\) at \(260^{\circ} \mathrm{C}\) and 7 bar and \(2860 \mathrm{kJ} / \mathrm{kg}\) at \(200^{\circ} \mathrm{C}\) and 4 bar. Use the open-system energy balance to calculate the exit steam velocity.

Air at \(38^{\circ} \mathrm{C}\) and \(97 \%\) relative humidity is to be cooled to \(14^{\circ} \mathrm{C}\) and fed into a plant area at a rate of \(510 \mathrm{m}^{3} / \mathrm{min}\) (a) Calculate the rate ( \(\mathrm{kg} / \mathrm{min}\) ) at which water condenses. (b) Calculate the cooling requirement in tons (1 ton of cooling \(=12,000 \mathrm{Btu} / \mathrm{h}\) ), assuming that the enthalpy of water vapor is that of saturated steam at the same temperature and the enthalpy of dry air is given by the expression $$ \hat{H}(\mathrm{k} \mathrm{J} / \mathrm{mol})=0.0291\left[T\left(^{\circ} \mathrm{C}\right)-25\right] $$

Horatio Meshuggeneh has his own ideas of how to do things. For instance, when given the task of determining an oven temperature, most people would use a thermometer. Being allergic to doing anything most people would do, however, Meshuggeneh instead performs the following experiment. He puts acopper bar with a mass of 5.0 kg in the oven and puts an identical bar in a well- insulated 20.0-liter vessel containing 5.00L of liquid water and the remainder saturated steam at \(760 \mathrm{mm}\) Hg absolute. He waits long enough for both bars to reachthermal equilibrium with their surroundings, then quickly takes the first bar out of the oven, removes the second bar from the vessel, drops the first bar in its place, covers the vessel tightly, waits for the contents to come to equilibrium, and notes the reading on a pressure gauge built into the vessel. The value he reads is 50.1 mm Hg. He then uses the facts that copper has a specific gravity of 8.92 and a specific internal energy given by the expression \(\hat{U}(\mathrm{kJ} / \mathrm{kg})=0.36 T\left(^{\circ} \mathrm{C}\right)\) to calculate the oven temperature. (a) The Meshuggeneh assumption is that the bar can be transferred from the oven to the vessel without any heat being lost. If he makes this assumption, what oven temperature does Meshuggeneh calculate? How many grams of water evaporate in the process? (Neglect the heat transferred to the vessel wall- -i.e., assume that the heat lost by the bar is transferred entirely to the water in the vessel. Also, remember that you are dealing with a closed system once the hot bar goes into the vessel.) (b) In fact, the bar lost 8.3 kJ of heat between the oven and the vessel. What is the true oven temperature? (c) The experiment just described was actually Meshuggeneh's second attempt. The first time he tried it, the final gauge pressure in the vessel was negative. What had he forgotten to do?

Water from a reservoir passes over a dam through a turbine and discharges from a \(70-\mathrm{cm}\) ID pipe at a point 55 m below the reservoir surface. The turbine delivers 0.80 MW. Calculate the required flow rate of water in \(\left.\mathrm{m}^{3} / \mathrm{min} \text { if friction is neglected. (See Example } 7.7-3 .\right)\) If friction were included, would a higher or lower flow rate be required? (Note: The equation you will solve in this problem has multiple roots. Find a solution less than \(2 \mathrm{m}^{3} / \mathrm{s}\).)

Steam produced in a boiler is frequently "wet"-that is, it is a mist composed of saturated water vapor and entrained liquid droplets. The quality of a wet steam is defined as the fraction of the mixture by mass that is vapor. A wet steam at a pressure of 5.0 bar with a quality of 0.85 is isothermally "dried" by evaporating the entrained liquid. The flow rate of the dried steam is \(52.5 \mathrm{m}^{3} / \mathrm{h}\). (a) Use the steam tables to determine the temperature at which this operation occurs, the specific enthalpies of the wet and dry steams, and the total mass flow rate of the process stream. (b) Calculate the heat input (kW) required for the evaporation process. (c) Suppose leaks developed in the feed pipe to the dryer and in the dryer exit pipe. Speculate on what you would see at each location.
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